MATHS MCQ

$a+(n-1) d=0$

$\Rightarrow 50+(n-1)(-2)=0 \Rightarrow n=26$

So $S_{26}=\frac{26}{2}[2 \times 50+25 \times(-2)]=650$

.Hence ${2^{n - 1}} > 100$ are in A.P.

and ${m^{th}}$ mean between $2a,\;b$ is $2a + \frac{{m(b - 2a)}}{{n + 1}}$......$(ii)$

Accordingly, $a + \frac{{m(2b - a)}}{{n + 1}} = 2a + \frac{{m(b - 2a)}}{{n + 1}}$

$ \Rightarrow $ $m(2b - a) = a(n + 1) + m(b - 2a)$

$ \Rightarrow $ $a(n - m + 1) = bm$

$ \Rightarrow $ $\frac{a}{b} = \frac{m}{{n - m + 1}}$.

Hence a worker can do ${\left( {\frac{1}{{150n}}} \right)^{th}}$ part of the work in a day.
Accordingly,

$[150 + 146 + 142 + ....... + {\rm{upto}}\;(n + 8)\,{\rm{terms}}] \times \frac{1}{{150n}} = 1$

$ \Rightarrow $$n = 17$

Therefore number of total days in completion $ = 17 + 8 = 25$.

New value of $\Sigma {x_i} = 1900 - 55 - 45$$ = 1800$, $n = 48$

New mean $ = \frac{{1800}}{{48}}$$ = 37.5$.

Then $(a - d) + a + (a + d) = 12$ and $(a - d)\,a\,(a + d) = 28$

==> $3a = 12$ and $a\,({a^2} - {d^2}) = 28$

==> $a = 4$ and $a\,({a^2} - {d^2}) = 28$

==> $16 - {d^2} = 7$

$\Rightarrow d = \pm \,3$.

$ \Rightarrow $ $({a_1} + {a_{24}}) + ({a_5} + {a_{20}}) + ({a_{10}} + {a_{15}}) = 225$

$ \Rightarrow $ $3({a_1} + {a_{24}}) = 225$

$ \Rightarrow $${a_1} + {a_{24}} = 75$

( In an $A.P.$ the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term)

${a_1} + {a_2} + ...... + {a_{24}} = \frac{{24}}{2}({a_1} + {a_{24}}) = 12 \times 75 = 900$.

and $n = n$, then ${S_1} + {S_2} + ....... + {S_m} = \frac{1}{2}mn(mn + 1)$

$\left[ {{\rm{Using}}\;S\; = \frac{m}{2}(a + l).\;{\rm{Since}}\;{S_1},\;{S_2},\;{S_3},......{S_m}\;{\rm{form}}\;{\rm{an}}\;{\rm{A}}{\rm{.P}}{\rm{.}}} \right]$

He pays $10\%$ annual interest on remaining amount

$\therefore $ Money given in first year

$ = 1000 + \frac{{10000 \times 10}}{{100}} = {\rm{Rs}}.2000$

Money given in second year $= 1000 +$ interest of$ (10000 -1000)$ with interest rate $10\%$ per annum $ = 1000 + \frac{{9000 \times 10}}{{100}} = {\rm{Rs}}.\,1900$

Money paid in third year = Rs. $1800$ etc.

So money given by Jairam in $10$ years will be Rs. $2000$, Rs. $1900$, Rs. $1800$, Rs. $1700$,....,

which is in arithmetic progression, whose first term $a = 2000$and $d = - 100$

Total money given in $10$ years = sum of $10$ terms of arithmetic progression

$ = \frac{{10}}{2}[2(2000) + (10 - 1)( - 100)]$= Rs. $15500$

Therefore, total money given by Jairam

$ = 5000 + 15500 = {\rm{Rs}}{\rm{. }}\,{\rm{20500}}{\rm{.}}$

$\angle C = x + {20^o}$and $\angle D = x + {30^o}$

So, we know that $\angle A + \angle B + \angle C + \angle D = 2\pi $

Putting these values, we get

$({x^o}) + ({x^o} + {10^o}) + ({x^o} + {20^o}) + ({x^o} + {30^o}) = {360^o}$

$ \Rightarrow x = {75^o}$

Hence the angles of the quadrilateral are ${75^o},\;{85^o},\;{95^o},\;{105^o}$.

Trick : In these type of questions, students should satisfy the conditions through options.

Here $(b)$ satisfies both the conditions

$i.e.$ angles are in $A.P.$ with common difference ${10^o}$ and sum of angles is ${360^o}$.

Then ${b^2} - {a^2} = {c^2} - {b^2}$

$ \Rightarrow $ $(b - a)(b + a) = (c - b)(c + b)$

==> $\frac{{b - a}}{{c + b}} = \frac{{c - b}}{{b + a}}$

==> $\frac{{(b - a)(a + b + c)}}{{(c + a)(b + c)}} = \frac{{(c - b)(a + b + c)}}{{(a + b)(c + a)}}$

$ \Rightarrow $ $\frac{{{b^2} + bc - ac - {a^2}}}{{(c + a)(b + c)}} = \frac{{{c^2} + ac - ab - {b^2}}}{{(a + b)(c + a)}}$

$ \Rightarrow $ $\frac{b}{{c + a}} - \frac{a}{{b + c}} = \frac{c}{{a + b}} - \frac{b}{{c + a}}$

Hence $\frac{a}{{b + c}},\;\frac{b}{{c + a}},\;\frac{c}{{a + b}}$ be in $A.P.$

==> $\frac{{2c}}{b} = \frac{b}{a} + \frac{a}{c}$

$ \Rightarrow \frac{{2c}}{b} = \frac{{bc + {a^2}}}{{ac}}$

==> $2a{c^2} = {b^2}c + b{a^2}$

$\therefore \,{a^2}b,\,{c^2}a$ and ${b^2}c$ are in $A.P.$

Add $1$ to each term, we get

$\frac{{a + b + c}}{{b + c}},\frac{{b + c + a}}{{c + a}},\frac{{c + a + b}}{{a + b}}$ are in $A.P.$

Divide each term by $(a + b + c),$

$\frac{1}{{b + c}},\frac{1}{{c + a}},\frac{1}{{a + b}}$ are in $A.P.$

Hence $b + c,\,\,c + a,\,\,a + b$ are in $H.P.$

which is given in question

Therefore, $\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}}$ are in $A. P.$

==> ${\tan ^{ - 1}}\left( {\frac{{2y}}{{1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{x + z}}{{1 - xz}}} \right)$

==> $\frac{{2y}}{{1 - {y^2}}} = \frac{{x + z}}{{1 - xz}}$

But $2y = x + z$

$1 - {y^2} = 1 - xz$

==> ${y^2} = xz$

$xyz$ are both in $G.P. $ and $A.P.$, 

$x = y = z$.

$11^{th}$ term of $A.P. =$ $a + 10d$

$21^{st}$ term of $A.P. = a+ 20d$

$2(a + 10d) = 7(a + 20d)$

==> $2a + 20d = 7a + 140d$

$5a + 120d = 0$

==> $a + 24d = 0$

Hence $25^{th}$ term is $0.$

${A_1} + {A_4} = 8$ ..$(i)$

and ${A_2}.\,{A_3} = 15$ ..$(ii)$

The sum of terms equidistant from the beginning and end is constant and is equal to

sum of first and last terms.

Hence, ${A_2} + {A_3} = {A_1} + {A_4} = 8$ ..$(iii)$

From $(ii)$ and $(iii),$

${A_2} + \frac{{15}}{{{A_2}}} = 8$

==> $A_2^2 - 8{A_2} + 15 = 0$

${A_2} = 3\,\,{\rm{or}}\,\,5$ and ${A_3} = 5\,\,\,{\rm{or}}\,\,{\rm{3}}$.

As we know, ${A_2} = \frac{{{A_1} + {A_3}}}{2}$

==> ${A_1} = 2{A_2} - {A_3}$

==> ${A_1} = 2 \times 3 - 5 = 1$ and ${A_4} = 8 - {A_1} = 7$

Hence the series is, $1, 3, 5, 7.$

So that least number of series is $1.$

$\therefore $ $\frac{1}{{r + q}} - \frac{1}{{p + q}} = \frac{1}{{q + r}} - \frac{1}{{r + p}}$

$ \Rightarrow $ $\frac{{p + q - r - p}}{{(r + p)(p + q)}} = \frac{{r + p - q - r}}{{(q + r)(r + p)}}$

$ \Rightarrow $ $\frac{{q - r}}{{p + q}} = \frac{{p - q}}{{q + r}}$ or ${q^2} - {r^2} = {p^2} - {q^2}$

$\therefore $ $2{q^2} = {r^2} + {p^2}$

Therefore ${p^2},\;{q^2},\;{r^2}$ are in $A.P.$

==> $x = \frac{{u + v}}{2},y = \frac{{u - v}}{2}$,

$\therefore f(u,v) = \left( {\frac{{u + v}}{2}} \right).\left( {\frac{{u - v}}{2}} \right)$

Now,$\frac{{f(x,y) + f(y,x)}}{2} = \frac{{\left( {\frac{{x + y}}{2}.\frac{{x - y}}{2}} \right) + \left( {\frac{{y + x}}{2}.\frac{{y - x}}{2}} \right)}}{2} = 0$.

therefore, ${a_2} - {a_1} = {a_4} - {a_3} = ....... = {a_{2n}} - {a_{2n - 1}} = d$

Here $a_1^2 - a_2^2 + a_3^2 - a_4^2 +$$ ....... + a_{2n - 1}^2 - a_{2n}^2$

$ = ({a_1} - {a_2})({a_1} + {a_2}) + ({a_3} - {a_4})({a_3} + {a_4}) +$$ ...... + ({a_{2n - 1}} - {a_{2n}})({a_{2n - 1}} + {a_{2n}})$

$ = - d({a_1} + {a_2} + ....... + {a_{2n}}) = - d\left\{ {\frac{{2n}}{2}({a_1} + {a_{2n}})} \right\}$

Also we know ${a_{2n}} = {a_1} + (2n - 1)d$$ \Rightarrow $$d = \frac{{{a_{2n}} - {a_1}}}{{2n - 1}}$

$ \Rightarrow $ $ - d = \frac{{{a_1} - {a_{2n}}}}{{2n - 1}}$.

$\therefore $ Therefore the sum is

= $\frac{{n({a_1} - {a_{2n}}).({a_1} + {a_{2n}})}}{{2n - 1}} = \frac{n}{{2n - 1}}(a_1^2 - a_{2n}^2)$.

and common difference $= d$

Let $S = \frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + .......... + \frac{1}{{{a_n}{a_{n + 1}}}}$

$⇒$  $S = \frac{1}{d}\,\left\{ {\frac{d}{{{a_1}{a_2}}} + \frac{d}{{{a_2}{a_3}}} + ...... + \frac{d}{{{a_n}\,\,{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\,\left\{ {\frac{{{a_2} - {a_1}}}{{{a_1}{a_2}}} + \frac{{{a_3} - {a_2}}}{{{a_2}{a_3}}} + ...... + \frac{{{a_{n + 1}} - {a_n}}}{{{a_n}\,\,\,{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\left\{ {\frac{1}{{{a_1}}} - \frac{1}{{{a_2}}} + \frac{1}{{{a_2}}} - \frac{1}{{{a_3}}} + ....... + \frac{1}{{{a_n}}} - \frac{1}{{{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\left\{ {\frac{1}{{{a_n}}} - \frac{1}{{{a_{n + 1}}}}} \right\} = \frac{1}{d}\left\{ {\frac{{{a_{n + 1}} - {a_1}}}{{{a_1}{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\left( {\frac{{nd}}{{{a_1}{a_{n + 1}}}}} \right) = \frac{n}{{{a_1}{a_{n + 1}}}}$.

Trick: Check for $n = 2$.

$ = \frac{{100}}{2}(1 + 100) = 50(101) = 5050$

Let ${S_1} = 3 + 6 + 9 + 12 + ......... + 99$

=$3(1 + 2 + 3 + 4 + ......... + 33)$

=$3.\frac{{33}}{2}(1 + 33) = 99 \times 17 = 1683$

Let ${S_2} = 5 + 10 + 15 + ........ + 100$

= $5(1 + 2 + 3 + ........ + 20)$

= $5.\frac{{20}}{2}(1 + 20) = 50 \times 21 = 1050$

Let ${S_3} = 15 + 30 + 45 + ........ + 90$

= $15(1 + 2 + 3 + ........ + 6)$

= $15.\frac{6}{2}(1 + 6) = 45 \times 7 = 315$

Required sum =$S - {S_1} - {S_2} + {S_3}$

= $5050 - 1683 - 1050 + 315= 2632.$

First term of the series $a = \frac{1}{x} + y$,

Second term =$\frac{2}{x} + y$

$d = \left( {\frac{2}{x} + y} \right) - \left( {\frac{1}{x} + y} \right) = \frac{1}{x}$

Sum of $r$ terms of the series

$ = \frac{r}{2}\left[ {2\left( {\frac{1}{x} + y} \right) + (r - 1)\frac{1}{x}} \right]$

$ = \frac{r}{2}\left[ {\frac{2}{x} + 2y + \frac{r}{x} - \frac{1}{x}} \right]$

$ = \frac{{{r^2} - r + 2r}}{{2x}} + ry$

$ = \left\{ {\frac{{r{\mkern 1mu} (r + 1)}}{{2x}}} \right\} + ry$.

$ \Rightarrow $$\frac{1}{{{{\log }_x}\sqrt 3 \,}} + \frac{1}{{{{\log }_x}\sqrt[4]{3}}} + \frac{1}{{{{\log }_x}\sqrt[6]{3}}} + ... + \frac{1}{{{{\log }_x}\sqrt[{16}]{3}}} = 36$

$ \Rightarrow $ $\frac{1}{{(1/2){{\log }_x}3}} + \frac{1}{{(1/4){{\log }_x}3}} + \frac{1}{{(1/6){{\log }_x}3}} + ..... + \frac{1}{{(1/16){{\log }_x}3}} = 36$

$ \Rightarrow $ $({\log _3}x)(2 + 4 + 6 + ..... + 16) = 36$

$ \Rightarrow $ $({\log _3}x)\frac{8}{2}[2 + 16] = 36$

$ \Rightarrow $${\log _3}x = \frac{1}{2}$

$ \Rightarrow $$x = {3^{1/2}}$

$ \Rightarrow x = \sqrt 3 $.

$ = \frac{n}{2}\{ 4a + 4nd - 2d - 2a - nd + d\} = \frac{n}{2}\{ 2a + (3n - 1)d\} $

$ = \frac{1}{3}.\frac{{3n}}{2}\{ 2a + (3n - 1)d\} = \frac{1}{3}{S_{3n}}$.

This is an $AP$ with first term $13$ and common difference $4$.

Let the number of terms be $n$.

Then $97 = 13 + (n - 1)4$

$ \Rightarrow $ $4n = 88$

$ \Rightarrow $ $n = 22$

Therefore the sum of the numbers

$ = \frac{{22}}{2}[13 + 97] = 11(110) = 1210$.

$\log a + \log \left( {\frac{{{a^2}}}{b}} \right) + \log \left( {\frac{{{a^3}}}{{{b^2}}}} \right) + \log \left( {\frac{{{a^4}}}{{{b^3}}}} \right) + ...... + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$

This is an $A.P.$ with first term $\log a$

and the common difference $\log \left( {\frac{{{a^2}}}{b}} \right) - \log a = \log \left( {\frac{a}{b}} \right)$

Therefore the sum of $n$ terms is

$\frac{n}{2}\left[ {\log a + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)} \right] = \frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$.

Trick : Check for $n = 1,\;2$.

