Sequences and Series

General term of the sequence,

$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & \mathrm{T}_{\mathrm{r}}=\frac{\frac{1}{2}\left[\left(\mathrm{r}^2+\mathrm{r}-1\right)-\left(\mathrm{r}^2-\mathrm{r}-1\right)\right]}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & =\frac{1}{2}\left[\frac{1}{\mathrm{r}^2-\mathrm{r}-1}-\frac{1}{\mathrm{r}^2+\mathrm{r}-1}\right] \end{aligned}$

Sum of 10 terms,

$\sum_\limits{\mathrm{r}=1}^{10} \mathrm{~T}_{\mathrm{r}}=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}$

The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.

For the first GP, with first term $a$ and third term $b$:

• The third term is given by $t_3 = ar^2 = b$, leading to $r^2 = \frac{b}{a}$.
• The eleventh term is $t_{11} = ar^{10} = a\left(\frac{b}{a}\right)^5$.

For the second GP, with first term $a$ and fifth term $b$:

• The fifth term is $T_5 = ar^4 = b$, yielding $r^4 = \frac{b}{a}$ and $r = \left(\frac{b}{a}\right)^{1/4}$.
• The $p^{\text{th}}$ term is thus $T_p = ar^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$.

Equating the eleventh term of the first GP to the $p^{\text{th}}$ term of the second GP gives: $a\left(\frac{b}{a}\right)^5 = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}$.

Simplifying, we find that $5 = \frac{p-1}{4}$, leading to $p = 21$, which is the solution.

$\begin{aligned} &\begin{aligned} & \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \\ & 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\\ & \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \\ & 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)} \end{aligned}\\ &\text { From (1) and (2) } a=-8, d=5 \end{aligned}$

$\begin{aligned} & S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\ & =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\ & =405-10 \\ & =395 \end{aligned}$

$\log _{\mathrm{e}} \mathrm{a}, \log _{\mathrm{e}} \mathrm{b}, \log _{\mathrm{e}} \mathrm{c}$ are in A.P.

$\therefore \mathrm{b}^2=\mathrm{ac}$ ..... (i)

Also

$\begin{aligned} & \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text { are in A.P. } \\ & \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\ & \frac{b}{c}=\frac{3}{2} \end{aligned}$

Putting in eq. (i) $b^2=a \times \frac{2 b}{3}$

$\begin{aligned} & \frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{2} \\ & \mathrm{a}: \mathrm{b}: \mathrm{c}=9: 6: 4 \end{aligned}$

Let $r^{\prime}$th term of the GP be $a^{n-1}$. Given,

$\begin{aligned} & 2 a_r=a_{r+1}+a_{r+2} \\ & 2 a r^{n-1}=a r^n+a r^{n+1} \\ & \frac{2}{r}=1+r \\ & r^2+r-2=0 \end{aligned}$

Hence, we get, $r=-2$ (as $r \neq 1$)

So, $\mathrm{S}_{20}-\mathrm{S}_{18}=$ (Sum upto 20 terms) $-$ (Sum upto 18 terms) $=\mathrm{T}_{19}+\mathrm{T}_{20}$

$\mathrm{~T}_{19}+\mathrm{T}_{20}=\mathrm{ar}^{18}(1+\mathrm{r})$

Putting the values $\mathrm{a}=\frac{1}{8}$ and $\mathrm{r}=-2$;

we get $T_{19}+T_{20}=-2^{15}$

$\begin{aligned} & a+a r+a r^2+a r^3+\ldots .+a r^{63} \\ & =7\left(a+a r^2+a r^4 \ldots .+a r^{62}\right) \\ & \Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2} \\ & r=6 \end{aligned}$

$\begin{aligned} & a_6=2 \Rightarrow a+5 d=2 \\ & a_1 a_4 a_5=a(a+3 d)(a+4 d) \\ & =(2-5 d)(2-2 d)(2-d) \\ & f(d)=8-32 d+34 d^2-20 d+30 d^2-10 d^3 \\ & f^{\prime}(d)=-2(5 d-8)(3 d-2) \end{aligned}$

