Maths M.C.Q (1 Marks)

are in $AP$ i.e. $1,5,9,13, \ldots, 77$.

Hence $a_{1}+a_{2}+a_{3}+\ldots+a_{18}=5+9+13+\ldots 18$ terms $=702$

$A _{1}= a + d$

$A _{ n }=100- d$

$\Rightarrow \frac{ A _{1}}{ A _{ n }}=\frac{1}{7} \Rightarrow \frac{ a + d }{100- d }=\frac{1}{7}$

$\Rightarrow 7 a+8 d=100$

$\Rightarrow 7\, a +8\left(\frac{100- a }{ n +1}\right)=100$........$(1)$

$\because a + n =33$.........(2)

$Now,\,by\, Eq. (1) and (2)$

$7 n^{2}-132 n-667=0$

$n =23$ and $n =\frac{-29}{7}$ $reject.$

$\Rightarrow 2 a _{1}+( n -1)=\frac{384}{ n } \ldots-(1)$

$\sum \limits_{ i =1}^{ n / 2} a _{2 i }=\frac{ n }{4}\left[2 a _{1}+2+\left(\frac{ n }{2}-1\right) 2\right]=120$

$2 a _{1}+ n =\frac{480}{ n } \cdots-(2)$

From equation ($2$) and ($1$)

$1=\frac{480}{ n }-\frac{384}{ n }$

$n =480-384=96$

Now, $24=2^{2} \cdot 3$

$\rightarrow \alpha$ is not the multiple of $2$ or $3$

Sum of values of $\alpha$

$= S ( U )-\{ S$ (multiple of $2$)$+ S$ (multiple of$3$ )

- $S$ (multiple of $6$)

$=(1+2+3+\ldots . .100)-(2+4+6 \ldots .+100)-(3$

$+6+\ldots . .99)+(6+12+\ldots .+96)$

$=\frac{100 \times 101}{2}-50 \times 51-\frac{33}{2} \times(3+99)+\frac{16}{2}(6+96)$

$=5050-2550-1683+816=1633$

$=\sum_{ i =1}^{20}\left( x _{ i - i }\right)^{2}$

$=\sum_{ i =1}^{20}\left( x _{ i }\right)^{2}+( i )^{2}-2 x _{ i } i$

Now $=\sum_{i=1}^{20}\left(x_{i}\right)^{2}=\frac{9\left(1-\left(\frac{1}{4}\right)\right)^{20}}{1-\frac{1}{4}}=12\left(1-\frac{1}{2^{40}}\right)$

$\sum_{i=1}^{20} i^{2}=\frac{1}{6} \times 20 \times 21 \times 41=2870$

$\sum_{ i =1}^{20} x _{ i } i = s =3+2.3 \frac{1}{2}+3.3 \frac{1}{2^{2}}+4.3 \frac{1}{2^{3}}+\ldots \ldots AGP$

$=6\left(2-\frac{22}{2^{20}}\right)$

$\overline{ x }=\frac{12-\frac{12}{2^{40}}+2870-12\left(2-\frac{22}{2^{20}}\right)}{20}$

$\overline{ x }=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right) \times \frac{1}{20}$

${[\overline{ x }]=142 }$

$\frac{6}{3^{12}}+\frac{10}{3^{11}}\left(\frac{6^{11}-1}{6-1}\right)$

$=2^{12} \cdot 1 ; m . n =12$

$\left(2^{30 \times 61} 4^{\frac{ n ( n +1)}{2}}\right) \frac{1}{60+ n }=2^{\frac{225}{8}}$

$2^{1830+ n ^{2}+ n }=2^{\frac{(225)(60+ n )}{8}}$

$=8 n ^{2}-217 n +1140=0$

$n =20, \frac{57}{8}$

$\sum_{ k =1}^{ n } nk - k ^{2}=\frac{ n ^{2}( n +1)}{2}-\frac{ n ( n +1)(2 n +1)}{6}$