$ \Rightarrow $ $\frac{{2a + (m - 1)d}}{{2a + (n - 1)d}} = \frac{m}{n}$

$ \Rightarrow $ $\frac{{a + \frac{1}{2}(m - 1)d}}{{a + \frac{1}{2}(n - 1)d}} = \frac{m}{n}$

$ \Rightarrow $ $an + \frac{1}{2}(m - 1)nd = am + \frac{1}{2}(n - 1)md$

$ \Rightarrow $ $a(n - m) + \frac{d}{2}[mn - n - mn + m] = 0$

$ \Rightarrow $ $a(n - m) + \frac{d}{2}(m - n) = 0$

$ \Rightarrow $ $ a = \frac{d}{2}$ or $d = 2a$

So, required ratio, $\frac{{{T_m}}}{{{T_n}}} = \frac{{a + (m - 1)d}}{{a + (n - 1)d}} = \frac{{a + (m - 1)2a}}{{a + (n - 1)2a}}$

$ = \frac{{1 + 2m - 2}}{{1 + 2n - 2}} = \frac{{2m - 1}}{{2n - 1}}$.

Trick : Replace $m$ by $2m - 1$ and $n$ by $2n - 1$.

Obviously if ${S_m}$ is of degree $2$, then ${T_m}$ is of $1$.  $i.e.$ linear.

$\therefore $ $\sin d\,\{ co{\rm{sec}}\;{a_1}co{\rm{sec}}\;{a_2} + ..... + {\rm{cosec}}\;{a_{n - 1}}{\rm{cosec}}\;{a_n}\} $

$ = \frac{{\sin ({a_2} - {a_1})}}{{\sin {a_1}.\;\sin {a_2}}} + ...... + \frac{{\sin ({a_n} - {a_{n - 1}})}}{{\sin {a_{n - 1}}\sin {a_n}}}$

$ = (\cot {a_1} - \cot {a_2}) + (\cot {a_2} - \cot {a_3}) + .... + (\cot {a_{n - 1}} - \cot {a_n})$

$ = \cot {a_1} - \cot {a_n}$.

${T_{11}}$ and $T{'_{11}}$ be the respective ${11^{th}}$ terms, then

$\frac{{{S_n}}}{{S{'_n}}} = \frac{{\frac{n}{2}[2a + (n - 1)d]}}{{\frac{n}{2}[2a' + (n - 1)d']}} = \frac{{7n + 1}}{{4n + 27}}$

$ \Rightarrow $ $\frac{{a + \frac{{(n - 1)}}{2}d}}{{a' + \frac{{(n - 1)}}{2}d'}} = \frac{{7n + 1}}{{4n + 27}}$

Now put $n = 21$,

we get $\frac{{a + 10d}}{{a' + 10d'}} = \frac{{{T_{11}}}}{{T{'_{11}}}} = \frac{{148}}{{111}} = \frac{4}{3}$.

Note : If ratio of sum of $n$ terms of two $A.P.'s$ are given in terms of $n$ and ratio of their ${p^{th}}$ terms are to be found then put $n = 2p - 1$.

Here we put $n = 11 \times 2 - 1 = 21$.

==> $\frac{{\frac{n}{2}[2{a_1} + (n - 1){d_1}]}}{{\frac{n}{2}[2{a_2} + (n - 1){d_2}]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{2\left[ {{a_1} + \left( {\frac{{n - 1}}{2}} \right)\,{d_1}} \right]}}{{2\left[ {{a_2} + \left( {\frac{{n - 1}}{2}} \right)\,{d_2}} \right]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{{a_1} + \left( {\frac{{n - 1}}{2}} \right)\,{d_1}}}{{{a_2} + \left( {\frac{{n - 1}}{2}} \right)\,{d_2}}} = \frac{{2n + 3}}{{6n + 5}}$

Put $n = 25$ then $\frac{{{a_1} + 12{d_1}}}{{{a_2} + 12{d_2}}} = \frac{{2(25) + 3}}{{6(25) + 3}}$

==> $\frac{{{T_{{{13}_1}}}}}{{{T_{{{13}_2}}}}} = \frac{{53}}{{155}}$.

Putting $n = 1,\;2,\;3,\;.............,$ we get,

${S_1} = A + B,\,{S_2} = 2A + 4B,\,\,{S_3} = 3A + 9B$

Therefore ${T_1} = {S_1} = A + B,\;{T_2} = {S_2} - {S_1} = A + 3B,$

${T_3} = {S_3} - {S_2} = A + 5B$,

Hence the sequence is $(A + B),(A + 3B),\;(A + 5B),...$

Here $a = A + B$ and common difference $d = 2B$.

and $3 + 10 + 17 + 24 + ......$….. $(ii)$

Now from $(i),$ ${m^{th}}$ term $ = (2m + 61)$

and ${m^{th}}$ term of $(ii)$ series $ = (7m - 4)$

Under condition,

$ \Rightarrow 7m - 4 = 2m + 61$

$\Rightarrow 5m = 65$

$\Rightarrow m = 13$.

$ \Rightarrow $$n\theta = N\pi + (m\theta )$

$ \Rightarrow $ $\theta = \frac{{N\pi }}{{n - m}}$,

putting $N = 1,\;2,\;3.........,$ we get

$\frac{\pi }{{n - m}},\;\frac{{2\pi }}{{n - m}},\;\frac{{3\pi }}{{n - m}}.........$

which are obviously in $A.P.$

Since common difference $d = \frac{\pi }{{n - m}}$.

$ = \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ......\infty } \right)$
+ $\left( {1 + \frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ......\infty } \right)$

$ = \left( {\frac{1}{{1 - (1/2)}}} \right) + \left( {\frac{1}{{1 - (1/3)}}} \right) = 2 + \frac{3}{2} = \frac{7}{2}$.

$\therefore $$a + b = 3,\;ab = p$

$c,\;d$ are roots of ${x^2} - 12x + q = 0$

$\therefore $$c + d = 12,\;cd = q$

$a,\;b,\;c,\;d$ are in GP.

$\therefore $$\frac{b}{a} = \frac{d}{c}$$ \Rightarrow $$\frac{{a + b}}{{a - b}} = \frac{{c + d}}{{c - d}}$

$ \Rightarrow $$\frac{{{{(a - b)}^2}}}{{{{(a + b)}^2}}} = \frac{{{{(c - d)}^2}}}{{{{(c + d)}^2}}}$$ \Rightarrow $$1 - \frac{{4ab}}{{{{(a + b)}^2}}} = 1 - \frac{{4cd}}{{{{(c + d)}^2}}}$

$ \Rightarrow $$\frac{{ab}}{{{{(a + b)}^2}}} = \frac{{cd}}{{{{(c + d)}^2}}}$$ \Rightarrow $$\frac{p}{9} = \frac{q}{{144}}$

$ \Rightarrow $$\frac{p}{1} = \frac{q}{{16}}$$ \Rightarrow $$\frac{p}{q} = \frac{1}{{16}}$$ \Rightarrow $$\frac{{p + q}}{{q - p}} = \frac{{17}}{{15}}$.

Trick : Let $a = 1,\;b = 2,\;c = 4,\;d = 8$, then
$p = 2,\;q = 32$ $ \Rightarrow $ $(q + p):(q - p) = 17:15$.

$⇒$  $2b = a + c,b -a = c -b$

${a^2},{b^2},{c^2}$ are in $H.P.$

$\frac{1}{{{b^2}}} - \frac{1}{{{a^2}}} = \frac{1}{{{c^2}}} - \frac{1}{{{b^2}}}$ $ \Rightarrow \frac{{{a^2} - {b^2}}}{{{a^2}{b^2}}} = \frac{{{b^2} - {c^2}}}{{{b^2}{c^2}}}$

$⇒$  $(a - b)[{c^2}(a + b) - {a^2}(b + c)] = 0$,

$[\because \,(b - c) = (a - b)]$

$⇒$  $a = b$ or ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$

$⇒$  ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$

$⇒$  $ac\,(c - a) = b\,({a^2} - {c^2})$

$⇒$  $ac = - b\,(c + a)$

$⇒$  $ - ac = b.2b$

$⇒$  ${b^2} = ( - a/2)\,c$,

$\therefore - a/2,b,c$ are in $G.P.$

==> $\frac{1}{{y - x}},\frac{1}{{2(y - a)}},\frac{1}{{y - z}}$ are in $A.P.$

==> $\frac{1}{{2(y - a)}} - \frac{1}{{(y - x)}} = \frac{1}{{y - z}} - \frac{1}{{2(y - a)}}$

==> $\frac{{y - x - 2y + 2a}}{{(y - x)}} = \frac{{2y - 2a - y + z}}{{(y - a) - (z - a)}}$

$ \Rightarrow \frac{{ - x - y + 2a}}{{(y - x)}} = \frac{{y + z - 2a}}{{(y - z)}}$

$ \Rightarrow \frac{{(x - a) + (y - a)}}{{(x - a) - (y - a)}} = \frac{{(y - a) + (z - a)}}{{(y - a) - (z - a)}}$

==> $\frac{{(x - a)}}{{(y - a)}} = \frac{{(y - a)}}{{(z - a)}}$

$(x - a),(y - a),(z - a)$ are in $G.P.$

$\therefore \,\,{G_1} = {p^{2/3}}\,\,{q^{1/3}},\,\,{G_2} = {p^{1/3}}\,\,\,{q^{2/3}}$

$\therefore \,\frac{{G_1^2}}{{{G_2}}} + \frac{{G_2^2}}{{{G_1}}} = \frac{{{p^{4/3}}\,\,{q^{2/3}}}}{{{p^{1/3}}\,\,{q^{2/3}}}} + \frac{{{p^{2/3}}\,{q^{4/3}}}}{{{p^{2/3}}\,{q^{1/3}}}}$

$ = p + q = 2 \times \,\left( {\frac{{p + q}}{2}} \right)\, = 2A$.

Then the middle term of the $A.P.$ is $\frac{{a + b}}{2}$.

The middle term of the $G.P.$ is $\sqrt {ab} $.

The middle term of the $H.P.$ is $\frac{{2ab}}{{a + b}}$.

Obviously, these terms are in $G.P.$

Then ${T_p} = a + (p - 1)d,\;$

${T_q} = a + (q - 1)d$

and ${T_r} = a + (r - 1)d$.

If ${T_p},\;{T_q},\;{T_r}$ are in $G.P.$

Then its common ratio $R = \frac{{{T_q}}}{{{T_p}}} = \frac{{{T_r}}}{{{T_q}}} = \frac{{{T_q} - {T_r}}}{{{T_p} - {T_q}}}$

$ = \frac{{[a + (q - 1)d] - [a + (r - 1)d]}}{{[a + (p - 1)d] - [a + (q - 1)d]}} = \frac{{q - r}}{{p - q}}$

Similarly, we can show that $R = \frac{{q - r}}{{p - q}} = \frac{{r - s}}{{q - r}}$

Hence $(p - q),\;(q - r),\;(r - s)$ be in $G.P.$

$⇒$  ${(a + 2c)^2} - {(2b)^2} = {a^2} + 4{c^2}$

$⇒$  ${a^2} + 4ac + 4{c^2} - 4{b^2} = {a^2} + 4{c^2}$

$⇒$  $4ac - 4{b^2} = 0$

$⇒$  ${b^2} = ac$

Hence $a, b, c$ are in $G.P.$

$= 0.037+0.000037+0.0000000037+…….$

= $\frac{{37}}{{{{10}^3}}} + \frac{{37}}{{{{10}^6}}} + \frac{{37}}{{{{10}^9}}} + ......$

= $37\left[ {\frac{1}{{{{10}^3}}} + \frac{1}{{{{10}^6}}} + \frac{1}{{{{10}^9}}} + ....} \right]$

= $37\left[ {\frac{{1/{{10}^3}}}{{1 - 1/{{10}^3}}}} \right] $

$= 37\left[ {\frac{1}{{{{10}^3}}}.\frac{{{{10}^3}}}{{999}}} \right]$ = $\frac{{37}}{{999}}$.

$= 0.5 + 0.073 + 0.00073$

$= 0.5 +$ $\frac{{73}}{{1000}} + \frac{{73}}{{100000}} + ....$

= $0.5 + 73\left[ {\frac{1}{{1000}} + \frac{1}{{100000}} + .....} \right]$

= $0.5 + 73\left[ {\frac{{1/1000}}{{1 - \frac{1}{{100}}}}} \right]$

= $0.5 + \frac{{73}}{{1000}}.\frac{{100}}{{99}} = \frac{5}{{10}} + \frac{{73}}{{990}}$

= $\frac{{495 + 73}}{{990}} = \frac{{568}}{{990}}$.

and $\frac{{{a^2}}}{{1 - {r^2}}} = 3$ .....(ii)

From (i) and (ii), $\frac{a}{{1 + r}} = 1$

$\Rightarrow a = 1 + r$

From (i), $\frac{{1 + r}}{{1 - r}} = 3 $

$\Rightarrow r = \frac{1}{2}$ , from (i), $ a = 3/2$

So, first term $= 3/2$ and common ratio $= 1/2.$

$\therefore $${\left( {\frac{1}{5}} \right)^{{{\log }_{\sqrt 5 }}\left( {\frac{1}{2}} \right)}} = {\left( {\frac{1}{5}} \right)^{{{\log }_5}\left( {\frac{1}{4}} \right)}} = {5^{ - {{\log }_5}{4^{ - 1}}}} = {5^{{{\log }_5}4}} = 4$.

${({S_2})_\infty } = \frac{{{a^2}}}{{1 - {r^2}}} = 3$

or ${a^2} = 3\,(1 - {r^2})$ or $9\,{(1 - r)^2} = 3\,(1 - {r^2})$ [by $(i)$]

or $3\,(1 - 2r + {r^2}) = 1 - {r^2}$ or $2{r^2} - 3r + 1 = 0$

or $(r - 1)\,(2r - 1) = 0$,

$\therefore $$r = 1,\frac{1}{2}$

If $r = 1,$ then $a = 3(1 - 1) = 0$ which is impossible.

If $r = \frac{1}{2},$then $a = 3\,\left( {1 - \frac{1}{2}} \right) = 3/2$

So first series is $3/2, 3/4, 3/8, 3/16,.....$

$ = 0.14 + 0.00189 + 0.00000189 + .......$

$ = \frac{{14}}{{100}} + 189\left[ {\frac{1}{{{{10}^5}}} + \frac{1}{{{{10}^8}}} + ....\infty } \right]$

$ = \frac{7}{{50}} + 189\,\left[ {\frac{{1/{{10}^5}}}{{1 - (1/{{10}^3})}}} \right]$$ = \frac{7}{{50}} + 189\,\left[ {\frac{1}{{{{10}^5}}} \times \frac{{{{10}^3}}}{{999}}} \right]$

$ = \frac{7}{{50}} + \frac{{189}}{{999 \times 100}}$

$ = \frac{7}{{50}} + \frac{7}{{3700}} = \frac{7}{{50}} + \frac{7}{{25 \times 148}}$$ = \frac{{21}}{{148}}$.

and $\frac{{{a^2}}}{{1 - {r^2}}} = \frac{a}{{1 - r}}.\frac{a}{{1 + r}} = y$ …..$(ii)$

$ \Rightarrow $ $y = x.\frac{a}{{1 + r}} = x.\frac{{x(1 - r)}}{{1 + r}}$

$ \Rightarrow $$\frac{y}{{{x^2}}} = \frac{{1 - r}}{{1 + r}}$

$ \Rightarrow $ $\frac{{{x^2}}}{y} = \frac{{1 + r}}{{1 - r}}$

$ \Rightarrow $$\frac{{{x^2}}}{y}(1 - r) = 1 + r$ 

$\Rightarrow r[1+ \frac{{x^2}}{{y}}] = -1 + \frac{{x^2}}{{y}}$

$\Rightarrow r=\frac{{{x^2} + y}}{{{x^2} - y}}$

then $xy = {x^2} - {x^3} + {x^4} - ......\infty $

Adding, $y + xy = x + 0 + 0...... + 0$

$ \Rightarrow $$x - xy = y $

$\Rightarrow x(1 - y) = y$

$\Rightarrow x = \frac{y}{{1 - y}}$.

Aliter : $y = \frac{x}{{1 - ( - x)}} $

$\Rightarrow y = \frac{x}{{1 + x}}$

$ \Rightarrow $$y + yx = x$

$\Rightarrow x = \frac{y}{{1 - y}}$.