$\mathrm{d}=\frac{8}{5}$

$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}$

This is A.P. with common difference

$\begin{aligned} & d_1=-1+\frac{1}{4}=-\frac{3}{4} \\ & -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20 \end{aligned}$

This is also A.P. $\mathrm{a}=-129 \frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}$

Required term $=$

$\begin{aligned} & -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\ & =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{aligned}$

$4,9,14,19, \ldots$, up to $25^{\text {th }}$ term

$\mathrm{T}_{25}=4+(25-1) 5=4+120=124$

$3,6,9,12, \ldots$, up to $37^{\text {th }}$ term

$\mathrm{T}_{37}=3+(37-1) 3=3+108=111$

Common difference of $\mathrm{I}^{\text {st }}$ series $\mathrm{d}_1=5$

Common difference of $\mathrm{II}^{\text {nd }}$ series $\mathrm{d}_2=3$

First common term $=9$, and their common difference $=15\left(\operatorname{LCM}\right.$ of $\mathrm{d}_1$ and $\left.\mathrm{d}_2\right)$ then common terms are $9,24,39,54,69,84,99$

Now, we have the following relations :

Arithmetic progression :

Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$, we can say that $a$, $A_1$, $A_2$, and $b$ are in an arithmetic progression. This means there are three equal intervals between $a$ and $b$, which are represented by the common difference $d$.

To find the value of $d$, we can use the following equation :

$ b - a = 3d $

From this equation, we can find the value of $d$ :

$ d = \frac{b - a}{3} $

$ A_1 = a + \frac{b - a}{3} = \frac{2a + b}{3} $

$ A_2 = \frac{a + 2b}{3} $

$ A_1 + A_2 = a + b $

Geometric progression :

$ a, G_1, G_2, G_3, b \text{ are in G.P. } $

$ r = \left(\frac{b}{a}\right)^{\frac{1}{4}} $

$ G_1 = \left(a^3b\right)^{\frac{1}{4}} $

$ G_2 = \left(a^2b^2\right)^{\frac{1}{4}} $

$ G_3 = \left(ab^3\right)^{\frac{1}{4}} $

We have the expression :

$ G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^3b + a^2b^2 + ab^3 + \left(a^3b\right)^{\frac{1}{2}}\cdot\left(ab^3\right)^{\frac{1}{2}} $

Simplify the expression :

$ a^3b + a^2b^2 + ab^3 + ab(a^2b^2) $

Factor out $ab$:

$ ab(a^2 + ab + b^2 + a^2b^2) $

Combine the terms :

$ ab(a^2 + 2ab + b^2) $

Rewrite the expression using the sum of squares :

$ ab(a + b)^2 $

Now, recall that $A_1 + A_2 = a + b$. Substitute this into the expression :

$ G_1 \cdot G_3 \cdot (A_1 + A_2)^2 $

Given the conditions :

1. $a_6 + a_8 = 2 \Rightarrow a r^5 + a r^7 = 2$
2. $a_3 \cdot a_5 = \frac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4 = \frac{1}{9} \Rightarrow a r^3 = \frac{1}{3}$

From this, we can form the equation $\frac{r^2}{3} + \frac{r^4}{3} = 2$, which simplifies to $r^4 + r^2 = 6$.

This can be factored to give $\left(r^2 - 2\right)\left(r^2 + 3\right) = 0$, yielding $r^2 = 2$ (since $r^2$ cannot be $-3$ for real $r$).

So, we have $r = \sqrt{2}$.

Substituting $r = \sqrt{2}$ into the equation $a r = \frac{1}{6}$, we get $a = \frac{1}{6\sqrt{2}}$.

Now, we find the value of $6(a_2+a_4)(a_4+a_6)$:

$6(a_2+a_4)(a_4+a_6) = 6\left(a r + a r^3\right)\left(a r^3 + a r^5\right)$

$= 6\left(\frac{1}{6\sqrt{2}} + \frac{1}{3\sqrt{2}}\right)\left(\frac{1}{3\sqrt{2}} + \frac{2}{3\sqrt{2}}\right)$

$= 6 \cdot \frac{1}{2} \cdot 1 = 3$.