$=1330$

$\left( A _{4}\right)^{4}=\frac{1}{1296}$

$A _{4}=\frac{1}{6}.....(1)$

$A _{2}+ A _{4}=\frac{7}{36}$

$A _{2}=\frac{1}{36}.....(2)$

$A _{6}=1$

$A _{8}=6$

$A _{10}=36$

$A _{6}+ A _{8}+ A _{10}=43$

$3 a _{2}+ a _{3}=2 a _{4}$

$3 ar + ar ^{2}=2 ar ^{3}$

$3+ r =2 r ^{2}$

$2 r ^{2}- r -3=0$

$r =-1$ and $r =\frac{3}{2}$

$a _{2}+ a _{4}=2 a _{3}+1$

$ar + ar ^{3}=2 ar ^{2}+1$

$a \left( r + r ^{3}-2 r ^{2}\right)=1$

$a\left(\frac{3}{2}+\frac{27}{8}-\frac{18}{4}\right)=1$

$a=\frac{8}{3}$

$When \;r =-1, a =-\frac{1}{4}\; (rejected, a _{1} > 0)$

$r =\frac{2}{3}, a =\frac{8}{3}(\text { selected })$

Now

$a_{2}+a_{4}+2 a_{5}$

$=\frac{8}{3} \times \frac{3}{2}+\frac{8}{3} \times \frac{27}{8}+2 \times \frac{8}{3} \times \frac{81}{16}$

$=4+9+27=40$

$\left(1-\frac{2}{3}\right)+\left(1-\frac{4}{9}\right)+\left(1-\frac{8}{27}\right)+\left(1-\frac{16}{81}\right) \ldots .100 \text { terms }$

$100-\left[\frac{2}{3}+\left(\frac{2}{3}\right)^{2}+\ldots\right]$

$100-\frac{2}{3}\left(1-\left(\frac{2}{3}\right)^{100}\right)$

$100-2\left(1-\left(\frac{2}{3}\right)^{100}\right)$

$S =98+2\left(\frac{2}{3}\right)^{100}$

$\Rightarrow[ S ]=98$

$\geq 7\left(\frac{\alpha^{2}}{2^{7}}\right)^{\frac{1}{7}}$

$\frac{7 \cdot(\alpha)^{2 / 7}}{2}=14$

$\left(\alpha^{2}\right)^{1 / 7}=2^{2}$

$\alpha=\left(2^{2}\right)^{7 / 2}=2^{7}$

$\alpha=128$

$=21\left[10\,ar +100\,ar ^{2}\right]$

$=21 . a _{11}$

$\frac{x+x+x+y+y}{5} \geq\left(x^{3} \cdot y^{2}\right)^{\frac{1}{5}}$

$\frac{3 x +2 y }{5} \geq\left(2^{15}\right)^{\frac{1}{5}}$

$(3 x +2 y )_{\min }=40$

$\frac{S}{2} =\frac{a_{1}}{2^{2}}+\frac{a_{2}}{2^{3}}+\ldots$

$\frac{S}{2}=\frac{a_{1}}{2}+d\left(\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots\right)$

$\frac{S}{2}=\frac{a_{1}}{2}+d\left(\frac{\frac{1}{4}}{1-\frac{1}{2}}\right)$

$\therefore S=a_{1}+d=a_{2}=4$

Or $4 a_{2}=16$

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

$S=\sum\limits_{n=2}^{\infty} \frac{n(n-1)}{2\left(7^{n}\right)}$

$S=\frac{1}{7^{2}}+\frac{3}{7^{3}}+\frac{6}{7^{4}}+\frac{10}{7^{5}}+\frac{15}{7^{5}}+\ldots \ldots$

$\frac{S}{7}=\frac{1}{7^{3}}+\frac{3}{7^{4}}+\frac{6}{7^{5}}+\frac{10}{7^{6}}+\ldots$

$6 \frac{S}{7}=\frac{1}{7^{2}}+\frac{2}{7^{3}}+\frac{3}{7^{4}}+\frac{4}{7^{5}}+\ldots$

$6 \frac{S}{7^{2}}=\frac{1}{7^{3}}+\frac{2}{7^{4}}+\frac{3}{7^{5}}+\ldots$

$6 \frac{S}{7} \cdot \frac{6}{7}=\frac{1}{7^{2}}+\frac{1}{7^{3}}+\ldots=\frac{1 / 7^{2}}{1-1 / 7}$

$6 \times 6 \frac{S}{7^{2}}=\frac{1}{7 \times 6}$

$S =\frac{7}{6^{3}}=\frac{7}{216}$

Alternate

$a _{ n +2}=2 a _{ n +1}- a _{ a }+1$

$\Rightarrow \frac{ a _{ n +2}}{7^{ n +2}}=\frac{2}{7} \frac{ a _{ n +1}}{7^{ n +1}}-\frac{1}{49} \frac{ a _{ n }}{7^{ n }}+\frac{1}{7^{ n +2}}$