$ \Rightarrow $ $8 = \frac{2}{{r(1 - r)}}\left( {\;a = \frac{2}{r}} \right)$

$ \Rightarrow $ $4r(1 - r) = 1 $

$\Rightarrow 4r - 4{r^2} - 1 = 0$

$ \Rightarrow $ $4{r^2} - 4r + 1 = 0$

$\Rightarrow \left( {r - \frac{1}{2}} \right)(4r - 2) = 0$

$\Rightarrow r = \frac{1}{2}$

So first term $a = 4$.

$x = \frac{1}{{1 - a}}$

$\Rightarrow a = \frac{{x - 1}}{x}$ and $y = \frac{1}{{1 - b}}$

$\Rightarrow b = \frac{{y - 1}}{y}$

$\therefore $$1 + ab + {a^2}{b^2} + ..........\infty = \frac{1}{{1 - ab}}$

$ = \frac{1}{{1 - \frac{{x - 1}}{x}.\frac{{y - 1}}{y}}} = \frac{{xy}}{{x + y - 1}}$.

$A = 1 + [{r^z} + {r^{2z}} + {r^{3z}} + ........\infty ]$

We know that sum of infinite $G.P.$ is

${S_\infty } = \frac{a}{{1 - r}}( - 1 < r < 1)$

Therefore, $A = 1 + \left[ {\frac{{{r^z}}}{{1 - {r^z}}}} \right]$

$\Rightarrow A = \frac{{1 - {r^z} + {r^z}}}{{1 - {r^z}}}$

$\therefore $ $A = \frac{1}{{1 - {r^z}}}$

$\Rightarrow 1 - {r^z} = \frac{1}{A} $

$\Rightarrow {r^z} = \frac{{A - 1}}{A}$

Hence $r = {\left[ {\frac{{A - 1}}{A}} \right]^{1/z}}$.

Under conditions, we get $\frac{a}{r}\;.\;a\;.\;ar = 216$

$ \Rightarrow $ $a = 6$

And sum of product pair wise $ = 156$

$ \Rightarrow $ $\frac{a}{r}\;.\;a + \frac{a}{r}\;.\;ar + a\;.\;ar = 156$

$ \Rightarrow $ $r = 3$

Hence numbers are $2, 6, 18.$

Trick : Since $2 \times 6 \times 18 = 216$ (as given) and no other option gives the value.

Given series can be written as,

$ = \frac{1}{3}[9 + 99 + 999 + ........ + n\,\,{\rm{terms]}}$

$ = \frac{1}{3}\left[ {(10 - 1) + ({{10}^2} - 1) + ({{10}^3} - 1) + .... + n\,{\rm{terms}}} \right]$

$ = \frac{1}{3}\left[ {10 + {{10}^2} + .... + {{10}^n}} \right]$$ - \frac{1}{3}\left[ {1 + 1 + 1 + .... + n\,{\rm{terms}}} \right]$

$ = \frac{1}{3}\,.\,\frac{{10\,({{10}^n} - 1)}}{{10 - 1}} - \frac{1}{3}.n\,$ $ = \frac{1}{3}\left[ {\frac{{{{10}^{n + 1}} - 10}}{9} - n} \right]$

$ = \frac{1}{3}\,\left[ {\frac{{{{10}^{n\, + \,1}} - 9n - 10}}{9}} \right]$ $ = \frac{1}{{27}}[{10^{n\, + \,1}} - 9n - 10]$.

$ = \frac{{1.({{10}^{91}} - 1)}}{{10 - 1}} = \frac{{{{({{10}^{13}})}^7} - 1}}{{{{10}^{13}} - 1}} \times \frac{{{{10}^{13}} - 1}}{{10 - 1}}$

$ = [{({10^{13}})^6} + {({10^{13}})^5} + {({10^{13}})^4} + ......1]$

$({10^{12}} + {10^{11}} + ...... + 1)$

It is the product of two integers and hence not prime.

$a{r^{n - 1}} = 128$ …..$(ii)$

and common ratio $r = 2$ …..$(iii)$

From $(iii), (i)$ and $(ii)$

we get $a{2^{n - 1}} = 128$  …..$(iv)$

and $\frac{{a({2^n} - 1)}}{{2 - 1}} = 255$ .....$(v)$

Dividing $(v)$ by $(iv)$

we get $\frac{{{2^n} - 1}}{{{2^{n - 1}}}} = \frac{{255}}{{128}}$

$ \Rightarrow $$2 - {2^{ - n + 1}} = \frac{{255}}{{128}}$

$ \Rightarrow $${2^{ - n}} = {2^{ - 8}}$

$ \Rightarrow $$n = 8$

Putting $n = 8$ in equation $(iv),$

we have $a\;.\;{2^7} = 128 = {2^7}$or $a = 1$.

then under condition,

${T_n} = {T_{n - 1}} + {T_{n - 2}}$

$ \Rightarrow $ $a{r^{n - 1}} = a{r^{n - 2}} + a{r^{n - 3}}$

$ \Rightarrow $ $a{r^{n - 1}} = a{r^{n - 1}}{r^{ - 1}} + a{r^{n - 1}}{r^{ - 2}}$

$ \Rightarrow $ $1 = \frac{1}{r} + \frac{1}{{{r^2}}}$

$ \Rightarrow $ ${r^2} - r - 1 = 0$

$ \Rightarrow $ $r = \frac{{1 \pm \sqrt {1 + 4} }}{2} = \frac{{1 + \sqrt 5 }}{2}$

Taking only $(+)$ sign . $(\because \;r > 1)$

= $0.2 + 0.034 + 0.00034 + 0.0000034 +.............$                                         

= $0.2 + \frac{{34}}{{1000}} + \frac{{34}}{{100000}} + \frac{{34}}{{10000000}} + .....\infty $

$ = \frac{2}{{10}} + 34\left[ {\frac{1}{{{{10}^3}}} + \frac{1}{{{{10}^5}}} + \frac{1}{{{{10}^7}}} + ........\infty } \right]$

$ = \frac{2}{{10}} + 34\left[ {\frac{{1/{{10}^3}}}{{1 - 1/1000}}} \right] = \frac{2}{{10}} + 34 \times \frac{1}{{1000}} \times \frac{{100}}{{99}}$

$ = \frac{2}{{10}} + \frac{{34}}{{990}} = \frac{{232}}{{990}}$.

$\therefore $${S_{100}} = a\left( {\frac{{1 - {r^{100}}}}{{1 - r}}} \right) = \frac{9}{{10}}\left( {\frac{{1 - \frac{1}{{{{10}^{100}}}}}}{{1 - \frac{1}{{10}}}}} \right) = 1 - \frac{1}{{{{10}^{100}}}}$.

$ \Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} $

$\Rightarrow {r^{n - 1}} = {r^n}(1 + r)$

$ \Rightarrow $ ${r^2} + r - 1 = 0$

$ \Rightarrow r = \frac{{ - 1 \pm \sqrt {1 + 4} }}{2} = \frac{{ - 1 \pm \sqrt 5 }}{2}$

Since, each term is $ + ve$.

Hence common ratio is $\frac{{\sqrt 5 - 1}}{2}$.

${r^{p + q - 1 - p + q + 1}} = \frac{m}{n}$

$\Rightarrow r = {\left( {\frac{m}{n}} \right)^{1/(2q)}}$

and $a = \frac{m}{{{{\left( {\frac{m}{n}} \right)}^{(p + q - 1)/2q}}}}$

Now ${p^{th}}$ term $ = a{r^{p - 1}} = \frac{m}{{{{\left( {\frac{m}{n}} \right)}^{(p + q - 1)/2q}}}}{\left( {\frac{m}{n}} \right)^{(p - 1)/2q}}$

$ = m{\left( {\frac{m}{n}} \right)^{(p - 1)/2q - (p + q - 1)/(2q)}} = m{\left( {\frac{m}{n}} \right)^{ - 1/2}} = {m^{1 - 1/2}}{n^{1/2}}$

$ = {m^{1/2}}{n^{1/2}} = \sqrt {mn} $.

Aliter : As we know each term in a $G.P.$ is geometric mean of the terms equidistant from it.

Here ${(p + q)^{th}}$and ${(p - q)^{th}}$ terms are equidistant from ${p^{th}}$ term

$i.e.$ at a distance of $q$.

Therefore, ${p^{th}}$ term will be $G.M.$ of ${(p + q)^{th}}$ and ${(p - q)^{th}}$

$i.e.$ $\sqrt {mn} $.

$ \Rightarrow $ ${b^2}(a - c) = ac(a - c)$

$ \Rightarrow $${b^2}a - {b^2}c = {a^2}c - a{c^2}$

$ \Rightarrow $ $a({b^2} + {c^2}) = c({a^2} + {b^2})$.

Trick : Put $a = 1,\;b = 2,\;c = 4$ and check the alternates.

==> ${({a_1}.{a_2}.{a_3})^{1/3}} \ge \frac{3}{{(1/{a_1} + 1/{a_2} + 1/{a_3})}}$

==> $({a_1}.\,{a_2}.{a_3})\, \ge \,\frac{{27}}{{{{\left( {1/{a_1} + 1/{a_2} + 1/{a_3}} \right)}^3}}}$

$({a_1}.\,{a_2}.{a_3})\,{\left( {\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}}} \right)^3}$$ \ge 27$.

${t_1} = \log 25$; ${s_1} = {\left[ {\log \,\,5/3} \right]^1}$

${t_2} = \log \frac{{125}}{3}$; ${s_2} = {\left[ {\log \,5/3} \right]^2}$

${t_3} = \log \frac{{625}}{9}$; ${s_3} = {\left[ {\log \,5/3} \right]^3}$

Clearly ${t_n}$ is an $A.P.$ and ${s_n}$ is $G.P.$

$a + ar + a{r^2} = 14$$ \Rightarrow a\,(1 + r + {r^2}) = 14$ …..$(i)$

and $2\,(ar + 1) = (a + 1) + (a{r^2} - 1)$

$a\,({r^2} - 2r + 1) = 2$ ….. $(ii)$

Put the value of a from $(i)$ to $(ii),$

==> $2{r^2} - 5r + 2 = 0$

==> $r = 2,\frac{1}{2}\,\,{\rm{and}}\,\,a = 2,8$

$\therefore $ Numbers are $2, 4, 8$ or $8, 4, 2$. So lowest term in series is $2$.

Now $G_1^3 + G_2^3 = 72$.

Also option $(b)$ gives this value

$i.e.$  $2 \times 2 \times 4 \times \frac{9}{2} = 72$.

if $\alpha ,\;\beta $ are roots of quadratic equation, then quadratic equation is

${x^2} - x(\alpha + \beta ) + \alpha \beta = 0$ ......$(i)$

$A.M. = \frac{{\alpha + \beta }}{2} = 8$

$ \Rightarrow $$\alpha + \beta = 16$ ......$(ii)$

and $G.M. = \sqrt {\alpha \beta } = 5$

$ \Rightarrow $$\alpha \beta = 25$ ......$(iii)$

So the required quadratic equation will be ${x^2} - 16x + 25 = 0$.

Now $A.M.$$ = \frac{{a + b}}{2}$

and $G.M.$$ = \sqrt {ab} $

Under conditions,

$A.M.=G.M.+ 2$

$ \Rightarrow $$\frac{{a + b}}{2} = \sqrt {ab} + 2$ ......$(i)$

and $\frac{a}{b} = \frac{4}{1}$

$ \Rightarrow $$a = 4b$ .....$(ii)$

From $(ii)$ and $(i),$ we get $a = 16$ and $b = 4$.

Condition $I$ : $\frac{a}{r} + a + ar = 14 $

$\Rightarrow a\left( {\frac{1}{r} + 1 + r} \right) = 14$..(i)

Condition $II$ : $\frac{a}{r} + 1,\;a + 1$ and $ar - 1$ will be in $A.P.$,

then $2(a + 1) = \frac{a}{r} + 1 + ar - 1 = \frac{a}{r}(1 + {r^2})$ ......(ii)

From (i) and (ii), we get $a = 4$ and $r = 2$.

So, required numbers are $2, 4, 8.$

Hence greatest number is $8.$

$\frac{{a + b}}{2} > \sqrt {ab} ,\;\frac{{b + c}}{2} > \sqrt {bc} $

and $\frac{{a + c}}{2} > \sqrt {ac} $.

Multiplying these inequalities, we get

$(a + b)(b + c)(c + a) > 8abc$.

$\frac{{x + y}}{{2(\sqrt {xy} )}} = \frac{p}{q}$…..$(i)$

$\frac{{{x^2} + {y^2} + 2xy}}{{4xy}} = \frac{{{p^2}}}{{{q^2}}}$

$\frac{{{x^2} + {y^2} + 2xy - 4xy}}{{4xy}} = \frac{{{p^2} - {q^2}}}{{{q^2}}}$

$\frac{{{{(x - y)}^2}}}{{4xy}} = \frac{{{p^2} - {q^2}}}{{{q^2}}}$

$\frac{{x - y}}{{2\sqrt {xy} }} = \frac{{\sqrt {{p^2} - {q^2}} }}{q}$…..$(ii)$

Equation $(i)$ is divided by $(ii),$

Then $\frac{{x + y}}{{x - y}} = \frac{p}{{\sqrt {{p^2} - {q^2}} }}$;

$\frac{x}{y} = \frac{{p + \sqrt {{p^2} - {q^2}} }}{{p - \sqrt {{p^2} - {q^2}} }}$.

$\therefore {T_2} = a + d,\,\,$

${T_{10}} = a + 9d,$

${T_{34}} = a + 33d$

$\therefore {(a + 9d)^2} = (a + d)(a + 33d)$

==> ${a^2} + 81{d^2} + 18ad = {a^2} + ad + 33ad + 33{d^2}$

Put $a = 1$

$ \Rightarrow 1 + 81{d^2} + 18d = 1 + d + 33d + 33{d^2}$

==> $48{d^2} - 16d = 0$

$ \Rightarrow 16d(3d - 1) = 0$

==> $d = 0,\,\,d = 1/3$.

$a + d + a + a - d = 27$

==> $3a = 27$

==> $a = 9$

Now, $(a + d - 1),(a - 1),(a - d + 3)$are in $G.P.$

$ \Rightarrow $${(a - 1)^2} = (a + d - 1)(a - d + 3)$

$ \Rightarrow 64 = (8 + d)(12 - d)$

$ \Rightarrow 64 = - {d^2} + 4d + 96$

$ \Rightarrow {d^2} - 4d - 32 = 0$

$ \Rightarrow {d^2} - 8d + 4d - 32 = 0$

==> $(d - 8)(d + 4) = 0$,

$\therefore d = - 4,\,8$

Series is $5, 9, 13$ (for $d = -4$) and $17, 9, 1$ (for $d = 8$)

Decreasing $A.P.$ is $17, 9, 1.$

Then ${(a - d)^2},\;{a^2},\;{(a + d)^2}$ are in $G.P.$

$\therefore $${a^4} = {(a - d)^2}{(a + d)^2}$

$ \Rightarrow $${d^4} - 2{a^2}{d^2} = 0$

$ \Rightarrow $$d = 0,\; \pm \sqrt 2 a$

Hence $d$ has three values.

So $\frac{a}{r}.\;a.\;ar = 512$

$ \Rightarrow $${a^3} = {8^3}$

$ \Rightarrow $$a = 8$

From second condition, we get $\frac{a}{r} + 8,\;a + 6,\;ar$ will be in $A.P.$

$ \Rightarrow $ $2(a + 6) = \frac{a}{r} + 8 + ar$

$ \Rightarrow $$28 = 8\left\{ {\frac{1}{r} + 1 + r} \right\}$

$ \Rightarrow $ $\frac{1}{r} + r + 1 = \frac{7}{2}$

$ \Rightarrow $$\frac{1}{r} + r - \frac{5}{2} = 0$

$ \Rightarrow $ ${r^2} - \frac{5}{2}r + 1 = 0$

$ \Rightarrow $$2{r^2} - 5r + 2 = 0$

$ \Rightarrow $ $(2r - 1)(r - 2) = 0$

$ \Rightarrow $$r = \frac{1}{2},\;r = 2$ $(\because \;r > 1)$

$ \Rightarrow $ $r = 2$.

