Maths M.C.Q (1 Marks)

$T _{2}+ T _{6}=\frac{25}{2} \Rightarrow \operatorname{ar}\left(1+ r ^{4}\right)=\frac{25}{2}$

$a^{2} r^{2}\left(1+r^{4}\right)^{2}=\frac{625}{4}$ $....(1)$

$T _{3} \cdot T _{5}=25 \Rightarrow\left( ar ^{2}\right)\left( ar ^{4}\right)=25$

$a^{2} r^{6}=25$ $....(2)$

On dividing $(1)$ by $(2)$

$\frac{\left(1+r^{4}\right)^{2}}{r^{4}}=\frac{25}{4}$

$4 r^{8}-17 r^{4}+4=0$

$\left(4 r^{4}-1\right)\left(r^{4}-4\right)=0$

$r^{4}=\frac{1}{4}, 4 \Rightarrow r^{4}=4$

(an increasing geometric series) $a ^{2} r ^{6}=25 \Rightarrow\left( ar ^{3}\right)^{2}=25$

$T _{4}+ T _{6}+ T _{8}= ar ^{3}+ ar ^{5}+ ar ^{7}$

$=\operatorname{ar}^{3}\left(1+ r ^{2}+ r ^{4}\right)$

$=5(1+2+4)=35$

$\mathrm{a}_{\mathrm{r}}, 2 \mathrm{a}, \mathrm{ar} \rightarrow \mathrm{A} \cdot \mathrm{P} \Rightarrow 4 \mathrm{a}=\frac{\mathrm{a}}{\mathrm{r}}+\mathrm{ar} \Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}=4$

$\mathrm{r}=2 \pm \sqrt{3}$

$4^{\text {th }}$ form of G.P $=3 \mathrm{r}^{2} \Rightarrow \mathrm{ar}^{2}=3 \mathrm{r}^{2} \Rightarrow \mathrm{a}=3$

$\mathrm{r}=2+\sqrt{3}, \mathrm{a}=3, \mathrm{~d}=2 \mathrm{a}-\frac{\mathrm{a}}{\mathrm{r}}=3 \sqrt{3}$

$\mathrm{r}^{2}-\mathrm{d}=(2+\sqrt{3})^{2}-3 \sqrt{3}$

$=7+4 \sqrt{3}-3 \sqrt{3}$

$=7+\sqrt{3}$

$a_{1}+2 b_{1}=15....(1)$

$c_{3}=a_{3}+b_{3}=a_{1}-6+4 b_{1}=13$

$a_{1}+4 b_{1}=19....(2)$

from $(1)\, \,(2) b_{1}=2, a_{1}=11$

$\sum_{k=1}^{10} c_{k}=\sum_{k=1}^{10}\left(a_{k}+b_{k}\right)=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$

$=\frac{10}{2}(2 \times 11+9 \times(-3))+\frac{2\left(2^{10}-1\right)}{2-1}$

$=5(22-27)+2(1023)$

$=2046-25=2021$

Now given $\frac{5}{4}=\frac{ t _{ p }+ t _{ q }}{2}, t = t _{ p } t _{ q }$ where

$t _{ r }=-16\left(-\frac{1}{2}\right)^{ r -1}$

so $\frac{5}{4}=-8\left[\left(-\frac{1}{2}\right)^{ p -1}+\left(-\frac{1}{2}\right)^{ q -1}\right]$

$1=256\left(-\frac{1}{2}\right)^{ p + q -2} \Rightarrow 2^{ p + q -2}=(-1)^{ p + q -2} 2^{8}$