$\Rightarrow \sum\limits_{ n =2}^{\infty} \frac{ a _{ n +2}}{7^{ n +2}}=\frac{2}{7} \sum\limits_{ n =2}^{\infty} \frac{ a _{ n +1}}{7^{ n +1}}-\frac{1}{49} \sum\limits_{ n =2}^{\infty} \frac{ a _{ n }}{7^{ n }}+\sum\limits_{ n =2}^{\infty} \frac{1}{7^{ n +2}}$

Let $\sum\limits_{ n =2}^{\infty} \frac{ a _{ n }}{7^{ n }}= p$

$\Rightarrow\left( p -\frac{ a _{2}}{7^{2}}-\frac{ a _{3}}{7^{3}}\right)=\frac{2}{7}\left( p -\frac{ a _{2}}{7^{2}}\right)-\frac{1}{49} p +\frac{1 / 7^{4}}{1-\frac{1}{7}}$

$\because a _{2}=1, a _{3}=3$

$\Rightarrow p -\frac{1}{49}-\frac{3}{343}=\frac{2}{7} p -\frac{2}{7^{3}}-\frac{ p }{49}+\frac{1}{6.7^{3}}$

$\Rightarrow p =\frac{7}{216}$

$\frac{S}{6}=\frac{1}{6}+\frac{5}{6^{2}}+\frac{12}{6^{3}}+\frac{22}{6^{4}}+\ldots$

on subtraction

$\frac{5}{6} S=1+\frac{4}{6}+\frac{7}{6^{2}}+\frac{10}{6^{3}}+\frac{13}{6^{4}}+\ldots$

$\frac{5}{36} S=1+\frac{4}{6^{2}}+\frac{7}{6^{3}}+\frac{10}{6^{4}}+\frac{13}{6^{5}}+\ldots$

on subtraction

$\frac{25}{36} S=1+\frac{3}{6}+\frac{3}{6^{2}}+\frac{3}{6^{3}}+\ldots=\frac{8}{5}$

$S=\frac{288}{125}$

$S =2+\frac{6}{7}+\frac{12}{7^{2}}+\frac{20}{7^{3}}+\frac{30}{7^{4}}+\ldots \ldots \ldots$

$\frac{S}{7}=\frac{2}{7}+\frac{6}{7^{2}}+\frac{12}{7^{3}}+\frac{20}{7^{4}}+\ldots \ldots \ldots$

$\Rightarrow  \frac{6 S }{7}=2+\frac{4}{7}+\frac{6}{7^{2}}+\frac{8}{7^{3}}+\frac{10}{7^{4}}+\ldots \ldots$

$\Rightarrow \frac{6 S }{7^{2}}=\quad \frac{2}{7}+\frac{4}{7^{2}}+\frac{6}{7^{3}}+\frac{8}{7^{4}}+\ldots \ldots \ldots$

$\Rightarrow  \frac{6 S }{7}\left(1-\frac{1}{7}\right)=2+\frac{2}{7}+\frac{2}{7^{2}}+\frac{2}{7^{3}}+\ldots \ldots \ldots$

$\Rightarrow  \frac{6^{2} S }{7^{2}}=\frac{2}{1-\frac{1}{7}}=\frac{2}{6} \times 7$

$\Rightarrow S= \frac{2 \times 7^{3}}{6^{3}} \Rightarrow 4 S =\frac{7^{3}}{3^{3}}=\left(\frac{7}{3}\right)^{3}$

$3 S =1 \cdot 3^{1}+2.3^{2} \ldots \ldots \ldots \ldots \ldots \ldots+9 \times 3^{9}+10 \times 3^{10}$

$-2 S =\left(1 \cdot 3^{0}+3^{1}+3^{2} \ldots 3^{9}\right)-10.3^{10}$

$S =5 \times 3^{10}-\left(\frac{3^{10}-1}{4}\right)$

$S =\frac{20.3^{10}-3^{10}+1}{4}=\frac{19.3^{10}+1}{4}$

$\frac{4-2}{2.3 .4}+\frac{5-3}{3.4 .5}+\ldots . .+\frac{102-100}{100.101 .102}=\frac{2 k }{101}$

$\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\ldots .+\frac{1}{100.101}-\frac{1}{101.102}=\frac{2 k }{101}$