Hence required numbers are $4,\;8,\;16$.

Trick : Check for $(a)$ $2 + 8,\;4 + 6,\;8$ are not in $A.P.$

$(b)$ $4 + 8,\;8 + 6,\;16\;i.e.\;12,\;14,\;16$ are in $A.P.$

Applying componendo and dividendo, we get

$\frac{{2a}}{{2bx}} = \frac{{2b}}{{2cx}} = \frac{{2c}}{{2dx}}$

$ \Rightarrow $${b^2} = ac$ and ${c^2} = bd$

$ \Rightarrow $$a,\;b,\;c$ and $b,\;c,\;d$ are in $G.P.$

Therefore, $a,\;b,\;c,\;d$ are in $G.P.$

$ \Rightarrow $${({\log _z}x)^2} = {\log _x}y \times {\log _y}z = {\log _x}z = \frac{1}{{{{\log }_z}x}}$

$ \Rightarrow $${({\log _z}x)^3} = 1$

$ \Rightarrow $$z = x$

Also, we can show $z = x = y = 4$.

or $\frac{1}{2}(a + b) = 2\sqrt {ab} $

or $\frac{{a + b}}{{2\sqrt {ab} }} = \frac{2}{1}$

$ \Rightarrow $$\frac{{a + b + 2\sqrt {ab} }}{{a + b - 2\sqrt {ab} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1}$

$ \Rightarrow $ $\frac{{{{(\sqrt a + \sqrt b )}^2}}}{{{{(\sqrt a - \sqrt b )}^2}}} = \frac{3}{1}$

$ \Rightarrow $ $\frac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }} = \frac{{\sqrt 3 }}{1}$

$ \Rightarrow $ $\frac{a}{b} = {\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)^2}$

$ \Rightarrow $ $\frac{a}{b} = \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }}$

or $a:b = (2 + \sqrt 3 ):(2 - \sqrt 3 )$.

So, ${b^2} = ac$…..(i)

$x = \frac{{a + b}}{2}$…..(ii)

$y = \frac{{b + c}}{2}$…..(iii)

Now $\frac{a}{x} + \frac{c}{y} = \frac{{2a}}{{a + b}} + \frac{{2c}}{{b + c}} = \frac{{2(ab + bc + 2ca)}}{{ab + ac + {b^2} + bc}}$

$ = \frac{{2(ab + bc + 2ca)}}{{(ab + ac + ac + bc)}} = 2$,$\left\{ {\because \;{b^2} = ac} \right\}$.

Trick : Let $a = 1,\;b = 2,\;c = 4,$ then obviously $x = \frac{3}{2}$

and $y = 3$, then $\frac{1}{{3/2}} + \frac{4}{3} = 2$.

and geometric mean $G = \sqrt {ab} $

Then $A - G = \frac{{a + b}}{2} - \sqrt {ab} $$ = \frac{{a + b - 2\sqrt {ab} }}{2}$

$ = \frac{{{{(\sqrt a )}^2} + {{(\sqrt b )}^2} - 2(\sqrt a )(\sqrt b )}}{2} = {\left[ {\frac{{\sqrt a - \sqrt b }}{{\sqrt 2 }}} \right]^2}$

${\rm{A}}{\rm{.M}}{\rm{.}}\, \ge {\rm{G}}{\rm{.M}}{\rm{.}}$

$\left( {\frac{{{x_1} + {x_2} + {x_3} + ...... + {x_n}}}{n}} \right)\, \ge \,{({x_1}.{x_2}.{x_3}.......{x_n})^{\frac{1}{n}}}$

$ = {(1)^{\frac{1}{n}}} = 1$

${x_1} + {x_2} + {x_3} + ........ + {x_n} \ge n$

${x_1} + {x_2} + {x_3} + ....... + {x_n}$ can never be less than $n$.

${T_n} = n{(2n + 1)^2} = 4{n^3} + 4{n^2} + n$

$\therefore $ ${S_n} = \sum\limits_1^n {{T_n}} = \sum\limits_1^n {(4{n^3} + 4{n^2} + n)} $

$ = 4\sum\limits_1^n {{n^3}} + 4\sum\limits_1^n {{n^2}} + \sum\limits_1^n n $

$ = 4{\left\{ {\frac{n}{2}(n + 1)} \right\}^2} + \frac{4}{6}n(n + 1)(2n + 1) + \frac{n}{2}(n + 1)$

$ = \frac{n}{6}(n + 1)(6{n^2} + 14n + 7)$.

$\frac{1}{5}{S_n} = {\rm{ }}\frac{1}{5} + \frac{2}{{{5^2}}} + \frac{3}{{{5^3}}} + ....... + \frac{n}{{{5^n}}}$

Subtracting,

$\left( {1 - \frac{1}{5}} \right){S_n} = 1 + \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + ...... + {\rm{upto}}\;n\;{\rm{terms}}\; - \frac{n}{{{5^n}}}$

$ \Rightarrow $$\frac{4}{5}{S_n} = \frac{{1 - \frac{1}{{{5^n}}}}}{{\frac{4}{5}}} - \frac{n}{{{5^n}}}$

$ \Rightarrow $${S_n} = \frac{{25}}{{16}} - \frac{{4n + 5}}{{16 \times {5^{n - 1}}}}$.

$S = 2 + 4 + 7 + 11 + 16 + ........{T_{n - 1}} + {T_n}$

Subtracting, we get

$0 = 2 + \left\{ {2 + 3 + 4 + ........ + ({T_n} - {T_{n - 1}})} \right\} - {T_n}$

$ \Rightarrow $${T_n} = 1 + (1 + 2 + 3 + 4 + ......{\rm{upto}}\;n\;{\rm{terms}})$

$ \Rightarrow $$1 + \frac{1}{2}n(n + 1) = \frac{{2 + {n^2} + n}}{2} $

$= \frac{{{n^2} + n + 2}}{2}$.

${S_n} = 12 + 16 + 24 + 40 + ..... + {T_n}$

Again ${S_n} = \,12 + 16 + 24 + ...... + {T_n}$

On subtraction

$0 = (12 + 4 + 8 + 16 + ...$+ upto $n$ terms) -${T_n}$

or ${T_n} = 12 + [4 + 8 + 16 + ... + {\rm{upto }}(n - 1)$ terms]

$ = 12 + \frac{{4({2^{n - 1}} - 1)}}{{2 - 1}} = {2^{n + 1}} + 8$

On putting $n = 1,\,2,\,3......$

${T_1} = {2^2} + 8$, ${T_2} = {2^3} + 8$, ${T_3} = {2^4} + 8......etc.$

${S_n} = {T_1} + {T_2} + {T_3} + .... + {T_n}$

$ = ({2^2} + {2^3} + {2^4} + ....{\rm{upto}}\,\,\,n\,\,\,{\rm{terms)}}$

$ + (8 + 8 + 8 + ......$upto $n$ terms)

$ = \frac{{{2^2}({2^n} - 1)}}{{2 - 1}} + 8n = 4({2^n} - 1) + 8n.$

Again $S = {\rm{ }}2 + 4 + 7 + 11 + ....... + {T_{n - 1}} + {T_n}$

Subtracting, we get

$0 = 2 + \left\{ {2 + 3 + 4 + 5 + .....({T_n} - {T_{n - 1}})} \right\} - {T_n}$

${T_n} = 2 + \frac{1}{2}(n - 1)(4 + \{ n - 2)1\} = \frac{1}{2}({n^2} + n + 2)$

Now $S = \Sigma {T_n} = \frac{1}{2}\Sigma ({n^2} + n + 2) $

$= \frac{1}{2}(\Sigma {n^2} + \Sigma n + 2\Sigma \,1)$

$ = \frac{1}{2}\left\{ {\frac{1}{6}n(n + 1)(2n + 1) + \frac{1}{2}n(n + 1) + 2n} \right\}$

$ = \frac{n}{{12}}\left\{ {(n + 1)(2n + 1 + 3) + 12} \right\}$

= $\frac{n}{6}\left\{ {(n + 1)(n + 2) + 6} \right\} $

$= \frac{n}{6}({n^2} + 3n + 8)$.

$ \Rightarrow $$x.S = x + 3{x^2} + 6{x^3} + .......\infty $

Subtracting $S(1 - x) = 1 + 2x + 3{x^2} + 4{x^3} + .......\infty $

$ \Rightarrow $$x(1 - x)S = x + 2{x^2} + 3{x^3} + .......\infty $

Again subtracting,

$ \Rightarrow $$S[(1 - x) - x(1 - x)] = 1 + x + {x^2} + {x^3} + ........\infty $

$ \Rightarrow $$S[(1 - x)(1 - x)] = \frac{1}{{1 - x}} $

$\Rightarrow S = \frac{1}{{{{(1 - x)}^3}}}$

$ \Rightarrow $ $\frac{1}{5}S = {\rm{ }}\frac{1}{5} + 4.\frac{1}{{{5^2}}} + 7.\frac{1}{{{5^3}}} + .........$

Subtracting $\left( {1 - \frac{1}{5}} \right)S = 1 + 3.\frac{1}{5} + 3.\frac{1}{{{5^2}}} + 3.\frac{1}{{{5^3}}} + ........$

$ = 1 + 3\left( {\frac{1}{5} + \frac{1}{{{5^2}}} + ......} \right)$

$ \Rightarrow $$\frac{4}{5}.S = 1 + 3.\frac{1}{5}\left( {\frac{1}{{1 - \frac{1}{5}}}} \right) = 1 + \frac{3}{4} = \frac{7}{4} \Rightarrow S = \frac{{35}}{{16}}$.

Aliter : Use direct formula ${S_\infty } = \frac{{ab}}{{1 - r}} + \frac{{dbr}}{{{{(1 - r)}^2}}}$

Here $a = 1,\;b = 1,\;d = 3,\;r = \frac{1}{5}$, therefore

${S_\infty } = \frac{1}{{1 - \frac{1}{5}}} + \frac{{3 \times 1 \times \frac{1}{5}}}{{{{\left( {1 - \frac{1}{5}} \right)}^2}}} = \frac{5}{4} + \frac{{\frac{3}{5}}}{{\frac{{16}}{{25}}}} = \frac{5}{4} + \frac{{15}}{{16}} = \frac{{35}}{{16}}$.

Aliter : Use $S = \left[ {1 + \frac{r}{{1 - r}} \times {\rm{diff}}{\rm{.}}\;{\rm{of}}\;{\rm{A}}{\rm{.P}}{\rm{.}}} \right]\frac{1}{{1 - r}}$

Hence sum of $2n$ terms is
$n$.

Let $s = 1 - \alpha \beta + {\alpha ^2}{\beta ^2}.......$

==> $s = \frac{1}{{1 + \alpha \beta }}$
$\alpha = \frac{1}{{{s_1}}} - 1,\,\,\beta = \frac{1}{{{s_2}}} - 1$;

$\therefore s = \frac{1}{{1 + \left( {\frac{1}{{{s_1}}} - 1} \right)\,\left( {\frac{1}{{{s_2}}} - 1} \right)}}$.

$s = \frac{{{s_1}{s_2}}}{{2{s_1}{s_2} + 1 - {s_1} - {s_2}}}$.

Rationalization of ${D^r}$

$\therefore S = (\sqrt 2 - \sqrt 1 ) + \left( {\sqrt 3 - \sqrt 2 } \right) + ... + \left( {\sqrt {{n^2}} - \sqrt {{n^2} - 1} } \right)$

$S = n -1.$
 

$= \frac{{\frac{{n(n + 1)}}{4}}}{{{{\left( {\frac{{n(n + 1)}}{2}} \right)}^2}}} $

$= \frac{1}{{n(n + 1)}} = \frac{1}{n} - \frac{1}{{n + 1}}$

${S_n} = \sum\limits_{}^{} {\left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)} $

$ = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ....... + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$

$ = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$.

$S = [2n + 2.2n + 3.2n + ...... + n.2n] - $

$[1 + 2.3 + 3.5 + .... + n.(2n - 1)]$

Let,${S_1} = 2n(1 + 2 + 3 + .... + n)$

= $\frac{{2n.n(n + 1)}}{2} = {n^2}(n + 1)$

and ${S_2} = 1 + 2.3 + 3.5 + ..... + n.(2n - 1)$

${T_n} = n(2n - 1) = 2{n^2} - n$

$\therefore {S_2} = \sum (2{n^2} - n)$

$ = 2\sum ({n^2}) - \sum (n)$

$ = \frac{{2n(n + 1)(2n + 1)}}{6} - \frac{{n(n + 1)}}{2}$

$\therefore S = {S_1} - {S_2} = {n^2}(n + 1) - \frac{{2n(n + 1)(2n + 1)}}{6} + \frac{{n(n + 1)}}{2}$

$ = n\,(n + 1)\left[ {n - \frac{{2n + 1}}{3} + \frac{1}{2}} \right]$

$ = n\,(n + 1)\left[ {\frac{{6n - 4n - 2 + 3}}{6}} \right]$

$ = \frac{{n\,(n + 1)(2n + 1)}}{6}$.

$= \left[ {\frac{{n(n + 1)}}{2}} \right]_{n = 20}^2 - \left[ {\frac{{n(n + 1)}}{2}} \right]_{n = 10}^2$

$= 44100 -3025 = 41075.$

and second factors of the terms of the given series $2,\;3,\;4,,\;........(n + 1)$

${n^{th}}$term of the given series $ = n(n + 1) = {n^2} + n$

Hence sum = $\Sigma {n^2} + \Sigma n = \frac{1}{6}n(n + 1)(2n + 1) + \frac{n}{2}(n + 1)$

$ = \frac{1}{6}n(n + 1)(2n + 1 + 3) = \frac{1}{3}n(n + 1)(n + 2)$.

${S_n} = n - \sum\limits_{n = 1}^n {{{\left( {\frac{1}{3}} \right)}^n}} $

$ = n - \frac{{\frac{1}{3}\left[ {1 - {{\left( {\frac{1}{3}} \right)}^n}} \right]}}{{\left( {1 - \frac{1}{3}} \right)}}$

$ = n - \frac{1}{2}(1 - {3^{ - n}}) = n + \frac{1}{2}({3^{ - n}} - 1)$.

$1\;.\;3\;.\;5 + 2\;.\;5\;.\;8 + 3\;.\;7\;.\;11 + ....... + n(2n + 1)(3n + 2)\;$

So, ${T_n} = n(2n + 1)(3n + 2) = n\,[6{n^2} + 4n + 3n + 2]$

${T_n} = 6{n^3} + 7{n^2} + 2n$

Now, sum $ = 6\Sigma {n^3} + 7\Sigma {n^2} + 2\Sigma n$

$ = 6{\left[ {\frac{1}{2}n(n + 1)} \right]^2} + 7\left[ {\frac{1}{6}n(n + 1)(2n + 1)} \right] + 2\left[ {\frac{1}{2}n(n + 1)} \right]$

$ = \frac{1}{6}n(n + 1)[9n\,(n + 1) + 7(2n + 1) + 6]$

$ = \frac{1}{6}n\,(n + 1)[9{n^2} + 9n + 14n + 7 + 6]$

$ = \frac{{n\,(n + 1)(9{n^2} + 23n + 13)}}{6}$.

${S_n} = \Sigma ({T_n}) = \Sigma \,6\,\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]$$ = 6\left[ {1 - \frac{1}{{n + 1}}} \right]$

${S_n} = \frac{{6n}}{{n + 1}}$.