hence $p+q=10$

$\frac{2}{b}=\frac{1}{a}+6$

$\frac{1}{8 a^{2}}=\frac{1}{a}+6$

$\frac{1}{a^{2}}-\frac{8}{a}-48=0$

$\frac{1}{a}=12,-4 \Rightarrow a=\frac{1}{12},-\frac{1}{4}$

$a=\frac{1}{12}, a>0$

$b=16 a^{2}=\frac{1}{9}$

$\Rightarrow \quad 72(a+b)=6+8=14$

$S=1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots$

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

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

$\frac{2 S}{3}=\frac{4}{3}+\frac{4}{3^{2}}+\frac{4}{3^{3}}+\ldots$

$S=\frac{3}{2}\left(\frac{4 / 3}{1-1 / 3}\right)=3$

Now, $\ell=(3)^{\log _{0.25}\left(\frac{1 / 3}{1-1 / 3}\right)}$

$\ell=3^{\log _{(1 / 4))}\left(\frac{1}{2}\right)}=3^{1 / 2}=\sqrt{3}$

$\Rightarrow \ell^{2}=3$

$\frac{1}{5} \mathrm{~S} =\frac{7}{5^{2}}+\frac{9}{5^{3}}+\frac{13}{5^{4}}+\ldots$

On subtracting

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

$S=\frac{7}{4}+\frac{1}{10}\left(1+\frac{2}{5}+\frac{3}{5^{2}}+\ldots\right)$

$S=\frac{7}{4}+\frac{1}{10}\left(1-\frac{1}{5}\right)^{-2}$

$=\frac{7}{4}+\frac{1}{10} \times \frac{25}{16}=\frac{61}{32}$

$\Rightarrow 160 \mathrm{~S}=5 \times 61=305$

$\mathrm{T}_{\mathrm{r}}=(2 \mathrm{r}+1)^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$

$\mathrm{S}=\Sigma \mathrm{T}_{\mathrm{r}}$

$\mathrm{S}=\Sigma(2 \mathrm{r}+1){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\sum 2 \mathrm{r}^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+\sum^{n} \mathrm{C}_{\mathrm{r}}$

$\mathrm{S}=2\left(\mathrm{n} \cdot 2^{n-1}\right)+2^{n}=2^{n}(\mathrm{n}+1)$

$2^{n}(n+1)=2^{100} \cdot 101 \Rightarrow n=100$

$2\left[\frac{n-1}{2}\right]=2\left[\frac{99}{2}\right]=98$

Divide by $8^{\mathrm{n}}$ we get

$\frac{a_{n+2}}{8^{n}}=\frac{2 a_{n+1}}{8^{n}}+\frac{a_{n}}{8^{n}}$

$\Rightarrow 64 \frac{a_{n+2}}{8^{n+2}}=\frac{16 a_{n+1}}{8^{n+1}}+\frac{a_{n}}{8^{n}}$

$64 \sum_{n-1}^{\infty} \frac{a_{n+2}}{8^{n+2}}=16 \sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}}+\sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}$

$64\left(P-\frac{a_{1}}{8}-\frac{a_{2}}{8^{2}}\right)=16\left(P-\frac{a_{1}}{8}\right)+P$

$\Rightarrow 64\left(P-\frac{1}{8}-\frac{1}{64}\right)=16\left(P-\frac{1}{8}\right)+P$

$64 P-8-1=16 P-2+P$

$47 P=7$

$T_{r}=n r-r^{2}$

$\Rightarrow S_{n}=\sum_{r=1}^{n} T_{r}=\sum_{r=1}^{n}\left(n r-r^{2}\right)$

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

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}+1)}{6}$

Now $\sum_{r=4}^{\infty}\left(\frac{2 S_{n}}{n !}-\frac{1}{(n-2) !}\right)$

$=\sum_{r=4}^{\infty}\left(2 \cdot \frac{n(n-1)(n+1)}{6 . n(n-1)(n-2) !}-\frac{1}{(n-2) !}\right)$

$=\sum_{r=4}^{\infty}\left(\frac{1}{3}\left(\frac{n-2+3}{(n-2) !}\right)-\frac{1}{(n-2) !}\right)$

$=\sum_{r=4}^{\infty} \frac{1}{3} \cdot \frac{1}{(n-3) !}=\frac{1}{3}(e-1)$

$\mathrm{T}_{\mathrm{n}}=(3 \mathrm{n}+4)(2 \mathrm{n}+6)=2(3 \mathrm{n}+4)(\mathrm{n}+3)$

$=2\left(3 \mathrm{n}^{2}+13 \mathrm{n}+12\right)=6 \mathrm{n}^{2}+26 \mathrm{n}+24$