$\frac{1}{2.3}-\frac{1}{101.102}=\frac{2\,k }{101}$

$\therefore 2\,k =\frac{101}{6}-\frac{1}{102}$

$\therefore 34\,k =286$

$\sum_{x=1}^{20}\left((r+1)^{2}-2 r\right) r !$

$\sum_{x=1}^{20}((r+1)(r+1) !-r \cdot r !)-\sum_{r=1}^{20} r \cdot r !$

$\sum_{x=1}^{20}((r+1)(r+1) !-r \cdot r !)-\sum_{r=1}^{20}((r+1) !-r !)$

$=(21.21-1)-(\lfloor 21-1)$

$=20.21 !=22 !-2 \cdot 21 !$

$a_{ n +2}=3 a_{ n +1}-2 a_{ n +1}: n \geq 0$

$a_{ n +2}-a_{ n +1}=2\left(a_{ n +1}- a _{ n }\right)+1$

$\begin{array}{ll} n =0 & a_{2}-a_{1}=2\left(a_{1}-a_{0}\right)+1 \\ n =1 & a_{3}-a_{2}=2\left(a_{2}-a_{1}\right)+1 \\ n =2 & a_{4}-a_{3}=2\left(a_{3}-a_{2}\right)+1 \\ n = n & a_{ n +2}-a_{ n +1}=2\left(a_{ n +1}-a_{ n }\right)+1\end{array}$

$\left(a_{ n +2}-a_{1}\right)-2\left(a_{ n +1}-a_{0}\right)-( n +1)=0$

$a_{ n +2}=2 a_{ n +1}+( n +1)$

$n \rightarrow n -2$

$a_{ n }-2 a_{ n -1}= n -1$

Now $a _{25} a _{23}-2 a _{25} a _{22}-2 a _{23} a _{24}+4 a _{22} a _{24}$

$=\left(a_{25}-2 a_{24}\right)\left(a_{23}-2 a_{22}\right)=(24)(22)=528$

$T _{ n }=1\,n$

So, $\sum_{n=1}^{15} T_{n}=120$

$=\frac{3}{2}(222) \times 21=6993$

$\Rightarrow \frac{1}{2} \sum_{ k =1}^{10} \frac{\left( k ^{2}+ k +1\right)-\left( k ^{2}- k +1\right)}{\left( k ^{2}+ k +1\right)\left( k ^{2}- k +1\right)}$

$\Rightarrow \frac{1}{2}\left(\sum_{ k =1}^{10}\left(\frac{1}{\left( k ^{2}- k +1\right)}-\frac{1}{ k ^{2}+ k +1}\right)\right)$

$\Rightarrow \frac{55}{111}=\frac{ m }{ n }$

$m + n =166$

$T_{n}=\frac{n}{4 n^{4}+1}$

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

$=\frac{1}{4}\left[\frac{1}{2 n ^{2}-2 n +1}-\frac{1}{2 n ^{2}+2 n +1}\right]$

$S _{10}=\sum\limits_{ n =1}^{10} T _{ n }=\frac{1}{4}\left[\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\ldots \ldots \cdot \frac{1}{200+20+1}\right]$

$=\frac{1}{4}\left[1-\frac{1}{221}\right]=\frac{1}{4} \times \frac{220}{221}-\frac{55}{221}=\frac{ m }{ n }$

$m + n =55+221=276$

$A=\left(\frac{1}{2}+\frac{1}{2^{3}}+\ldots \infty\right)+\left(\frac{1}{4^{2}}+\frac{1}{4^{4}}+\ldots \ldots \infty\right)$

$A=\left(\frac{\frac{1}{2}}{1-\frac{1}{4}}+\frac{\frac{1}{16}}{1-\frac{1}{16}}\right)$

$\Rightarrow A =\frac{1}{2} \times \frac{4}{3}+\frac{1}{16} \times \frac{16}{15} \Rightarrow A =\frac{11}{15}$

$B=\frac{-1}{2}+\frac{1}{4^{2}}+\frac{-1}{2^{3}}+\frac{1}{4^{4}}+\ldots \ldots \infty$

$B =\left(\frac{-1}{2}+\frac{-1}{2^{3}}+\ldots \infty\right)+\left(\frac{1}{4^{2}}+\frac{1}{4^{4}}+\ldots . \infty\right)$