${\left\{ {\frac{n}{2}(n + 1)} \right\}^2} = {(1 + 2 + ...... + n)^2} = \sum\limits_1^n {{r^2}} + 2\sum\limits_{s < t} {st} $

$ \Rightarrow $$\sum\limits_{s < t} {st} = \frac{1}{2}\left\{ {\frac{{{n^2}{{(n + 1)}^2}}}{4} - \frac{{n(n + 1)(2n + 1)}}{6}} \right\}$

$ = \frac{n}{{24}}(n - 1)(n + 1)(3n + 2)$.

Trick : ${S_n} = 1\;.\;2 + 2\;.\;3 + 3\;.\;4 + ........ + (n - 1)\;.\;n$

Check by putting $(n - 1) = 1,\;2\;$

$i.e.,\;n = 2,\;3$ in the options.

So ${n^{th}}$ term of series is given by

${T_n} = \frac{{1 + 2 + 3 + ....... + n}}{n} = \frac{{\frac{1}{2}n(n + 1)}}{n} = \frac{{n + 1}}{2}$

$\therefore $${S_n} = \Sigma ({n^3}) + \Sigma (3{n^2}) + \Sigma (2n)$

${S_n} = {\left[ {\frac{{n\,(n + 1)}}{2}} \right]^2} + \frac{{3.n\,(n + 1)(2n + 1)}}{6} + \frac{{2.n\,(n + 1)}}{2}$

${S_n} = \frac{1}{4}n\,(n + 1)\,(n + 2)(n + 3)$.

$ = (1\;.\;4 + 2\;.\;5 + 3\;.\;6 + 4\;.\;7 + .........$upto $n$ terms) $-14$

$ = \Sigma n(n + 3) - 14 = \frac{1}{6}(2{n^3} + 12{n^2} + 10n) - 14$

$ = \left( {\frac{{2{n^3} + 12{n^2} + 10n - 84}}{6}} \right),\;$

where $n = 3,\;4,\;5.......$

Trick : ${S_1} = 18,\;{S_2} = 46$

Now put in options $(n - 2) = 1,\;2\;\;i.e.\;\;n = 3,\;4$

Obviously $(b)$ gives the values.

$ = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$.

Therefore ${S_n} = \frac{1}{2}\left\{ {\Sigma {n^2} + \Sigma n} \right\}$

$ = \frac{{n(n + 1)(n + 2)}}{6}$.

$ = {2^2}({1^2} + {2^2} + {3^2} + ....... + {n^2})$

$ = \frac{{4n(n + 1)(2n + 1)}}{6} = \frac{{2n(n + 1)(2n + 1)}}{3}$.

==> ${T_n} = \frac{{2n(n + 1)}}{2} - n = {n^2}$

$\therefore {S_n} = \sum\limits_{k = 1}^n {({k^2})} = \frac{{n(n + 1)(2n + 1)}}{6}$

Hence sum of $(n - 1)$ terms ${S_{n - 1}} = \frac{{(n - 1)\,n\,(2n - 1)}}{6}$.

Put $n = 1,\,2,\,3,....,(n + 1)$

${T_1} = 2\,\left[ {\frac{1}{1} - \frac{1}{2}} \right]\,,\,\,{T_2} = 2\,\left[ {\frac{1}{2} - \frac{1}{3}} \right]\,,........,$

${T_{n + 1}} = 2\left[ {\frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right]$

Hence sum of $(n + 1)$ terms $ = \sum\limits_{k = 1}^{n + 1} {{T_k}} $

$ \Rightarrow {S_{n + 1}} = 2\left[ {1 - \frac{1}{{n + 2}}} \right]$

$ \Rightarrow {S_{n + 1\,}}\, = \frac{{2(n + 1)}}{{(n + 2)}}$.

$= \frac{{\cos A\cos C - \sin A\sin C}}{{\cos A\cos C + \sin A\sin C}}$

==> $\frac{{1 - {{\tan }^2}B}}{{1 + {{\tan }^2}B}} $

$= \frac{{1 - \tan A\tan C}}{{1 + \tan A\tan C}}$

==> $1 + {\tan ^2}B - \tan A\tan C - \tan A\tan C{\tan ^2}B$

$ = 1 - {\tan ^2}B + \tan A\tan C - \tan A\tan C{\tan ^2}B$

==> $2{\tan ^2}B = 2\tan A\tan C $

$\Rightarrow {\tan ^2}B = \tan A\tan C$

Hence, $\tan A, \tan B$ and $\tan C$ will be in $G.P.$

$a_1, a_2, a_3, \ldots a_n$ are in A.P.

$a_1=2, a_2-a_1=5=d$

$a_k \leq 2021$

$a_1+(k-1) d \leq 2021$

$k \leq 405$

median of $a_1, a_2, \ldots, a_{405}$

is $a_{203}=a_1+(203-1) d=1011$

Work done by Team A in one day $=\frac{1}{12}$

work done by Team $B$ in one day $=\frac{1}{36}$

$\left(\frac{4}{12}\right)+2\left(\frac{1}{12}+\frac{1}{36}\right)+ n \times \frac{2}{36}=1$ $\frac{4}{12}+\frac{2}{12}+\frac{2}{36}+\frac{ n }{18}=1$

$\frac{1}{2}+\frac{2}{36}+\frac{ n }{18}=1$

$\frac{ n }{18}=1-\frac{20}{36}=\frac{16}{36}$

$n=8$

$\frac{\sin A \frac{\cos C}{\sin C}+\cos A}{\sin B \frac{\cos C}{\sin C}+\cos B}=\frac{\frac{\sin A \cos C+\sin C \cos A}{\sin C}}{\frac{\sin B \operatorname{Cos} C+\cos B \sin C}{\sin C}}$

$=\frac{\sin (A+C)}{\sin (B+C)}=\frac{b}{a}$

a, b, c $\rightarrow$ G.P.

$b =a r ; c =a r ^2$

$C-I$ If $r > 1$ then

$a+a r > a r^2 \Rightarrow r^2-r-1 < 0$

$r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ but $r > 1$

Hence $r \in\left(1, \frac{1+\sqrt{5}}{2}\right) \quad \ldots .( I ) \ldots$

$C-II$ $0 < r < 1$

$a r+a r^2 > a \Rightarrow r^2+r-1 > 0$

$r \in\left(-\infty, \frac{-1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)$

but $r \in(0,1)$

$r \in\left(\frac{\sqrt{5}-1}{2}, 1\right) \quad \ldots (II) ....$

$C-III$ when $r=1$ then $\Delta$ is equilateral

Hence $r \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$

$p =99 \quad q =101$

$p _1=\log \left(\frac{ p + q }{2}\right)=\log _{10} 100=2$

$q _1=\frac{1}{2}(\log p +\log q )=\frac{1}{2} \log 99 \times 101$

$=\log \sqrt{ pq }$

$\frac{ p + q }{2} \geq \sqrt{ pq }$

$\log _{10}\left(\frac{p+q}{2}\right) > \frac{1}{2} \log (p q)$

$p _1 > q _1 \quad \ldots . .( a )$

$\frac{ p _1+ q _1}{2}>\sqrt{ p _1 q _1}$

$\log \left(\frac{ p _1+ q _1}{2}\right) > \log \sqrt{ p _1 q _1}$

$p _2 > q _2$

so option $A$ is correct.

$1,2,3, \ldots . n$

$\frac{\frac{ n ( n +1)}{2}- k }{ n -1}=16$

$n ( n +1)-2 k =32( n -1)$

$n ^2+ n -2 k =32 n -32$

$n^2-31 n-2 k=-32$

$n ^2-31 n +32=2 k$

$k =\frac{ n ^2-31 n +32}{2}$

$1 \leq \frac{ n ^2-31 n +32}{2} \leq n$

$n ^2-31 n +30 \geq 0 \quad \quad n ^2-33 n +32 \leq 0$

$( n -30)( n -1) \geq 0 \quad \quad ( n -32)( n -1) \leq 0$

$n =31$

$k =\frac{ n ^2-31 n +32}{2}$

$k =\frac{(31)^2-(31)^2+32}{2}=16$

$k =16, n =31$

$n + k =47$

Since, $\angle D A B, \angle A B C$ and $\angle B C D$ are in $AP \therefore$ Let $\angle D A B=\theta-\alpha, \angle A B C=\theta$ and $\angle B C D=\theta+\alpha$

$\therefore$ Median of $\angle D A B, \angle A B C$ and $\angle B C D=\theta$

From point $E$ all the vertices are at equal distance.

$\therefore A B C D$ is cyclic.

and $\angle A D C=2 \pi-(\theta-\alpha+\theta+\theta+\alpha)$

$\quad=2 \pi-3 \theta$

and $\angle A D C+\angle A B C=\pi$

$\Rightarrow 2 \pi-3 \theta+\theta=\pi$

$\therefore \quad \theta=\frac{\pi}{2}$

Given,

$\begin{aligned} M^2=& 2^{66}-2^{46}+2^{32}+2^{30}-2^{16}+1 \\ M^2=& 2^{46}\left[\frac{2^{14}-1}{2-1}\right]+2^{32}+2^{16}\left[\frac{2^{14}-1}{2-1}\right]+1 \\ M^2=& 2^{46}\left[1+2+2^2+\ldots+2^{13}\right] \\ & \quad+2^{32}+2^{16}\left[1+2+2^2+\ldots+2^{13}\right]+1 \\ M^2=& \frac{2^{59}+2^{58}+\ldots+2^{46}}{14 \text { terms }}+2^{32} \end{aligned}$

$\underbrace{+2^{29}+2^{28}+\ldots+2^{16}}_{14 \text { terms }}+2^0$

Therefore, on base $2$ representation of $M^2$, there will be $30$ times digit $1$ .

$\frac{1+x^{2020}}{1+x^{2018}}=\frac{x^2\left(1+x^{2018}\right)+1-x^2}{1+x^{2018}}$

$=x^2+\frac{1-x^2}{1+x^{2018}}$

Put $x=2 \therefore\left[4+\frac{(-3)}{1+2^{2018}}\right]=3$

Put $x=3 \therefore\left[9-\frac{8}{1+3^{2018}}\right]=8$

Similarly for $x=4,\left[16-\frac{15}{1+4^{2018}}\right]=15$

For $x=5,\left[25-\frac{24}{1+5^{2018}}\right]=24$

For $x=6,\left[36-\frac{35}{1+6^{2018}}\right]=35$

$\therefore$ Required sum

$=3+8+15+24+35=85$

Given,

$a= 2^{101}\left(\frac{1}{101 !}\right)+2^{102}\left(\frac{1}{101 !}+\frac{1}{102 !}\right)$ $+2^{103}\left(\frac{1}{101 !}+\frac{1}{102 !}+\frac{1}{103 !}\right)$

$\quad\quad\quad\quad\quad\quad\quad+ \ldots+2^{200}\left(\frac{1}{101 !}+\frac{1}{102 !}+\ldots+\frac{1}{200 !}\right)$

$a= \frac{1}{101 !}\left(2^{101}+2^{102}+\ldots+2^{200}\right)$

$\quad\quad\quad\quad\quad\quad+\frac{1}{102 !}\left(2^{102}+2^{103}+\ldots+2^{200}\right)$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\ldots+\frac{1}{200 !}\left(2^{200}\right)$

$a=\sum_{n=101}^{200} \frac{1}{n !}\left(2^n+2^{n+1}+\ldots \ldots+2^{200}\right)$

$a= \sum_{n=101}^{200} \frac{1}{n !}\left(\frac{2^n\left(2^{201-n}-1\right)}{2-1}\right)$

$= \sum_{n=101}^{200} \frac{2^{201}-2^n}{n !}$

$a= b \Rightarrow \frac{ a }{b}=1$

The sum of first $n, n \geq 1$ terms of arithmetic progression with first term $2$ and common difference $4$, is

$S_n=\frac{n}{2}[4+(n-1) 4]=2 n^2$

So, the average of the first $n$ terms

$M_n=\frac{S_n}{n}=2 n$

Now, $\sum\limits_{n=1}^{10} M_n=2 \sum\limits_{n=1}^{10} n$

$=2 \times\left(\frac{10 \times 11}{2}\right)=110$

Given, $S_m=n$ and $S_n=m$

$S_m=\frac{m}{2}[2 a+(m-1) d]=n$

$S_n=\frac{n}{2}(2 a+(n-1) d)=m \text { (i) }$

On subtracting Eq.$(ii)$ from Eq.$(i)$, we get

$(m-n) a+(m-n)(m+n-1) \frac{d}{2}$

$=-(m-n)$

$2 a+(m+n-1) d=-2 \quad[m \neq n]$

$S_{m+n}=\frac{m+n}{2}(2 a+(m+n-1) d)$

$=\frac{m+n}{2}(-2)=-(m+n)$

We have,

$x^4+4 y^4+16 z^4+64=32 x y z$

We know $\quad AM \geq GM$

$\Rightarrow \frac{x^4+4 y^4+16 z^4+64}{4}$

$\geq\left(x^4 \times 4 y^4 \times 16 z^4 \times 64\right)^{1 / 4}$

$\Rightarrow \quad x^4+4 y^4+16 z^4+64 \geq 32 x y z$

$\Rightarrow \quad x=\pm 2 \sqrt{2}, y=\pm 2, z=\pm \sqrt{2}$

For $x, y, z$

For $x^4+4 y^4+16 z^4+64=32 x y z$

Either each of $x, y, z$ is positive $\rightarrow 1$ case

or two of $x, y, z$ are negative $\rightarrow 3$ cases

$\therefore 4$ cases of different $(x, y, z)$ triplets

$4$ possible $x+y+z$ values $(x \neq y \neq z)$.

We have,

$x_k \geq k^4+k^2+1, \forall k \in[1,2018]$

$N \sum \limits_{k=1}^{2018} k$

I. I $\left(\sum \limits_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum \limits_{k=1}^{2018} k x_k^2\right)$

We know

$\left(\frac{x_1+x_2+x_3+\ldots+x_n}{n}\right) \leq \frac{x_1^2+x_2^2+x_3^2+\ldots+x_n^2}{n}$

$\therefore\left[\begin{array}{c}x_1+\left(x_2+x_2\right)+\left(x_3+x_3+x_3\right) \\ +\ldots\left(x_{2018}+x_{2018 \ldots \ldots} 2018 \text { times }\right) \\ \hline 1+2+3+4+\ldots+2018\end{array}\right]^2$

$\leq \frac{x_1^2+\left(x_2^2+x_2^2\right)+\left(x_3^2+x_3^2+x_3^2\right) \ldots}{1+2+3+4+\ldots+2018}$

$\Rightarrow\left(\frac{x_1+2 x_2+3 x_3+\ldots+2018 x_{2018}}{1+2+3+4+\ldots+2018}\right)$

$\leq \frac{x_1^2+2 x_2^2+\ldots+2018 x_{2018}^2}{1+2+3+\ldots+2018}$

$\Rightarrow \frac{\left(\sum \limits_{k=1}^{2018} k x_k\right)^2}{\left(\sum \limits_{k=1}^{2018} k\right)^2} \leq \frac{\sum \limits_{k=1}^{2018} k x_k^2}{\sum \limits_{k=1}^{2018} k}=\left(\sum_{k=1}^{2018} k x_k\right)^2 \leq N \sum_{k=1}^{2018} k x_k^2$

Hence, $I$ is true.

II. $\left(\sum \limits_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum \limits_{k=1}^{2018} k^2 x_k^2\right)$

Similarly for $1$

$\left(\frac{x_1+2 x_2+3 x_3+\ldots+2018 x_{2018}}{2018}\right)^2$

$\leq\left(\frac{x_1^2+\left(2 x_1\right)+\ldots\left(2018 x_{2018}\right)}{2018}\right)$

$\Rightarrow \frac{\left(\sum \limits_{k=1}^{2018} k x_k\right)^2}{(2018)^2} \leq \frac{\sum \limits_{k=1}^{2018} k^2 x_k 2}{2018}$

$=\left(\sum \limits_{k=1}^{2018} k x_k\right)^2 \leq 2018 \sum \limits_{k=1}^{2018} k^2 x_k$

$N=\sum \limits_{k=1}^{2018} k=\frac{2018 \times 2019}{2}$

$\because\left(\sum \limits_{k=1}^{2018} k x_k\right)^2$ is always less than or equal to

$2018 \sum \limits_{k=1}^{2018} k^2 x_k^2$

$\therefore$ It will always be less than $N\left(\sum \limits_{h=1}^{2018} k^2 x_k\right)$ Hence,$I$ and $II$ both are true.