$\mathrm{~S}_{10}=\sum_{\mathrm{n}=1}^{10} \mathrm{~T}_{\mathrm{n}}=6 \sum_{\mathrm{n}=1}^{10} \mathrm{n}^{2}+26 \sum_{\mathrm{n}=1}^{10} \mathrm{n}+24 \sum_{\mathrm{n}=1}^{10} 1$

$=\frac{6(10 \times 11 \times 21)}{6}+26 \times \frac{10 \times 11}{2}+24 \times 10$

$=10 \times 11(21+13)+240$

$=3980$

$\text { Mean }=\frac{\mathrm{S}_{10}}{10}=\frac{3980}{10}=398$

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

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

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

$x=\frac{1}{2} \Rightarrow y=1-\ell n 2$

$e^{1+y}=e^{1+1-l n 2}$

$=e^{2-ln 2}=\frac{e^{2}}{2}$

$\Rightarrow \log _{10} x=6 \Rightarrow x=10^{6}$

Now,

$y=\left(\log _{10} x\right)+\left(\log _{10} x^{\frac{1}{3}}\right)+\left(\log _{10} x^{\frac{1}{9}}\right)+. . \infty$

$=\left(1+\frac{1}{3}+\frac{1}{9}+\ldots \infty\right) \log _{10} x$

$=\left(\frac{1}{1-\frac{1}{3}}\right) \log _{10} x=9$

So, $(x, y)=\left(10^{6}, 9\right)$

$\mathrm{~S}+\frac{1}{1-\mathrm{x}}=\frac{1}{1-\mathrm{x}}+\frac{1}{\mathrm{x}+1}+\ldots=\frac{2}{1-\mathrm{x}^{2}}+\frac{2}{1+\mathrm{x}^{2}}+\ldots$

$\mathrm{S}+\frac{1}{1-\mathrm{x}}=\frac{2^{101}}{1-\mathrm{x}^{2^{101}}}$

Put $\mathrm{x}=2$

$\mathrm{S}=1-\frac{2^{101}}{2^{2^{101}}-1}$

$=\frac{(2 n+1)^{2}+20 n+39}{4 \cdot(2 n+1) !}$

$=\frac{(2 n+1)^{2}+(2 n+1) \cdot 10+29}{4(2 n+1) !}$

$=\frac{1}{4}\left[\frac{(2 n+1)^{2}}{(2 n+1)(2 n) !}+\frac{(2 n+1) 10}{(2 n+1)(2 n) !}+\frac{29}{(2 n+1) !}\right]$

$=\frac{1}{4}\left[\frac{2 n+1}{(2 n) !}+\frac{10}{(2 n) !}+\frac{29}{(2 n+1) !}\right]$

$=\frac{1}{4}\left[\frac{1}{(2 n-1) !}+\frac{11}{(2 n) !}+\frac{29}{(2 n+1) !}\right]$

$S_{1}=\frac{1}{1 !}+\frac{1}{3 !}+\frac{1}{5 !}+\ldots=\frac{e-\frac{1}{e}}{2}$

$S_{2}=11\left[\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots\right]=11\left[\frac{e+\frac{1}{e}-2}{2}\right]$

$S_{3}=29\left[\frac{1}{3 !}+\frac{1}{5 !}+\frac{1}{7 !}+\ldots\right]=29\left[\frac{e-\frac{1}{e}-2}{2}\right]$

Now, $S =\frac{1}{4}\left[ S _{1}+ S _{2}+ S _{3}\right]$

$=\frac{1}{4}\left[\frac{ e }{2}-\frac{1}{2 e }+\frac{11 e }{2}+\frac{11}{2 e }+\frac{29 e }{2}-\frac{29}{2 e }-4\right]$

$=\frac{41 e }{8}-\frac{19}{8 e }-10$

$\frac{S}{3}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{7}{3^{3}}+\frac{12}{3^{4}}+\ldots$

$\frac{2 S}{3}=1+\frac{1}{3}+\frac{5}{3^{2}}+\frac{5}{3^{3}}+\frac{5}{3^{4}}+\ldots+$ up to infinite terms