$B =\frac{-\frac{1}{2}}{1-\frac{1}{4}}+\frac{\frac{1}{16}}{1-\frac{1}{16}}$

$\Rightarrow B =-\frac{1}{2} \times \frac{4}{3}+\frac{1}{16} \times \frac{16}{15}$

$B =-\frac{9}{15}$

$\frac{ A }{ B }=\frac{11}{15} \times \frac{15}{(-9)}$

$\frac{ A }{ B }=-\frac{11}{9}$

$K=2^{9}+2^{8} \cdot 3+2^{7} \cdot 3^{2}+\ldots . .+3^{9}$

$=\frac{2^{9}\left(\left(\frac{3}{2}\right)^{10}-1\right)}{\frac{3}{2}-1}=3^{10}-2^{10}$

Now,

$3^{10}-2^{10} =\left(3^{5}-2^{5}\right)\left(3^{5}+2^{5}\right)$

$=(211)(275)$

$=(35 \times 6+1)(45 \times 6+5)$

$=6  \lambda+5$

Remainder is $5 .$

$B=\sum_{i=1}^{10} \sum_{j=1}^{10} \max \{ i , j \}$

$A =\sum_{ j =1}^{10} \min ( i , 1)+\min ( j , 2)+\ldots \min ( i , 10)$

$=\underbrace{(1+1+1+\ldots+1)}_{19 \text { limes }}+\underbrace{(2+2+2 \ldots+2)}_{17 \text {dims }}+\underbrace{(3+3+3 \ldots+3)}_{15 \text { times }}$

$+\ldots (1) \;1$ times

$B =\sum_{ j =1}^{10} \max (i, 1)+\max ( j , 2)+\ldots \max ( i , 10)$

$=\underbrace{(10+10+\ldots+10)}_{19 \text { times }}+\underbrace{(9+9+\ldots+9)}_{17 \text {times }}+\ldots+11 \text { times }$

$A + B =20(1+2+3+\ldots+10)$

$=20 \times \frac{10 \times 11}{2}=10 \times 110=1100$

Roots are $p+\sqrt{q}, p-\sqrt{q}$ absolute difference between roots $2 \sqrt{q}$.

Now, $\left|f\left(a_{i}\right)\right|=500$

Let $a_{1}, a_{2}, a_{3}, a_{4} a_{r} a_{1} a+d, a+2 d, a+3 d$

$\left|f\left(a_{4}\right)\right|=500$

$\left|\left(a_{1}-p\right)^{2}-q\right|=500$

$\Rightarrow\left(a_{1}-p\right)^{2}-q=500$

$\Rightarrow \frac{9}{4} d^{2}-q=500$

$\text { and }\left|f\left(a_{1}\right)\right|^{2}=\left|f\left(a_{2}\right)\right|^{2}$

$\left(\left(a_{1}-p\right)^{2}-q\right)^{2}=\left(\left(a_{2}-p\right)^{2}-q\right)^{2}$

$\left(\left(a_{1}-p\right)^{2}-\left(a_{2}-p\right)^{2}\right)\left(\left(a_{1}-p\right)^{2}-q+\left(a_{2}-p\right)^{2}-q\right)=0$

