Maths M.C.Q (1 Marks)

${\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

$N=13^2+14^2+15^2+16^2+\ldots+30^2$

$N=\left(1^2+2^2+3^2+4^2+\ldots+30^2\right)$

$\Rightarrow N=5 \times 31 \times 61-2 \times 13 \times 25$

$=9455-650$

$=8805$

$\therefore \quad 8000 < N < 9000$