$\Rightarrow S=\frac{13}{4}$

So, $a _{ n }=\sqrt{2} a _{ n +1}, a _{1}=12$

$\Rightarrow a _{ n }=12 \times\left(\frac{1}{\sqrt{2}}\right)^{ n -1}$

Now, $\left(a_{n}\right)^{2}<1 \Rightarrow \frac{144}{2^{(n-1)}}<1$

$\Rightarrow 2^{(n-1)}>144$

$\Rightarrow n -1 \geq 8$

$\Rightarrow n \geq 9$

$A=\left(1-\frac{1}{2}\right)^{n}+\left(1-\frac{3}{4}\right)^{n}+\left(1-\frac{7}{8}\right)^{n}+\left(1-\frac{15}{16}\right)^{n}+\left(1-\frac{31}{32}\right)^{n}$

$A=\frac{1}{2^{n}}+\frac{1}{4^{n}}+\frac{1}{8^{n}}+\frac{1}{16^{n}}+\frac{1}{32^{n}}$

$A=\frac{1}{2^{n}}\left(\frac{1-\left(\frac{1}{2^{n}}\right)^{5}}{1-\frac{1}{2^{n}}}\right) \Rightarrow A=\frac{\left(1-\frac{1}{2^{5 n}}\right)}{\left(2^{n}-1\right)}$

$\left(2^{n}-1\right) A=1-\frac{1}{2^{5 n}},$ Given $63 A =1-\frac{1}{2^{30}}$

Clearly $5 n=30$

$n=6$

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

$S=\frac{1}{4}\left(1-\frac{1}{101}\right)=\frac{1}{4}\left(\frac{100}{101}\right)=\frac{25}{101}$

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

$=2\left(\mathrm{x}^{2}+\mathrm{x}^{3}+\mathrm{x}^{4}+\ldots \infty\right)-\left(\frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{x}^{3}}{3}+\frac{\mathrm{x}^{4}}{4}+\ldots \infty\right)$

$=\frac{2 x^{2}}{1-x}-(\ell n(1-x)-x)$

$\Rightarrow t=\frac{2 x^{2}}{1-x}+x-\ell n(1-x)$

$\Rightarrow t=\frac{x(1+x)}{1-x}-\ell n(1-x)$

Let $S _{1}=2+3+6+11+18+27+\ldots .+ T _{ n }$

$S _{1}=2+3+6+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+ T _{ n }$

$T _{ n }=2+1+3+5+\ldots \ldots+ n$ terms

$T _{ n }=2+( n -1)^{2}$

$S _{1}=\Sigma T _{ n }=2 n +\frac{( n -1) n (2 n -1)}{6}$

$\Rightarrow S _{ n }( x )=\left(2 n +\frac{ n ( n -1)(2 n -1)}{6}\right) \log _{ a } x$

$S _{24}( x )=1093$ (Given)

$\log _{ a } x \left(48+\frac{23.24 .47}{6}\right)=1093$

$\log _{ a } x =\frac{1}{4} \quad \ldots (1)$

$S _{12}(2 x )=265$

$S_{12}(2 x)=265$

$\log _{ a }(2 x )\left(24+\frac{11.12 .23}{6}\right)=265$

$\log _{ a } 2 x =\frac{1}{2} \quad \ldots (2)$

$(2) -(1)$

$\log _{ a } 2 x -\log _{ a } x =\frac{1}{4}$

$\log _{ a } 2=\frac{1}{4} \Rightarrow a =16$

$S =\sum_{ x =0}^{99}(100- x )(100+ x )=\sum 100^{2}- x ^{2}$

$=100^{3}-\frac{99 \times 100 \times 199}{6}$

$\alpha=3 \quad \beta=1650$

slope $=\frac{1650}{3}=550$

difference of the given A.P. Where $d>0$

$\overline{ X }= a +\frac{0+ d +2 d +\ldots+10 d }{11}$

$=a+5 d$

$\Rightarrow$ varience $=\frac{\Sigma\left(\bar{X}-x_{i}\right)^{2}}{11}$

$\Rightarrow 90 \times 11=\left(25 d^{2}+16 d^{2}+9 d^{2}+4 d^{2}\right) \times 2$