$\Rightarrow \frac{9}{4} d^{2}-q+\frac{d^{2}}{4}-q=0$

$2 q=\frac{10 d^{2}}{4} \Rightarrow q=\frac{5 d^{2}}{4}$

$\Rightarrow d^{2}=\frac{4 q}{5}$

From equation $(1)$ $\frac{9}{4} \cdot \frac{4 \cdot q}{5}-q=500$

$\frac{4 q}{5}=500$

$\frac{4 q}{5}=500$

and $2 \sqrt{q}=2 \times \frac{50}{2}=50$

Last term $=199=\ell$

If $3$ term

$a, a+d, a+2 d$

$a _{ a }=\ell= a +( n -1) d$

$d _{ i }=\frac{\ell- a }{ n - l }$

$n \rightarrow$ number of terms

$n =3, d _{1}=\frac{199-100}{2}$

$=\frac{99}{2} \notin I$

$n =4, d _{2}=\frac{99}{3}=33 \in I$

$n =10, d _{3}=\frac{99}{9}=11 \in I$

$n =12, d _{4}=\frac{99}{11}=9 \in I$

$\therefore \sum d _{ i }=33+11+9=53$

$p + r =8 a$

$pr =2 a$

$\frac{1}{ p }+\frac{1}{ r }=4$

$\frac{2}{ q }=4$

$q =\frac{1}{2}$

$p =\frac{1}{5}$

$x^{2}+12 b x+6 b=0$

$q+s=-12 b$

$q s=6 b$

$\frac{1}{q}+\frac{1}{s}=-2$

$\frac{2}{r}=-2$

$r=-1$

$s=\frac{-1}{4}$

Now,$\frac{1}{ a }-\frac{1}{ b }=\frac{2}{ pr }-\frac{6}{ qs }=38$

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

$S_{n}=\frac{n[n(n+2)+1]}{(n+2)}$

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

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

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

Now $\frac{1}{26}+\sum \limits_{n=1}^{50}\left[\left(n^{2}-n\right)-2\left(\frac{1}{n+2}-\frac{1}{n+1}\right)\right]$

$=\frac{1}{26}+\left[\frac{50 \times 51 \times 101}{6}-\frac{50 \times 51}{2}-2\left(\frac{1}{52}-\frac{1}{2}\right)\right]$

$=41651$

$a _{2}= a _{1}+2=3$

$a _{3}= a _{2}+2=5$

$a _{4}= a _{2}+2=7$

$a _{ n }=2 n -1$

$b _{2}= a _{1}+ b _{1}=4$

$b _{3}= a _{3}+ b _{2}=9$

$b _{4}= a _{4}+ b _{3}=16$

$b _{ n }= n ^{2}$

$\sum_{ n =1}^{15} a _{ n } b _{ n }$

$\sum_{ n =1}^{15}(2 n -1) n ^{2}$

$\sum_{ n =1}^{15}\left(2 n ^{3}- n ^{2}\right)$

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

$Put _{ n }=15$

$=\frac{2 \times 225 \times 16 \times 16}{4}-\frac{15 \times 16 \times 31}{6}$

$=27560$

Series will satisfy

$a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4}, a_{4} a_{5}$

$\frac{1.2}{2.2} 2.3 \quad 2.4$

$a_{n}+\frac{1}{a_{n+1}}=\frac{a_{n+2}-\frac{1}{a_{n+1}}}{a_{n+2}}$

$=1-\frac{1}{a_{n+1} a_{n+2}}$

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

$=\frac{2 r+1}{2(r+1)}$

Now proof is given by

$=\prod_{r=1}^{30} \frac{(2 r+1)}{2(r+1)}$

$=\frac{(1 \cdot 3 \cdot 5 \cdot \ldots \ldots \cdot 61)}{2^{30} \cdot(2 \cdot 3 \cdot \ldots \ldots \cdot 31)}$

$\Rightarrow \frac{(1 \cdot 3 \cdot 5 \cdot \ldots \ldots \ldots \cdot 61)}{\mid 31 \cdot 2^{30}} \times \frac{2^{30} \times \underline{30}}{2^{30} \times \underline{30}}$

$=\frac{\lfloor 61}{2^{60}|31 \cdot| 30}$

$\alpha=-60$

Let $2^{\mathrm{x}}=\mathrm{t}$

$\log _{3}(t-5)^{2}=\log _{3} 2\left(t-\frac{7}{2}\right)$

$(t-5)^{2}=2 t-7$

$t^{2}-12 t+32=0$

$(t-4)(t-8)=0$

$\Rightarrow 2^{x}=4 \text { or } 2^{x}=8$

$X=2 \text { (Rejected) }$

$\text { Or } x=3$

Given $\mathrm{S}_{3 \mathrm{n}}=3 \mathrm{~S}_{2 \mathrm{n}}$

$\Rightarrow \frac{3 n}{2}[2 a+(3 n-1) d]=3 \frac{2 n}{2}[2 a+(2 n-1) d]$

$\Rightarrow 2 a+(3 n-1) d=4 a+(4 n-2) d$

$\Rightarrow 2 a+(n-1) d=0$

$\text { Now } \frac{S_{a n}}{S_{2 n}}=\frac{\frac{4 n}{2}[2 a+(4 n-1) d]}{\frac{2 n}{2}[2 a+(2 n-1) d]}=\frac{2[\underbrace{2 a+(n-1) d}_{-0}+3 n d]}{[\underbrace{2 a+(n-1) d}_{-0}+n d]}$