We have,

$I$. $a, b,, c, d, e$ are angle of convex pentagon in degree.

$II$. $a \leq b \leq c \leq d \leq e$

$IIl$. $a, b, c, d, e$ are in $AP$

$a + b + c + d + e=540^{\circ}$

Let $a=\alpha$, common difference $=D$

$\therefore \quad \frac{5}{2}(2 a+4 D)=540^{\circ}$

$a+2 D=108$ and $a+4 D<180^{\circ}$

$[\because$ interior angle of polygon is

less than $\left.180^{\circ}\right]$

$\therefore \quad 108^{\circ}-2 D+4 D < 180^{\circ}$

$2 D < 180^{\circ}-108^{\circ}$

$0 < D < 36$

$\therefore$ Total 36 types are possible.

Given, $a_1+a_2+a_3+a_4=0$ and $\quad a_1^2+a_2^2+a_3^2+a_4^2=1$

It is possible only

when, $a_1=a_2=\frac{1}{2}$ and $a_3=a_4=-\frac{1}{2}$ $\therefore\left(a_1-a_2\right)^2+\left(a_2-a_3\right)^2+\left(a_3-a_4\right)^2 +\left(a_4-a_1\right)^2$

$\left(\frac{1}{2}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{1}{2}\right)^2+\left(-\frac{1}{2}+\frac{1}{2}\right)^2+\left(-\frac{1}{2}-\frac{1}{2}\right)^2$

$0+1+0+1=2$

The value lies between $(1.5,2.5)$.

Given $a_1=i+\frac{1}{i}$ for $i=1,2,3, \ldots, 20$

$p=\frac{1}{20}\left(a_1+a_2+a_3+\ldots+a_{20}\right)$

$q=\frac{1}{20}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3} \ldots+\frac{1}{a_{20}}\right)$

Clearly, $q > 0$.

Let $\quad q < \frac{22-p}{21}$

To prove $q+\frac{p}{21}<\frac{22}{21}$

$\therefore q+\frac{p}{21}=\frac{1}{20}\left[\frac{1}{\alpha_1}+\frac{1}{a_2}+\frac{1}{\alpha_3} \ldots \frac{1}{a_2}\right] +\frac{1}{20} \frac{1}{21}\left[a_1+a_2+a_3 \ldots a_{20}\right]$

$=\frac{1}{20}\left[\sum \frac{i}{i^2+1}+\frac{1}{21}\left(\sum i+\sum \frac{1}{i}\right)\right]$

$=\frac{1}{20}\left[\frac{1}{2}+\sum \limits_{i=2}^{20} \frac{i}{i+1}+\frac{20 \times 21}{2}+\sum \limits_{i=1}^{20} \frac{1}{21}\right]$

$=\frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\sum_{i=2}^{20} \frac{i}{i^2+1}+\frac{1}{21}+\sum \limits_{i=1}^{20} \frac{1}{i}\right]$

$ < \frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\frac{2}{5} \sum \limits_{i=2}^{20} 1+\frac{1}{21} \sum \limits_{i=1}^{20}\right]$

$ < \frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\frac{2}{5} \times 19+\frac{1}{21} \times 20\right]$

$ < \frac{1}{2}+\frac{1}{20}[1+8+1] < \frac{1}{2}+\frac{1}{2} < 1 < \frac{22}{21}$

$\because q \in\left(0, \frac{22-p}{21}\right)$

We have, $S=\sum \limits_{n=0}^{\infty} \frac{ c _n}{10^{2 n}}$

$a_{0}=a_1=1$ and $a_j=20 a_{j-1}-108 a_{j-2}, j \geq 2$

$a_n=20 a_{n-1}-108 a_{n-2}$

$\frac{a_n}{10^{2 n}}=\frac{20 a_{n-1}}{10^{2 n}}-\frac{108 a_{n-2}}{10^{2 n}}$

$=\frac{20}{100} \frac{a_{n-1}}{10^{2(n-1)}}-\frac{108}{10^4}-\frac{a_{n-2}}{10^{2(n-2)}}$

$\sum \limits_{n=2}^{\infty} \frac{a_n}{10^{2 n}}=\frac{1}{51 a_{n-1}^{2(n-1)}}-\frac{27}{2500}-\frac{\Sigma a_{n-2}}{10^{2(n-2)}}$

$S-1-\frac{1}{10}=\frac{1}{5}(S-1)-\frac{27}{2500} S$

$S\left(1-\frac{1}{5}+\frac{27}{2500}\right)=1+\frac{1}{100}-\frac{1}{5}$

$S\left(\frac{2500-500+27}{2500}\right)=\frac{100+1-20}{100}$

$S=\frac{81 \times 25}{2027}$

$S=\frac{2025}{2027}$

$a=2025$

We have,

$x^4+y^4+z^4+1=4 x y z$

$AM \geq GM$

$\therefore \frac{x^4+y^4+z^4+1}{4} \geq\left(x^4 \cdot y^4 \cdot z^4 \cdot 1\right)^{1 / 4}$

$\Rightarrow \quad \frac{4 x y z}{4} \geq|x y z| \Rightarrow x y z > 0$

$\text { It is possible }(1,1,1)(-1,-1,1)(-1,1,-1)$

$(1,-1,-1)$

So number of triplet $(x, y, z)=4$

Here,

$S_1=a^2+\left(\frac{a}{2}\right)^2+\left(\frac{a}{4}\right)^2+\ldots$

$S_2=\left(\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{2 \sqrt{2}}\right)^2+\ldots$

$S_1=a^2+\frac{a^2}{4}+\frac{a^2}{16}+\ldots=\frac{a^2}{1-\frac{1}{4}}=\frac{4 a^2}{3}$

$S_2=\frac{a^2}{2}+\frac{a^2}{8}+\frac{a^2}{32}+\ldots=\frac{a^2 / 2}{1-\frac{1}{4}}=\frac{4 a^2}{6}$

$\therefore \quad \frac{S_1}{S_2}=\frac{\frac{4 a^2}{3}}{4 a^2 / 6}=2$

Let,$P(x)=1+x^2+x^4+x^6+\ldots+x^{22}$

$Q(x)=1+x+x^2+x^3+\ldots+x^{11}$

$P(x)=\frac{x^{24}-1}{x^2-1}$

$\left[\because a+a r+a r^2+\ldots+a r^{n-1}=\frac{a\left(r^n-1\right)}{r-1}\right]$

Similarly, $Q(x)=\frac{x^{12}-1}{x-1}$

$\therefore \quad \frac{P(x)}{Q(x)}=\left(\frac{x^{24}-1}{x^2-1}\right)\left(\frac{x-1}{x^{12}-1}\right)=\frac{x^{12}+1}{x+1}$

Now, $\left(1+x^{12}\right)$ is divided by $(x+1)$, then by remainder theorem remainder is $2$

Now, $P(x)$ is divided by $Q(x)$, then

remainder $=2\left(1+x^2\right)\left(1+x^4+x^8\right)$

$=2\left(1+x^2+x^4+\ldots+x^{10}\right)$

We have, $a_1, a_2, a_3, \ldots, a_{100}$ be non-zero real number and

$a_1+a_2+a_3+\ldots+a_{100}=0$

$a_i \cdot 2^{a_i} > a_i$ and $a_i \cdot 2^{-a_i} < a_i$

$\therefore \sum \limits_{i=1}^{100} a_1 \cdot 2^{a i} > \sum \limits_{i=1}^{100} a_i \text { and } \sum \limits_{i=1}^{100} a_1 \cdot 2^{-a_i} < \sum \limits_{i=1}^{100} a_i$

$\Rightarrow \sum \limits_{i=1}^{100} a_1 \cdot 2^{a_i} > 0 \text { and } \sum \limits_{i=1}^{100} a_1 \cdot 2^{-a_i} < 0$

Hence, option $(a)$ is correct.

We have,

$n+2 n+3 n+\ldots+99 n$ is a perfect square

$n(1+2+\ldots+99), \frac{n \times 99 \times 100}{2}$

$n \times 11 \times 9 \times 2 \times 25$

$=(3)^2 \times(5)^2 \times 2 \times 11 \times n$ is a perfect square

$\therefore n$ must be 22 .

$\therefore \quad n^2=(22)^2=484$

Number of digit of $n^2$ is $3 .$

Let the number of houses be $x, x+2, x+4, x+6, x+8, x+10, \ldots$ 6 th number of house is $a$.

$\because x+10=a \Rightarrow x=a-10$

$\therefore x > 10$

Now, $\quad S_n=\frac{n}{2}(2 x+(n-1) 2)$

$S_n=n(x+n-1)$

$\Rightarrow 170=n(a-10+n-1)$

$\Rightarrow n^2+(a-11) n-170=0$

$\Rightarrow n=-(a-11) \pm \sqrt{(a-11)^2+680}$

$\Rightarrow \quad n=\frac{(11-a) \pm \sqrt{(a-11)^2+680}}{2}$

$n \geq 6$

$\Rightarrow \quad n=\frac{(11-a) \pm \sqrt{(a-11)^2+680}}{2} \geq 6$

$\Rightarrow \quad a \leq \frac{800}{24} \leq 33.33$

$\because \quad 12 \leq a \leq 32$

$a=12,14,16,18, \ldots$

When, $a=18, n=10$, then $S_n=170$ $\because \quad a=18$

We have,

$C_1, C_2, C_3, \ldots, C_n$ be circle with radii

$r_1, r_2, \ldots, r_n$ respectively. $C_i$ and $C_{i+1}$ touch externally $X$-axis and $y=2 \sqrt{2} x+10$ are tangent of each circle.

Slope of line $y=2 \sqrt{2} x+10$ is $2 \sqrt{2}$

$\therefore \quad \tan 2 \theta =2 \sqrt{2}$

$\frac{2 \tan \theta}{1-\tan ^2 \theta} =2 \sqrt{2}$

$\sqrt{2} \tan ^2 \theta+\tan \theta-\sqrt{2}=0$

$(\sqrt{2} \tan \theta-1)(\tan \theta+\sqrt{2}) =0$

$\tan \theta=\frac{1}{\sqrt{2}} \tan \theta \neq-\sqrt{2}$

$\sin \theta=\frac{1}{\sqrt{3}}$

$\operatorname{In} \Delta P Q M, \sin \theta=\frac{Q M}{P Q}$

$\Rightarrow \quad P Q=\sqrt{3} Q M \Rightarrow P Q=\sqrt{3} r_1$

$\operatorname{In} \triangle P R N$,

$\sin \theta=\frac{R N}{P R}=\frac{r_2}{P Q+r_1+r_2}=\frac{r_2}{\sqrt{3} r_1+r_1+r_2}$

$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{r_2}{(\sqrt{3}+1) r_1+r_2}$

$\Rightarrow r_2+(\sqrt{3}+1) r_1=\sqrt{3}-r_2$

$\Rightarrow \quad \frac{r_2}{r_1}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$r_1, r_2, r_3$ a geometric progression with common ratio $2+\sqrt{3}$.

We have, $C_0$ be a circle of radius $1.$ $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1}$.

Let $a_0, a_1, a_2, a_3, \ldots, a_n$ be the length of sides of square inscribed in circle $C_0, C_1, C_2, \ldots, C_n$ and $r_0, r_1, r_2, \ldots, r_n$ be radius of circle.

$2 a_0^2 =4$

$a _0^2 =2$

$\pi r_1^2 =a_0^2$

$r_1^2 =\frac{2}{\pi}$

$2 a _1^2 =\left(2 r_1\right)^2=4 r_1^2$

$a _1^2 =\frac{4}{\pi}$

$\pi r_2^2 = c _1^2$

$\Rightarrow r_2^2=\frac{4}{\pi^2}$

Similarly, $r_n^2=\frac{2^2}{\pi^n}$

Now, $\sum \limits_{i=0}^{\infty}$ Area $\left(C_{i j}\right)$

$=\pi\left(-1+\frac{2}{\pi}+\frac{2^2}{\pi^2}+\frac{2^3}{\pi^3}+\ldots\right)$

$=\pi\left(\frac{1}{1-\frac{2}{\pi}}\right)$

$\left[S_{\infty}=1+r+r^2+\ldots=\frac{1}{1-r}\right]$

$=\frac{\pi^2}{\pi-2}$ 

We have,

$f_\sigma(x) =a_n x^{n-1}+a_{n-1} x^{n-2}+\ldots+a_2 x+a_1$

$\sigma =\left(a_1, a_2, a_3, \ldots, a_n\right) \text { of }(1,2,3, \ldots, n)$

$S_\sigma=$ Sum of roots of $f_\sigma(x)=0$

$S=\Sigma S_\sigma$

$S=-\left[\frac{\lambda-a_n}{a_n}+\frac{\lambda-a_{n-1}}{a_{n-1}}+\ldots+\frac{\lambda- a _1}{ \alpha _1}\right]$

$\forall \lambda=a_1+a_2+a_3+\ldots+a_n$

$S=-\left[\left(a_1+a_2+a_3+\ldots+a_n\right)\right.$

$\left.\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\right)-n\right]$

$S=n-\left[\begin{array}{r}\left(a_1+a_2+a_3+\ldots+a_n\right) \\ \left.\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\right)\right]\end{array}\right.$

From $AM \geq HM$

$\frac{ a _1+a_2+ a _3+\ldots+a_n}{n} \geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}}$

$\Rightarrow\left(a_1+a_2+a_3+\ldots+a_n\right)$

$\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\right) \geq n^2$

$S \leq-n(n-1)$

$\therefore \quad S \leq-n !$

Given, $\frac{1^3+2^3+3^3+\ldots+(2 n)^3}{1^2+2^2+3^2+\ldots+n^2}$

$=\frac{\frac{4 n^2(2 n+1)^2}{n(n+1)(2 n+1)}}{6}=\frac{6 n(2 n+1)}{n+1}$

$=\frac{12 n^2+6 n}{n+1}=(12 n-6)+\frac{6}{n+1}$

$\because \frac{6}{n+1} \text { is an integer if } n+1 \text { is factor of } 6$

$\because n+1=1,2,3,6 \Rightarrow n=1,2,5$

Sum of $n=1+2+5=8$

We have, $a x+9 y=5$ and $4 x+b y=3$ are parallel. $\therefore \quad \frac{a}{4}=\frac{9}{b} \Rightarrow a b=36$

$AM \geq GM$

$\therefore \quad \frac{a+b}{2} \geq \sqrt{a b}$

$a+b \geq 2 \sqrt{36}$

$a+b \geq 12$

Hence, least possible value of $a+b=12$.

We have,

$a, b, c, d, e$ are natural number and in AP.

Let $D$ is common difference of $AP$.

$\therefore$ Let $c=C$

$a =C-2 D$
$b =C-D$

$d =C+D$

$e =C+2 D$

$a+b+c+d+e=5 C$

and $b+c+d=3 C$

Given, $a+b+c+d+e$ is a cube of number

$\therefore 5 C=\lambda^3$

and $b+c+d$ is a square of number

$\therefore 3 C=u^2$

From Eqs.$(i)$ and $(ii)$, we get

$\frac{\lambda^3}{5}=\frac{u^2}{3}$

$\lambda^3$ and $u^2$ is a multiple of 15

$\therefore$ Smallest possible value of $\lambda=15$ and $u=45$

$\therefore \quad c=\frac{u^2}{3}=\frac{(45)^2}{3}=675$

$\therefore$ Number of digits $=3$

We have, $x+y+z=10$

Let three number $x+1, y+1, z+1$

$AM \geq GM$

$\quad \frac{(x+1)+(y+1)+(z+1)}{3} \geq [(x+1)(y+1)(z+1)]^{1 / 3}$

$\Rightarrow \frac{x+y+z+3}{3} \geq (x y z+x y+y z+x z+x+y+z+1)^{1 / 3}$

$\Rightarrow \quad\left(\frac{13}{3}\right)^3 \geq x y z+x y+y z+x z+11$

$\Rightarrow \quad \quad(y+1)(z+1)]^{1 / 3}$

Now, $x, y, z$ are integer.