$\Rightarrow d =\pm 3 \Rightarrow d =3$

$\mathrm{T}_{20}=\frac{1}{10}=\mathrm{a}+19 \mathrm{d} \quad \ldots .(\text {ii})$

$a=\frac{1}{200}=d$

Hence, $S_{200}=\frac{200}{2}\left[\frac{2}{200}+\frac{199}{200}\right]=\frac{201}{2}$

$S^{\prime}=\left(2^{10}\right) \frac{\left(\left(\frac{3}{2}\right)^{11}-1\right)}{\frac{3}{2}-1}=2^{11}\left(\frac{3^{11}}{2^{11}}-1\right)$

$S^{\prime}=3^{11}-2^{11}= S -2^{11}( Given )$

$\therefore S =3^{11}$

$\Rightarrow \mathrm{a}^{3}=27 \Rightarrow \mathrm{a}=3$

$S=\frac{3}{r}+3 r+3$

For ${r}>0$

$\frac{\frac{3}{r}+3 r}{2} \geq \sqrt{3^{2}} \quad($ By $A M \geq G M)$

$\Rightarrow \frac{3}{r}+3 r \geq 6$

For $r<0 \quad \frac{3}{r}+3 r \leq-6 \quad \ldots(2)$

From ( 1)  (2)

$\mathrm{S} \in(-\infty-3) \cup[9, \infty]$

$(A . M \geq G . M)$

$=\alpha-220 \beta$

$=11-\left(2^{2} \cdot 1+4^{2} \cdot 3+\ldots \ldots+20^{2} \cdot 19\right)$

$=11-2^{2} \cdot \sum_{ r =1}^{10} r ^{2}(2 r -1)=11-4\left(\frac{110^{2}}{2}-35 \times 11\right)$

$=11-220(103)$

$\Rightarrow \alpha=11, \beta=103$

$+(1 !+2 !+3 ! \ldots \ldots \ldots .$ upto 51 terms $)$

$\left[\because{ }^{n} p_{n-1}=n !\right]$

$ \therefore \quad S =$$(2 \times 1 !-3 \times 2 !+4 \times 3 ! \ldots+52.51 !)$

$+(1 !-2 !+3 ! \ldots \ldots \ldots .(51) !) $$=(2 !-3 !+4 ! \ldots \ldots .+52 !) $

$+(1 !-2 !+3 !-4 !+\ldots \ldots+(51) !) $$= 1 !+52 !$

$(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)$

$+\ldots \ldots \ldots$

By multiplying and dividing $x-y:$

$\frac{\left(x^{2}-y^{2}\right)+\left(x^{3}-y^{3}\right)+\left(x^{4}-y^{4}\right)+\ldots \ldots}{x-y}$

$=\frac{\left(x^{2}+x^{3}+x^{4}+\ldots \ldots\right)-\left(y^{2}+y^{3}+y^{4}+\ldots \ldots\right)}{x-y}$

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

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

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

$=\frac{1}{6} \times 20 \times 21 \times 22=1540$

$=\frac{1}{4}\left(2\left(\frac{7 \times 8}{2}\right)^{2}+3\left(\frac{7 \times 8 \times 15}{6}\right)+\frac{7 \times 8}{2}\right)$

$=504$

$b _{1}, b _{2}, \ldots, b _{ m } \rightarrow( CD = d +2)$

$a_{40}=a+39 d=-159$

$a_{100}=a+99 d=-399$

Subtract : $60 d =-240 \Rightarrow d =-4$

using equation (1)