$=\frac{6 n d}{n d}=6$

$\Rightarrow 2 a+9 d=106 \ldots .(1)$

$\text { and } S_{5}=140 \Rightarrow \frac{5}{2}\{2 a+4 d\}=140$

$\Rightarrow 2 a+4 d=56 \ldots . .(2)$

$\Rightarrow 5 d=50 \Rightarrow d=10 \Rightarrow a=8$

Now, $S_{20}-S_{6}=\frac{20}{2}\{2 a+19 d\}-\frac{6}{2}\{2 a+5 d\}$

$=14 a+175 d$

$=(14 \times 8)+(175 \times 10)$

$=1862$

$A P : 11,16,21,26,31,36$

Common terms : $16,256,4096$ only

$s=\log _{9} \times(2+3+\ldots+22)$

$s=\log _{9} x\left\{\frac{21}{2}(2+22)\right\}$

Given $252\,\log _{9} x=504$

$\Rightarrow \log _{9} x=2 \Rightarrow x=81$

$=\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]+\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]+\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]+\ldots+\left[\frac{1}{10^{2}}-\frac{1}{11^{2}}\right]$

$=1-\frac{1}{121}$

$=\frac{120}{121}$

$=4-5+5-6+6+\ldots-50+50=4$

and $2 y=\tan \frac{\pi}{9}+\tan \frac{5 \pi}{18}$

so, $x-2 y=\frac{1}{2}\left(\tan \frac{\pi}{9}+\tan \frac{7 \pi}{18}\right)$

$-\left(\tan \frac{\pi}{9}+\tan \frac{5 \pi}{18}\right)$

$\Rightarrow|x-2 y|=\left|\frac{\cot \frac{\pi}{9}-\tan \frac{\pi}{9}}{2}-\tan \frac{5 \pi}{18}\right|$

$=\left|\cot \frac{2 \pi}{9}-\cot \frac{2 \pi}{9}\right|=0$

$\left(\operatorname{as\,\,\,\,tan} \frac{5 \pi}{18}=\cot \frac{2 \pi}{9} ; \tan \frac{7 \pi}{18}=\cot \frac{\pi}{9}\right)$

$n$ should not be multiple of $2,3,5$ and $17 .$

Sum of all $n=(1+3+5 \ldots . .+99)-(3+9+15+21+\ldots . .+99)-(5+25+35+55+65$ $+85+95)-(17)$

$=2500-\frac{17}{2}(3+99)-365-17$

$=2500-867-365-17=1251$

$=\frac{1}{d} \sum_{n=1}^{20}\left(\frac{1}{a_{n}}-\frac{1}{a_{n}+d}\right)$

$\Rightarrow \frac{1}{d}\left(\frac{1}{a_{1}}-\frac{1}{a_{21}}\right)=\frac{4}{9} \text { (Given) }$

$\Rightarrow \frac{1}{\mathrm{~d}}\left(\frac{\mathrm{a}_{21}-\mathrm{a}_{1}}{\mathrm{a}_{1} \mathrm{a}_{21}}\right)=\frac{4}{9}$

$\Rightarrow \frac{1}{\mathrm{~d}}\left(\frac{\mathrm{a}_{1}+20 \mathrm{~d}-\mathrm{a}_{1}}{\mathrm{a}_{1} \mathrm{a}_{2}}\right)=\frac{4}{9}$$\Rightarrow \mathrm{a}_{1} \mathrm{a}_{2}=45 \ldots (1)$

Now sum of first $21$ terms $=\frac{21}{2}\left(2 \mathrm{a}_{1}+20 \mathrm{~d}\right)=189$

$\Rightarrow a_{1}+10 d=9 \ldots (2)$

For equation $(1) \,\&\,(2)$ we get

$a_{1}=3\,\&\, d=\frac{3}{5}$

OR

$a_{1}=15\, \&\, d=-\frac{3}{5}$

$\mathrm{So}, a_{6} \cdot \mathrm{a}_{16}=\left(\mathrm{a}_{1}+5 \mathrm{~d}\right)\left(\mathrm{a}_{1}+15 \mathrm{~d}\right)$

$\Rightarrow \mathrm{a}_{6} \mathrm{a}_{16}=72$

$\left(2 \mathrm{a}_{1}+9 \mathrm{~d}\right) \mathrm{p}=10\left(2 \mathrm{a}_{1}+(\mathrm{p}-1) \mathrm{d}\right)$