$2{b^3} = {a^3} + {c^3}$

${\left( {{{\log a} \over {\log c}}} \right)^2} = \left( {{{\log b} \over {\log a}}} \right)\left( {{{\log c} \over {\log b}}} \right)$

$ \Rightarrow {(\log a)^3} = {(\log c)^3}$

$ \Rightarrow \log a = \log c$

$ \Rightarrow a = c$

$ \Rightarrow a = b = c$

${T_1} = 2a,d = - {{3a} \over 5}$

${S_{20}} = - 444$

$ \Rightarrow {{20} \over 2}\left( {2(2a) + (19)\left( { - {{3a} \over 5}} \right)} \right) = - 444$

$ \Rightarrow 10{{(20a - 57a)} \over 5} = - 444$

$ \Rightarrow 37a = 222$

$ \Rightarrow a = 6$

$ \Rightarrow abc = {(6)^3} = 216$

$\sum\limits_{i = 1}^{25} {{a_i} = \sum {{{ - 2} \over {4{n^2} - 16n + 15}} = \sum {{{ - 2} \over {(2n - 5)(2n - 3)}}} } } $

$ = \sum\limits_{i = 1}^{25} {\left( {{1 \over {2n - 3}} - {1 \over {2n - 5}}} \right)} $

$ = \left[ {\left( {{1 \over { - 1}} - {1 \over { - 3}}} \right) + \left( {{1 \over 1} - {1 \over { - 1}}} \right) + \left( {{1 \over 3} - {1 \over 1}} \right)......} \right.$

$ = {1 \over {2(25) - 3}} + {1 \over 3} = {{50} \over {141}}$

Given,

${a_0} = {a_1} = 0$

and ${a_{n + 2}} = 3{a_{n + 1}} - 2{a_n} + 1$

For $n = 0,\,{a_2} = 3{a_1} - 2{a_0} + 1$

$ = 3\,.\,0 - 2\,.\,0 + 1$

$ = 1$

For $n = 1,\,{a_3} = 3{a_2} - 2{a_1} + 1$

$ = 3\,.\,1 - 2\,.\,0 + 1$

$ = 4$

For $n = 2,\,{a_4} = 3{a_3} - 2{a_2} + 1$

$ = 3\,.\,4 - 2\,.\,1 + 1$

$ = 11$

For $n = 3,\,{a_5} = 3{a_4} - 2{a_3} + 1$

$ = 3\,.\,11 - 2\,.\,4 + 1$

$ = 26$

For $n = 4,\,{a_6} = 3{a_5} - 2{a_4} + 1$

$ = 3\,.\,26 - 2\,.\,11 + 1$

$ = 57$

$\therefore$ ${S_n} = 1 + 4 + 11 + 26 + 57\, + \,....\, + \,{t_n}$

${S_n} = 1 + 4 + 11 + 26\, + \,....\, + \,{t_{n - 1}} + {t_n}$

$0 = 1 + 3 + 7 + 15 + 31\, + \,.....\, - {t_n}$

$ \Rightarrow {t_n} = 1 + 3 + 7 + 15 + 31\, + \,....$

Now, find the sum of the series,

${t_n} = 1 + 3 + 7 + 15 + 31\, + \,.....\, + \,{x_{n - 1}} + {x_n}$ .....(1)

${t_n} = $        $1 + 3 + 7 + 15\, + \,.....\, + \,{x_{n - 1}} + {x_n}$ ......(2)

Subtracting (2) from (1), we get

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$0 = 1 + 2 + 4 + 8 + 16\, + \,....\, + \,{x_n}$