$a+39(-4)=-159$

$a=156-159=-3$

$a_{70}=a+69 d=-3+69(-4)=-279=b_{100}$

$b_{100}=-279$

$b_{1}+99(d+2)=-279$

$b_{1}-198=-279 \Rightarrow b_{1}=-81$

$3^{4-\sin 2 \alpha}+3^{2 \sin 2 \alpha-1}=28$

Let $3^{2} \sin 2 \alpha=t$

$\frac{81}{t}+\frac{t}{3}=28$

$t=81,3$

$3^{2 \sin 2 \alpha}=3^{1}, 3^{4}$

$2 \sin 2 \alpha=1,4$

$\sin 2 \alpha=\frac{1}{2}, 2($ rejected $)$

First term $a=3^{2} \sin 2 \alpha-1$

$a=1$

Second term $=14$

$\therefore$ common difference $d=13$

$T_{6}=a+5 d$

$T _{6}=1+5 \times 13$

$T_{6}=66$

$\Rightarrow 300=1+(n-1) d$

$\Rightarrow \quad(n-1) d=299=13 \times 23$

since, $n \in[15,50]$

$\therefore n=24$ and $d=13$

$a_{n-4}=a_{20}=1+19 \times 13=248$

$\Rightarrow a_{n-4}=248$

$S_{n-4}=\frac{20}{2}\{1+248\}=2490$

$\Rightarrow\left( T _{1}+\ldots \ldots+ T _{25}\right)=\left( T _{26}+\ldots . .+ T _{40}\right)$

$\Rightarrow\left( T _{1}+\ldots . .+ T _{40}\right)=2\left( T _{1}+\ldots \ldots+ T _{25}\right)$

$\Rightarrow \frac{40}{2}[2 \times 3+(39 d )]=2 \times \frac{25}{2}[2 \times 2+24 d ]$

$\Rightarrow d=\frac{1}{6}$

$\Rightarrow\left(a_{1}+a_{11}\right) \times \frac{11}{2}=0$

$\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{11}=0$

$\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{1}+10 \mathrm{d}=0$

where d is common difference

$\Rightarrow \quad \mathrm{a}_{1}=-5 \mathrm{d}$

$a_{1}+a_{3}+a_{5}+\ldots \ldots+a_{23}$

$=\left(a_{1}+a_{23}\right) \times \frac{12}{2}=\left(a_{1}+a_{1}+22 d\right) \times 6$

$=\left(2 a_{1}+22\left(\frac{-a_{1}}{5}\right)\right) \times 6$

$=-\frac{72}{5} a_{1} \Rightarrow K=\frac{-72}{5}$

$\Rightarrow n \leq 14.71$

$\mathrm{n}=14$

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

$\because \operatorname{sum}=25 \Rightarrow \mathrm{a}=5$

Product $=2520$

$\left(25-4 d^{2}\right)\left(25-d^{2}\right)=504$

$4 \mathrm{d}^{4}-125 \mathrm{d}^{2}+121=0$

$\Rightarrow \mathrm{d}^{2}=1, \frac{121}{4}$

$\Rightarrow \mathrm{d}=\pm 1, \pm \frac{11}{2}$

$\mathrm{d}=\pm 1$ is rejected because none of the term can be $\frac{-1}{2}$

$\Rightarrow \mathrm{d}=\pm \frac{11}{2}$

$\Rightarrow$ AP will be $-6,-\frac{1}{2}, 5, \frac{21}{2}, 16$

Largest term is $16$

$=0$

$\Rightarrow\left(a^{2} p^{2}+2 a b p+b^{2}\right)+\left(b^{2} p^{2}+2 b c p+c^{2}\right)+$

$\left(c^{2} p^{2}+2 c d p+d^{2}\right)=0$

$\Rightarrow(a b+b)^{2}+(b p+c)^{2}+(c p+d)^{2}=0$

This is possible only when $a p+b=0$ and $b p+c=0$ and $c p+d=0$

$p =-\frac{ b }{ a }=-\frac{ c }{ b }=-\frac{ d }{ c }$

or $\frac{ b }{ a }=\frac{ c }{ b }=\frac{ d }{ c }$

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

Common ratio $=r>0$

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

$ar ^{5}+ ar ^{6}+ ar ^{7}=243$

$r^{4}\left(a r+a r^{2}+a r^{3}\right)=243$

$r^{4}(3)=243 \Rightarrow r=3$ as $r>0$

from (1)

$3 a+9 a+27 a=3$

$a=\frac{1}{13}$

$S_{50}=\frac{a\left(r^{50}-1\right)}{(r-1)}=\frac{1}{26}\left(3^{50}-1\right)$

$\alpha, \beta, \gamma, \delta$ in G.P.