$9 \mathrm{dp}=20 \mathrm{a}_{1}-2 \mathrm{pa}_{1}+10 \mathrm{~d}(\mathrm{p}-1)$

$9 \mathrm{p}=(20-2 \mathrm{p}) \frac{\mathrm{a}_{1}}{\mathrm{~d}}+10(\mathrm{p}-1)$

$\frac{\mathrm{a}_{1}}{\mathrm{~d}}=\frac{(10-\mathrm{p})}{2(10-\mathrm{p})}=\frac{1}{2}$

$\therefore \frac{\mathrm{a}_{11}}{\mathrm{a}_{10}}=\frac{\mathrm{a}_{1}+10 \mathrm{~d}}{\mathrm{a}_{1}+9 \mathrm{~d}}=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}=\frac{21}{19}$

$a=4$

and $\frac{c+b+b}{3}=\frac{7}{3}$

$c+2 b=7$

also $2 b=a+c$

$2 b-a+2 b=7$

$b=\frac{11}{4}$

now $4 x ^{2}+\frac{11}{4} x +1=0 (0=\alpha \,And \, \beta)$

$\alpha^{2}+\beta^{2}-\alpha \beta=(\alpha+\beta)^{2}-3 \alpha \beta$

$=\left(\frac{-11}{16}\right)^{2}-3\left(\frac{1}{4}\right)$

$=\frac{121}{256}-\frac{3}{4}=\frac{-71}{256}$

$\Rightarrow S _{2}- S _{1}=\frac{4 n }{2}[2 a +(4 n -1) d ]-\frac{2 n }{2}[2 a +(2 n -1)d]$

$=4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n$

$=2 n a+n d[8 n-2-2 n+1]$

$\Rightarrow 2 n a+n d[6 n-1]=1000$

$2 a+(6 n-1) d=\frac{1000}{n}$

Now, $S_{6 n}=\frac{6 n}{2}[2 a+(6 n-1) d]$

$=3 n \cdot \frac{1000}{ n }=3000$

$\left(4^{ x }-2\right)^{2}=10\left(4^{ x }+\frac{18}{5}\right)$

$\left(4^{ x }\right)^{2}+4-4\left(4^{ x }\right)-32=0$

$\left(4^{ x }-16\right)\left(4^{ x }+2\right)=0$

$4^{ x }=16$

$x =2$

$\left|\begin{array}{lll}3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0\end{array}\right|=3(-2)-1(0-4)+4(1)$

$=-6+4+4=2$

$\frac{\mathrm{a}}{1-\mathrm{r}}=15....(i)$

Series formed by square of terms:

$\mathrm{a}^{2}, \mathrm{a}^{2} \mathrm{r}^{2}, \mathrm{a}^{2} \mathrm{r}^{4}, \mathrm{a}^{2} r^{6} \ldots \ldots$

$\text { Sum }=\frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=150$

$\Rightarrow \frac{\mathrm{a}}{1-\mathrm{r}} \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150 \Rightarrow 15 \cdot \frac{\mathrm{a}}{1+\mathrm{r}}=150$

$\Rightarrow \frac{\mathrm{a}}{1+\mathrm{r}}=10 \ldots \ldots \ldots \text { (ii) }$

by $(i)$ and $(ii)$ $\mathrm{a}=12 ; \mathrm{r}=\frac{1}{5}$

Now series : $\mathrm{ar}^{2}, \mathrm{ar}^{4}, \mathrm{ar}^{6}$

$\text { Sum }=\frac{a r^{2}}{1-r^{2}}=\frac{12 \cdot(1 / 25)}{1-1 / 25}=\frac{1}{2}$

$a \frac{\left(r^{4}-1\right)}{r-1}=\frac{65}{12}......(1)$

$\frac{1}{a} \frac{\left(\frac{1}{r^{4}}-1\right)}{\frac{1}{r}-1}=\frac{65}{18}$

$\frac{1}{a r^{3}}\left(\frac{1-r^{3}}{1-r}\right)=\frac{65}{18}......(2)$

$\frac{(1)}{(2)} \Rightarrow a ^{2} r ^{3}=\frac{3}{2}$

and $\quad a^{3} \cdot r^{3}=1$

$ar =1$

$(\operatorname{ar})^{2} \cdot r =\frac{3}{2}$

$r=\frac{3}{2}, a=\frac{2}{3}$

So, third term $=\operatorname{ar}^{2}=\frac{2}{3} \times \frac{9}{4}$

$\alpha=\frac{3}{2}$

$2 \alpha=3$