$\therefore x y z+x y+y z+x z+11$ is also integer.

$\therefore\left(\frac{13}{3}\right)^3$ is also integer.

$\therefore \quad\left[\left(\frac{13}{3}\right)^3\right]=81\left[\because\left(\frac{13}{3}\right)^3=8137\right]$

$\therefore x y z+x y+y z+x z+11 \leq 81$

$\Rightarrow \quad x y z+x y+y z+x z \leq 70$

$\therefore$ Maximum value of $x y z+x y+y z+x z$ is

We have,

$\sum \limits_{k=1}^n\left(a k^3+b k^2+d k+d\right)=n^4, n \in N$

$a\left(\frac{n(n+1)}{2}\right)^2+b \frac{n(n+1)(2 n+1)}{6}$

$+c \frac{n(n+1)}{2}+d n=n^4$

$\Rightarrow \frac{a}{4} n^4+\left(\frac{2 a}{4}+\frac{2 b}{6}\right) n^3+\left(\frac{a}{4}+\frac{3 b}{6}+\frac{c}{2}\right)$

$n^2+\left(\frac{b}{6}+\frac{c}{2}+d\right) n=n^4$

$\therefore \frac{a}{4}=1, \frac{2 a}{4}+\frac{2 b}{6}=0, \frac{a}{4}+\frac{3 b}{6}+\frac{c}{2}=0$

$\frac{b}{6}+\frac{c}{2}+d=0$

On solving these equations, we get

$a=4, b=-6, c=4, d=-1$

$\therefore|a|+|b|+|d|+|d|=4+6+4+1=15$

Let the sides be $a, b, c$ which are in A.P. with $c$ as the smallest.

$\therefore c =10$

$\therefore a , b > 10$

$\therefore 2 b = a + c = a +10$

$\therefore b + c > a$

$\Rightarrow b +10 > a$

From eq $(1)$ and $(2)$:

$\Rightarrow b +10 > 2 b -10$

$\Rightarrow b < 20$

$\therefore 10 < b < 20$

$\therefore$ No. of possible values of $b=9$

$\therefore$ No. of triangles possible $=9$

We have,

$\frac{2^2+4^2+6^2+\ldots+(2 n)^2}{1^2+3^2+5^2+\ldots+(2 n-1)^2} > 101$

$\frac{\Sigma(2 n)^2}{\Sigma(2 n-1)^2}>\frac{101}{100}$

$\frac{4 \Sigma n^2}{\Sigma\left(4 n^2-4 n+1\right)} > \frac{101}{100}$

$\frac{4 \Sigma n^2}{4 \Sigma n^2-4 \Sigma n+\Sigma 1} > \frac{101}{100}$

$\frac{\frac{4(n)(n+1)(2 n+1)}{6}}{\frac{4 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n} > \frac{101}{100}$

$\frac{4 n(n+1)(2 n+1)}{n\left[4\left(2 n^2+3 n+1\right)-12 n-12+6\right]} > \frac{101}{100}$

$\frac{4(n+1)(2 n+1)}{8 n^2-2} > \frac{101}{100}$

$\frac{4(2 n+1)(n+1)}{2(2 n+1)(2 n-1)} > \frac{101}{100}$

$\frac{2 n+2}{2 n-1} > \frac{101}{100}$

$200 n+200 > 202 n-101$
$2 n < 301$

$n < \frac{301}{2}$

$\therefore$ Maximum value of $n=150$

Given,

$a_2=\frac{ a _1+ a _i}{2}$

$a_3=\frac{a_2+a_4}{2}$

$a_4=\frac{a_3+u_5}{2}$

$a_{2012}=\frac{a_{2011}+a_1}{2}$

$a_2+a_4+a_6+\ldots+a_{2012}=3018 \ldots$ (i)

$2 a_2+2 a_4+2 a_6+\ldots+2 a_{2012}=6036$

$\left(a_1+a_3\right)+\left(a_3+ a _5\right)+\left( a _5+a_7\right)+\ldots$

$+\left(a_{2011}+a_1\right)=6036$

$2\left(a_1+a_3+a_5+\ldots+a_{2011}\right)=6036$

$a_1+a_3+a_5+\ldots+a_{2011}=3018$

On adding Eqs.$(i)$ and $(ii)$,

$a_1+a_2+a_3+a_4+\ldots+a_{2011}+a_{2012}$

$=6036$

We have,

$\log _a b+\log _b a =c$

$AM \geq GM$

$\therefore \quad \frac{\log _a b+\log _b a}{2} \geq \sqrt{\log _a b \log _b a}$

$\Rightarrow \frac{c}{2} \geq \sqrt{\frac{\log b}{\log a} \times \frac{\log a}{\log b}} \Rightarrow \frac{c}{2} \geq 1 \Rightarrow c \geq 2$

$\therefore$ Smallest positive integer value of $c=2$.

We have,

Let two-digits numbers are $10 a+b$.

Given, $10 a+b$ is AM of $x$ and $y$

and $10 b+a$ is GM of $x$ and $y$.

$\therefore \quad \frac{x+y}{2}=10 a+b$

$\Rightarrow \quad \sqrt{x y}=10 b+a$

$\Rightarrow \quad x y=(10 b+a)^2$

$\Rightarrow(x+y)^2-(x-y)^2=4 x y$

$\therefore \quad(x-y)^2=(x+y)^2-4 x y$

$\Rightarrow \quad(x-y)^2=4(10 a+b)^2-4(10 b+a)^2$

$\Rightarrow \quad(x-y)^2=4(10 a+b+10 b+a)$

$(10 a+b-10 b-a)$

$\Rightarrow \quad(x-y)^2=4(11)(a+b) \cdot 9(a-b)$

$\Rightarrow \quad(x-y)^2=4 \times 11 \cdot(a+b) \cdot 9(a-b)$

$4 \times 11(a+b) \times 9(a-b)$ must be a perfect

square.

$\therefore \quad a+b=11, a-b=1$

On solving these equations, we get

$\therefore a=6, b=5$

$\therefore \quad x+y=2(10 a+b)$

$\Rightarrow \quad x+y=2(60+5)$

$\Rightarrow \quad x+y=130$

Let $S_n=i+2 i^2+3 i^3+\ldots+n i^n \ldots$ (i) $i S_n=i^2+2 i^2+\ldots(n-1) i^n+n i^{n+1} \ldots$ (ii)

On subtracting Eq. (ii) from Eq. (i), we get

$S_n(1-i)=i+i^2+i^3+\ldots+i^n-n i^{n+1}$

$\Rightarrow S_n(1-i)=\frac{i\left(1-i^n\right)}{1-i}-n i^{n+1}$

$\Rightarrow \quad S_n=\frac{i\left(1-i^n\right)}{-2 i}-\frac{n i^{n+1}}{1-i}$

$\Rightarrow \quad S_n=\frac{1-i^n}{-2}-\frac{n i^{n+i}(1+i)}{2}$.

Let $Z_1=\frac{1-i^n}{-2}$ and $Z_2=\frac{n i^{n+1}(1+i)}{2}$

$\therefore \quad\left|Z_1\right|=\frac{1}{\sqrt{2}}$ or 0 and $\left|Z_2\right|=\frac{n}{2} \sqrt{2}$

$\therefore \quad \frac{n}{2} \sqrt{2}=18 \sqrt{2} \Rightarrow n=36$

Let

$S=\frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+\frac{4^2+1}{4^2-1}+\ldots+\frac{(2011)^2+1}{(2011)^2-1}$

Here, $\quad T_r=\frac{r^2+1}{r^2-1}$

$T_r =\frac{r^2-1+2}{r^2-1}$

$=1+\frac{2}{r^2-1}=1+\frac{2}{(r-1)(r+1)}$

$T_r =1+\frac{1}{r-1}-\frac{1}{r+1}$

$S =\sum \limits_{r=2}^{2011} T_r$

$=\sum \limits_{r=2}^{2011}\left[1+\frac{1}{r-1}-\frac{1}{r+1}\right]$

$S=T_2+T_3+T_4+\ldots+T_{2011}$

$S =\left(1+\frac{1}{1}-\frac{1}{3}\right)+\left(1+\frac{1}{2}-\frac{1}{4}\right)$

$+\left(1+\frac{1}{3}-\frac{1}{5}\right)+\ldots+\left(1+\frac{1}{2010}-\frac{1}{2012}\right)$

$S =2010+1+\frac{1}{2}-\frac{1}{2012}-\frac{1}{2011}$

$S =2011+\frac{1}{2}-\left[\frac{1}{2011}+\frac{1}{2012}\right]$

$S$ is lie between $\left(2011,2011 \frac{1}{2}\right)$

Given, square base pyramid is incomplete.

The top layer $=13$ balls

There are $18$ layer completed.

So, total number of balls

$N=13^2+14^2+15^2+16^2+\ldots+30^2$

$N=\left(1^2+2^2+3^2+4^2+\ldots+30^2\right)$

$\Rightarrow N=5 \times 31 \times 61-2 \times 13 \times 25$

$=9455-650$

$=8805$

$\therefore \quad 8000 < N < 9000$

We have,

$\frac{2^2+4^2+6^2+\ldots+(2 n)^2}{1^2+3^2+5^2+\ldots+(2 n-1)^2} < 1.01$

$=\frac{\Sigma 4 n^2}{\sum\left(4 n^2-4 n+1\right)} < 1.01$

$= \frac{4 \frac{n(n+1)(2 n+1)}{6}}{4 \frac{n(n+1)(2 n+1)}{6}-4 \frac{n(n)(n+1)}{2}+n}$

$= \frac{4 \frac{n(n+1)(2 n+1)}{6}}{\frac{n(2 n+1)(2 n-1)}{3}}<101=\frac{2(n+1)}{2 n-1} < \frac{101}{100}$

$= 200 n+200 < 202 n-101$

$\Rightarrow 2 n > 301 \Rightarrow n > \frac{301}{2}=150.5$

$\therefore n > 151$

Let

$S_n=\left(1^2-1+1\right)(1 !)+ \left(2^2-2+1\right)(2 !)$

$+\ldots+\left(n^2-n+1\right)(n !)$

Here, $T_r=\left(r^2-r+1\right)(r !)$

$T_r=\left(r^2-1-r+2\right)(r !)$

$T_r=\left(r^2-1\right) r !-(r-2) r !$

$T_r=(r-1)(r+1) !-(r-2) r !$

$S_n= T_1+T_2+T_3+\ldots+T_n$
$S_n= (1-0)+(3 !-0)+(2 \cdot 4 !-1 \cdot 3 !)$

$\quad\quad+(3 \cdot 5 !-2 \cdot 4 !)+(4 \cdot 6 !)-3 \cdot 5 ! \ldots$

$\quad\quad+[(n-1)(n+1) !-(n-2)(n !)]$

$S_n= 1+(n-1)(n+1) !$

$S_n=(n-1)(n+1) !+1$

We have,

$\frac{1^2+2^2+3^2+4^2+\ldots+n^2}{1+2+3+4+\ldots+n}$

$=\frac{\frac{n(n+1)(2 n+1)}{6}}{n(n+1)}$

$=\frac{2 n+1}{2}=k(l \text { let })$

$\Rightarrow \quad n=\frac{3 k-1}{2}$

$\text { Now, } 1 \leq \frac{3 k-1}{2} \leq 100$

$\Rightarrow \quad 2 \leq 3 k-1 \leq 200$

$\Rightarrow \quad 2+1 \leq 3 k \leq 200+1$

$\Rightarrow \quad 3 \leq 3 k \leq 201$

$\Rightarrow \quad 1 \leq k \leq 201$

$\Rightarrow \quad 1 \leq k \leq 67$

Number of odd integer $=34$

Let $P(x)=1+x^2+x^4+\ldots+x^{2010}$

${c}P(x)=\frac{1-x^{2012}}{1-x^2}$

$P(x)=\frac{\left(1-x^{1006}\right)\left(1+x^{1006}\right)}{(1-x)(1+x)}$$P(x)=\left(1+x^{1006}\right)\left(\frac{1-x^{503}}{1-x}\right)\left(\frac{1+x^{503}}{1+x}\right)$

$\left.P(x)=\left(1+x^{1006}\right) 1+x+x^2+x^3+\ldots+x^{502}\right)$

$\left(1-x+x^2-x^3+\ldots+x^{502}\right)$

Thus, $P(x)$ is divisible by $1+x+x^2+\ldots x^{n-1}$

If $n-1=502 \Rightarrow n=503$

Let the sides of triangle are

$a, a r, a r^2 . \quad[\because$ sides of triangle in $GP ]$

Case I $r > 1$

We know sum of two sides is greater than third side.

$\therefore a+a r>a r^2 \Rightarrow r^2-r-1<0$

$\begin{array}{ll} \Rightarrow & r=\frac{1 \pm \sqrt{5}}{2} \\\Rightarrow & 1 < r < \frac{\sqrt{5}+1}{2}, r > 1\end{array}$

Case $II$ $0 < r < 1$

$\therefore a r^2+a r > a \Rightarrow r^2+r-1 > 0$

$\Rightarrow \quad r=\frac{-1 \pm \sqrt{5}}{2}$

$\Rightarrow \quad 1 > r > \frac{\sqrt{5}-1}{2}, 0 < r < 1$

$\therefore \quad r \in\left(\frac{\sqrt{5}-1, \sqrt{5}+1}{2}, \frac{2}{2}\right)$

If all trucks had packets of $1000\,g$, then total weight is

$1000(1+2 \left.+2^2+\ldots+2^9\right)$

$=1000\left(2^{10}-1\right)$

$=1000(1024-1)=1023000$

We have given total weight is $1022870$.

$\therefore 1022870 < 1023000$

Extra amount of weight

$=1023000-1022870$

$=130$

$130 =2^7+2^1=128+2$

$\therefore$ The trucks have the lighter bags are $2,8$

We have, $a_1, a_2, a_3, \ldots, a_n$ in an $AP$ and $a_1, a_2, a_4, a_8$ in GP.

Let $a_1=a, a_2=a+d, a _3=a+2 d$ and $a_1$, $a_2, a_4, a_8$ are in $GP$.

Let common ratio is $r$.

$\therefore \quad a_1=a, a_2=a r, a_4=a r^2, a_8=a r^3$

$\therefore a+ d =a r, a+3 d=a r^2, a+7 d=a r^3$

$\Rightarrow \quad 2 d =a r(r-1), 4 d=a r^2(r-1)$

$\Rightarrow \quad r=2$

We have, $2^n < k < 2^{n+1}, k \in N$

Number of integer between $2^n$ and $2^{n+1}$ is i.e $k=2^{n+1}-2^n-1$

First term $=2^n+1$

Last term $=2^{n+1}-1$

$\therefore S_n=\frac{2^{n+1}-2^n-1}{2}\left[2^n+1+2^{n+1}-1\right]$

$S_n=\frac{2^{n+1}-2^n-1}{2}\left(2^n\right)(1+2)$

$S_n=\frac{\left(2^n-1\right)\left(2^n\right) \cdot 3}{2}$

But $S_n=9 m, m \in I$

$\therefore \quad \frac{\left(2^n-1\right) 2^n \cdot 3}{2}=9 m$

$\Rightarrow \quad\left(2^n-1\right) 2^{n-1}=3\,m$

$\Rightarrow \quad 2^n\left(2^n-1\right)=6\,m$

It is possible when, $n$ is even.