$\alpha+\alpha r=3 \ldots .(1)$

$x^{2}-6 x+q=0<\frac{\gamma}{\delta}$

$\alpha r^{2}+\alpha r^{3}=6 \quad \ldots(2)$

$(2) \div(1)$

$r^{2}=2$

So, $\frac{2 q+p}{2 q-p}=\frac{2 r^{5}+r}{2 r^{5}-r}=\frac{2 r^{4}+1}{2 r^{4}-1}=\frac{9}{7}$

$=2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \ldots \infty$

$=2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \cdot \ldots \infty$

$=2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots \infty}{32}=(2)\left(\frac{1 / 4}{1-1 / 2}\right)=2^{1 / 2}$

$\Rightarrow \operatorname{ar}^{2} \frac{\left(\mathrm{r}^{200}-1\right)}{\left(\mathrm{r}^{2}-1\right)}=200$

$\sum_{n=1}^{100} a_{2 n}=100 \Rightarrow a_{2}+a_{4}+a_{6}+\ldots+a_{200}=100$

$\Rightarrow \frac{\operatorname{ar}\left(\mathrm{r}^{200}-1\right)}{\left(\mathrm{r}^{2}-1\right)}=100$

On dividing $\mathrm{r}=2$

on adding $a_{2}+a_{3}+a_{4}+a_{5}+\ldots+a_{200}+a_{201}=300$

$\Rightarrow \mathrm{r}\left(\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\ldots .+\mathrm{a}_{200}\right)=300$

$\Rightarrow \sum_{n=1}^{200} a_{n}=150$

$\mathrm{r}^{2} \mathrm{a}_{1}+\mathrm{r}^{2} \mathrm{a}_{2}=16$

$\Rightarrow \mathrm{r}^{2}=4 \Rightarrow \mathrm{r}=-2 \quad$ as $\mathrm{a}_{1}<0$

and $a_{1}+a_{2}=4$

$a_{1}+a_{1}(-2)=4 \Rightarrow a_{1}=-4$

$4 \lambda=(-4)\left(\frac{(-2)^{9}-1}{-2-1}\right)=(-4) \times \frac{513}{3}$

$\Rightarrow \lambda=-171$

$\Rightarrow \frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$

$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1+\left(\frac{\sin x+\cos x}{2}\right)}$

$\Rightarrow \min \left(2^{\sin x}+2^{\cos x}\right)=2^{1-\frac{1}{\sqrt{2}}}$

$\Rightarrow 460=\log _{7} x \cdot\left(\frac{21 \times 22}{2}-1\right)$

$\Rightarrow 460=230 \cdot \log _{7} x$

$\Rightarrow \log _{7} x=2 \Rightarrow x=49$

$d =\frac{243-3}{ m +1}=\frac{240}{ m +1}$

Now $3, G _{1}, G _{2}, G _{3}, 243$

$r=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=3$

$\therefore \quad A_{4}=G_{2}$

$\Rightarrow \quad a +4 d = ar ^{2}$

$3+4\left(\frac{240}{ m +1}\right)=3(3)^{2}$

$m=39$

$S _{ n }=\frac{ n }{2}\left(2 \times \frac{100}{5}+( n -1)\left(-\frac{2}{5}\right)\right)=188$

$n (100- n +1)=488 \times 5$

$n ^{2}-101 n +488 \times 5=0$

$n =61,40$

$T _{ n }= a +( n -1) d =\frac{100}{5}-\frac{2}{5} \times 60$

$=20-24=-4$

$4 a]+\left[x^{4}+k a+6 a\right]+\ldots \ldots 9$ terms

$\Rightarrow S =\left( x + x ^{2}+ x ^{3}+ x ^{4}+\ldots . .9\right.$ terms $)+( ka + ka$

$+ ka + ka +\ldots \ldots .9$ terms $)+(0+2 a +4 a +6 a +$$\ldots \ldots 9$ terms

$\Rightarrow S = x \left[\frac{ x ^{9}-1}{ x -1}\right]+9 ka +72 a$

$\Rightarrow S =\frac{\left( x ^{10}- x \right)+(9 k +72) a ( x -1)}{( x -1)}$

Compare with given sum, then we get, $(9 k +$72)$=45$

$\Rightarrow \quad k=-3$