We have,

$P(x)=1+x+x^2+x^3+x^4+x^5$

$P(x)=\frac{1-x^6}{1-x}$

${\left[\because a+a r+a r^2+\ldots+a r^n=\frac{a\left(-r^n\right)}{1-r}\right]}$

It has $5$ roots let $\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$ they are $6$ th roots of unity

Now,

$P\left(x^{12}\right)=1+x^{12}+x^{24}+x^{36}+x^{48}+x^{60}$

$\therefore P\left(x^{12}\right) =P(x) \cdot Q(x)+R(x)$

Here, $R(x)$ is a polynomial of maximum degree $4 .$

Put $x=\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$; we get

$R\left(\alpha_1\right)=6=R\left(\alpha_2\right) =R\left(\alpha_3\right)$

$=R\left(\alpha_4\right)=R\left(\alpha_5\right)$

$\therefore R(x)-6=0$ has 6 roots, which contradicts that $R(x)$ is maximum of degree $4 .$

$\therefore$ So it is an identity.

$\because \quad R(x)=6$

Let $a, b, c$ be the sides of a triangle.

$\therefore \quad a^2+b^2 \geq 2 a b \quad[\because AM \geq GM ]$

Similarly, $b^2+c^2 \geq 2 b c$

$c^2+a^2 \geq 2 a c$

$\Rightarrow 2\left(a^2+b^2+c^2\right) \geq 2(a b+b c+c a)$

$\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+a c} \geq 1$

$t \geq 1 \quad\left[\because \frac{a^2+b^2+c^2}{a b+b c+a c}=t\right]$

$\because a, b, c$ be the side of a triangle.

$\begin{aligned} a+b>c \\ & \quad|a|>|c-b| \\ a \end{aligned}$

$\Rightarrow \quad a^2 > (c-b)^2$

$\Rightarrow \quad a^2 > c^2+b^2-2 b c$

$\Rightarrow \quad b^2+c^2-a^2<2 b c$

Similarly, $a^2+b^2-c^2 < 2 a b$

and $\quad c^2+a^2-b^2 < 2 a c$

On adding Eqs.$(i), (ii)$ and $(iii)$, we get

$a^2+b^2+c^2 < 2(a b+b c+c a)$

$\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+c a} < 2$

$\therefore \quad t < 2$

$\therefore \quad\{x \in R \mid 1 \leq x < 2\}$

If we fix $j=1$, then $i$ range, we get $1+2+3+\ldots+(n-1)$

if we fix $j=2$, then $i$ range, we get $1+2+3+\ldots+(n-2)$

Similarly, we fix $j=n-1$, then $i$ range, We gret

if we put them all together we get a total of ( $n-1)$ numbers $1^{\prime}$ s,$(n-2)$ number of $2^{\prime} s . \ldots .1$ number of $(n-1)^{\prime}$ sthen sum equals

$\sum \limits_{k=1}^{n-1} k(n-k)=n \sum \limits_{k=1}^{n-1} k-\sum \limits_{k=1}^{n-1} k^2$

$=\frac{n(n)(n-1) \quad n(n-1)(2 n-1)}{2}$

$\left.=\frac{n(n-1)\left[n-\frac{2 n-1}{2}[n\right.}{3}\right]$

$=\frac{n(n-1)(n+1)}{2 \cdot 3}={ }^{n+1} C_3$

$ 2 \mathrm{a}+19 \mathrm{~d}=79$      $.............(1)$

$ \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 $

$ 2 \mathrm{a}+9 \mathrm{~d}=29$        $................(2)$   

From $(1)$ and $(2)$ a $=-8, d=5$

$ S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] $

$ =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] $

$ =405-10 $

$ =395$

$ p=A, q=A R, r=A R^2$

$ A x^2+A R x-A R^2=0$

$ x^2+R x-R^2=0 < \beta $

$ \because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} $

$ \therefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{-R}{-R^2}=\frac{3}{4} \Rightarrow R=\frac{4}{3} $

$ (\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta=R^2-4\left(-R^2\right)=5\left(\frac{16}{9}\right) $

$ =80 / 9$

$ \frac{a+b+c}{3}=8 \rightarrow b=8, a+c=16 $

$ 64=(a+1)(19-a)=19+18 a-a^2 $

$ a^2-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a>10) $

$ a=15, c=1, b=8 $

$ \left((a b c)^{1 / 3}\right)^3=a b c=120$

(sum of infinite terms of A.G.P $=\frac{\mathrm{a}}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1-\mathrm{r})^2}$ )

$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$

$ 17 \mathrm{~m}=25 \mathrm{~m}-4(1+2 \ldots 24) $

$ 8 \mathrm{~m}=\frac{4 \cdot 24 \cdot 25}{2}=150$

$=7\left(a+a r^2+a r^4 \ldots+a r^{62}\right) $

$\Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2}$

$r=6$

$ \quad \geq 2 \quad \geq 2 $

$ \mathrm{k} \geq 10$

$ \mathrm{A}_{\mathrm{k}}=-\mathrm{kd}[2 \mathrm{a}+(2 \mathrm{k}-1) \mathrm{d}] $

$ \mathrm{A}_3=-153 $

$ \Rightarrow 153=13 \mathrm{~d}[2 \mathrm{a}+5 \mathrm{~d}] $

$ 51=\mathrm{d}[2 \mathrm{a}+5 \mathrm{~d}] $

$ \mathrm{A}_5=-435 $

$ 435=5 \mathrm{~d}[2 \mathrm{a}+9 \mathrm{~d}] $

$ 87=\mathrm{d}[2 \mathrm{a}+9 \mathrm{~d}] $

$ (2)-(1) $

$ 36=4 \mathrm{~d}^2$

$ \mathrm{~d}=3, \mathrm{a}=1 $

$ \mathrm{a}_{17}-\mathrm{A}_7=49-[-7.3[2+39]]=910$

$ \frac{10}{2}[2 \mathrm{a}+(10-1) \mathrm{d}]=390 $

$ \Rightarrow 2 \mathrm{a}+9 \mathrm{~d}=78 $               $......(1)$

$ \frac{\mathrm{t}_{10}}{\mathrm{t}_5}=\frac{15}{7} \Rightarrow \frac{\mathrm{a}+9 \mathrm{~d}}{\mathrm{a}+4 \mathrm{~d}}=\frac{15}{7} \Rightarrow 8 \mathrm{a}=3 \mathrm{~d} $                        $......(2)$

$ \text { From }(1) \&(2) \quad \mathrm{a}=3 \& \mathrm{~d}=8 $

$ \mathrm{~S}_{15}-\mathrm{S}_5=\frac{15}{2}(6+14 \times 8)-\frac{5}{2}(6+4 \times 8) $

$ =\frac{15 \times 118-5 \times 38}{2}=790$

$2,5,8,11, \ldots ., 404$

$\operatorname{LCM}(4,3)=12$

$11,23,35, \ldots . \text { let }(403)$

$403=11+(n-1) \times 12$

$\frac{392}{12}=n-1$

$33 \cdot 66=n$

$n=33$

$\operatorname{Sum} \frac{33}{2}(22+32 \times 12)$

$=6699$

${n}{2}(6+(n-1) 4)=3 n+2 n^2-2 n $

$ =2 n^2+n $

$ \sum_{k=1}^n S_k=2 \sum_{k=1}^n K^2+\sum_{k=1}^n K $

$ =2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} $

$=n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]$

$=\frac{n(n+1)(4 n+5)}{6} $

Rightarrow $40<\frac{6}{n(n+1)} \sum_{k=1}^n S_k<42 $

$ 40<4 n+5<42 $

$ 35<4 n<37 $

$ n=9$

$\therefore \mathrm{b}^2=\mathrm{ac}$

Also

$\log _{\circ}\left(\frac{a}{2 b}\right), \log _{\circ}\left(\frac{2 b}{3 c}\right), \log _{\circ}\left(\frac{3 c}{a}\right)$ are in $A.P.$

$\left(\frac{2 b}{3 \mathrm{c}}\right)^2=\frac{\mathrm{a}}{2 \mathrm{~b}} \times \frac{3 \mathrm{c}}{\mathrm{a}} $

$ \frac{\mathrm{b}}{\mathrm{c}}=\frac{3}{2}$

Putting in eq. $(i)$ $b^2=a \times \frac{2 b}{3}$

$ \frac{a}{b}=\frac{3}{2}$

$ a: b: c=9: 6: 4$

This is $A.P$. with common difference

$ d_1=-1+\frac{1}{4}=-\frac{3}{4} $

$ -129 \frac{1}{4}, \ldots \ldots \ldots . . ., 19 \frac{1}{4}, 20$

This is also $A.P.$ $a=-129 \frac{1}{4}$ and $d=\frac{3}{4}$

Required term $=$

$ -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right)$

$ =-129-\frac{1}{4}+15-\frac{3}{4}=-115$

$\mathrm{T}_{25}=4+(25-1) 5=4+120=124$

$3,6,9,12, \ldots$, up to $37^{\text {th }}$ term

$\mathrm{T}_{37}=3+(37-1) 3=3+108=111$

Common difference of $\mathrm{I}^{\text {st }}$ series $\mathrm{d}_{\mathrm{l}}=5$

Common difference of $\mathrm{In}^{\text {nd }}$ series $\mathrm{d}_2=3$

First common term $=9$, and

their common difference $=15\left(\mathrm{LCM}\right.$ of $\mathrm{d}_1$ and $\mathrm{d}_2$ )

then common terms are

$9,24,39,54,69,84,99$

$ \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\infty=57 $

$ \frac{\mathrm{a}}{1-\mathrm{r}}=57 \ldots \ldots \ldots \ldots .(\mathrm{I}) $

$ \sum_{\mathrm{n}=0}^{\infty} \mathrm{a}^3 \mathrm{r}^{3 \mathrm{n}}=9747 $

$ \mathrm{a}^3+\mathrm{a}^3 \cdot \mathrm{r}^3+\mathrm{a}^3 \cdot \mathrm{r}^6+\ldots \ldots \ldots \infty=9746 $

$ \frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9746 \ldots \ldots \ldots \ldots(\mathrm{II}) $

$ \frac{(\mathrm{I})^3}{(\mathrm{II})} \Rightarrow \frac{\mathrm{a}^3}{\frac{(1-\mathrm{r})^3}{\mathrm{a}^3}}=\frac{57^3}{9717}=19 $

$ \text { On solving, } \mathrm{r}=\frac{2}{3} \text { and } \mathrm{r}=\frac{3}{2}(\text { rejected) } $

$ \mathrm{a}=19 $

$ \therefore \mathrm{a}+18 \mathrm{r}=19+18 \times \frac{2}{3}=31$

$ \mathrm{ar}+\mathrm{ar}^5=\frac{70}{3} $

$ \mathrm{~T}_3 \cdot \mathrm{T}_5=49 $

$ \mathrm{ar}^2 \cdot \mathrm{ar}^4=49 $

$ \mathrm{a}^2 \mathrm{r}^6=49 $

$ \mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3} $

$ \mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3} $

$ \frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t} $

$ \frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3} $

$ 3 \mathrm{t}^2-10 \mathrm{t}+3=0 $

$ \mathrm{t}=3, \frac{1}{3}$

Increasing $G.P$. $\mathrm{r}^2=3, \mathrm{r}=\sqrt{3}$

$ \mathrm{T}_4+\mathrm{T}_6+\mathrm{T}_8 $

$ =\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7 $

$ =\mathrm{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right) $

$ =7(1+3+9)=91$

$ f(\theta)=1+\frac{2 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta} $

$ f(\theta)=\frac{2 \cos ^2 \theta}{\cos ^4 \theta-\cos ^2 \theta+1}+1 $

$ f(\theta)=\frac{2}{\cos ^2 \theta+\sec ^2 \theta-1}+1 $

$ \left.f(\theta)\right|_{\min .}=1 $

$ f(\theta)_{\max }=3 $

$ S=\frac{64}{1-1 / 3}=96$

Area of second $\Delta=\frac{\sqrt{3} a^2}{4} \frac{a^2}{4}=\frac{\sqrt{3} a^2}{16}$

Area of third $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{64}$

sum of area $=\frac{\sqrt{3} a^2}{4}\left(1+\frac{1}{4}+\frac{1}{16} \ldots\right)$

$\mathrm{Q}=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{1}{\frac{3}{4}}=\frac{\mathrm{a}^2}{\sqrt{3}}$

perimeter of $1^{\text {st }} \Delta=3 \mathrm{a}$

perimeter of $2^{\text {nd }} \Delta=\frac{3 a}{2}$

perimeter of $3^{\text {rd }} \Delta=\frac{3 \mathrm{a}}{4}$

$ \mathrm{P}=3 \mathrm{a}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right) $

$ \mathrm{P}=3 \mathrm{a} \cdot 2=6 \mathrm{a} $

$ \mathrm{a}=\frac{\mathrm{P}}{6} $

$ \mathrm{Q}=\frac{1}{\sqrt{3}} \cdot \frac{\mathrm{P}^2}{36} $

$ \mathrm{P}^2=36 \sqrt{3} \mathrm{Q}$

Sum of any two sides $>$ third side

$ a+a r>a r^2, a+a r^2>a r, a r+a r^2>a $

$ r^2-r-1<0 $

$ r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) $

$ r^2-r+1>0$                    $............(1)$

always true

$ \mathrm{r}^2+\mathrm{r}-1>0 $

$ \mathrm{r} \in\left(-\infty,-\frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)$               $...............(2)$

Taking intersection of $(1)$, $(2)$

$\mathrm{r} \in\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$

As $\mathrm{r}>1$

$ r \in\left(1, \frac{1+\sqrt{5}}{2}\right) $

$ {[r]=1[-r]=-2} $

$ 3[r]+[-r]=1$

$ \mathrm{f}(\mathrm{x})=\mathrm{a}+\mathrm{b}-2 \mathrm{c}+\mathrm{b}+\mathrm{c}-2 \mathrm{a}+\mathrm{c}+\mathrm{a}-2 \mathrm{~b}=0$

$ \mathrm{f}(1)=0 $

$\therefore \alpha \cdot 1=\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}} $

$ \alpha=\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}} $

$ \text { If, }-1<\alpha<0 $

$ -1<\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}}<0$

$\mathrm{b}+\mathrm{c}<2 \mathrm{a} \text { and } \mathrm{b}>\frac{\mathrm{a}+\mathrm{c}}{2}$
therefore, $\mathrm{b}$ cannot be G.M. between $\mathrm{a}$ and $\mathrm{c}$.
If, $0<\alpha<1$
$0<\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}}<1$
$\mathrm{b}>\mathrm{c}$ and $\mathrm{b}<\frac{\mathrm{a}+\mathrm{c}}{2}$
Therefore, $\mathrm{b}$ may be the $G.M.$ between $\mathrm{a}$ and $\mathrm{c}$.

$ t_{11}  =a r^{10}=a\left(r^2\right)^5=a \cdot\left(\frac{b}{a}\right)^5 $

$2^{\text {nd }} \text { G.P. }  \Rightarrow T_1=a, T_5=a r^4=b $

$\Rightarrow r^4  =\left(\frac{b}{a}\right) \Rightarrow r=\left(\frac{b}{a}\right)^{1 / 4} $

$ T_p  =a r^{p-1}=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}} $

$ t_{11}  =T_p \Rightarrow a\left(\frac{b}{a}\right)^5=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}} $

$ \Rightarrow 5  =\frac{p-1}{4} \Rightarrow p=21$

$ {\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right]} $

$ \frac{\alpha}{\beta}=\frac{a+b}{2 \sqrt{a b}}=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right) $

$ =\frac{1}{2}\left(\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}+\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}\right) $

$ =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{2 \times \sqrt{2}}=\sqrt{2}$

$ p=a+7 d \quad \ldots \text { (ii) } $

$ q=a+12 d \quad \ldots \text { (iii) } $

$ p-2=d $                    $((ii)-(i))$

$ q-p=5 d $                 $((iii)-(ii))$

$ q-p=5(p-2) $

$ q=6 p-10 $

$ p^2=2(6 p-10) $

$ p^2-12 p+20=0 $

$ p=10,2 $

$ p=10 ; q=50 $

$ d=8 $

$ a=-46 $

$ 2,10,50,250,1250 $

$ a^4=a+(n-1) d $

$ 1250=-46+(n-1) 8 $

$ n=163$