$ \Rightarrow {x_n} = 1 + 2 + 4 + 8 + 16\, + \,.....\, + \,$ n terms

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

$ = {2^n} - 1$

$\therefore$ ${t_n} = \sum\limits_{n = 1}^n {{x_n}} $

$ = \sum\limits_{n = 1}^n {({2^n} - 1)} $

$ = \sum\limits_{n = 1}^n {{2^n} - \sum\limits_{n = 1}^n 1 } $

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

$ = {2^{n + 1}} - 2 - n$

${t_1} = {2^2} - 2 - 1 = 1 = {a_2}$

${t_2} = {2^3} - 2 - 2 = 4 = {a_3}$

${t_3} = {2^4} - 2 - 3 = 11 = {a_4}$

$\therefore$ ${a_{22}} = {t_{21}} = {2^{22}} - 2 - 21 = {2^{22}} - 23$

${a_{23}} = {t_{22}} = {2^{23}} - 2 - 22 = {2^{23}} - 24$

${a_{24}} = {t_{23}} = {2^{24}} - 2 - 23 = {2^{24}} - 25$

${a_{25}} = {t_{24}} = {2^{25}} - 2 - 24 = {2^{25}} - 26$

Now,

${a_{25}}{a_{23}} - 2{a_{25}}{a_{22}} - 2{a_{23}}{a_{24}} + 4{a_{22}} {a_{24}}$

$ = {a_{25}}({a_{23}} - 2{a_{22}}) - 2{a_{24}}({a_{23}} - 2{a_{22}})$

$ = ({a_{23}} - 2{a_{22}})({a_{25}} - 2{a_{24}})$

$ = [({2^{23}} - 24) - 2({2^{22}} - 23)][({2^{25}} - 26) - 2({2^{24}} - 25)]$

$ = [({2^{23}} - 24 - {2^{23}} + 46)][({2^{25}} - 26 - {2^{25}} + 50)]$

$ = (22)(24)$

$ = 528$

${a_{n + 2}} = {2 \over {{a_{n + 1}}}} + {a_n}$

$ \Rightarrow {a_n}{a_{n + 1}} + 1 = {a_{n + 1}}{a_{n + 2}} - 1$

$ \Rightarrow {a_{n + 2}}{a_{n + 1}} - {a_n}\,.\,{a_{n + 1}} = 2$

For

$\matrix{ {n = 1} & {{a_3}{a_2} - {a_1}{a_2} = 2} \cr {n = 2} & {{a_4}{a_3} - {a_3}{a_2} = 2} \cr {n = 3} & {{a_5}{a_4} - {a_4}{a_3} = 2} \cr {} & {\matrix{ . \cr . \cr . \cr . \cr } } \cr {n = n} & {{{{a_{n + 2}}{a_{n + 1}} - {a_n}{a_{n + 1}} = 2} \over {{a_{n + 2}}{a_{n + 1}} = 2n + {a_1}{a_2}}}} \cr } $

Now,

${{({a_1}{a_2} + 1)} \over {{a_2}{a_3}}}\,.\,{{({a_2}{a_3} + 1)} \over {{a_3}{a_4}}}\,.\,{{({a_3}{a_4} + 1)} \over {{a_4}{a_5}}}\,.\,.....\,.\,{{({a_{30}}{a_{31}} + 1)} \over {{a_{31}}{a_{32}}}}$

$ = {3 \over 4}\,.\,{5 \over 6}\,.\,{7 \over 8}\,.\,....\,.\,{{61} \over {62}}$

$ = {2^{ - 60}}\left( {{}^{61}{C_{31}}} \right)$

Let first term of G.P. be a and common ratio is r

Then, ${a \over {1 - r}} = 5$ ...... (i)

$a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}$

$\therefore$ ${r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3 \over 5}}}$

$\therefore$ Then, ${S_{21}} = {{21} \over 2}\left[ {2 \times 10ar + 20 \times 10a{r^2}} \right]$

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

$ = 21\,{a_{11}}$

$\because$ a₁, a₂, .... a_n be an A.P of natural numbers and

${{{S_5}} \over {{S_9}}} = {5 \over {17}} \Rightarrow {{{5 \over 2}[2{a_1} + 4d]} \over {{9 \over 2}[2{a_1} + 8d]}} = {5 \over {17}}$

$ \Rightarrow 34{a_1} + 68d = 18{a_1} + 72d$

$ \Rightarrow 16{a_1} = 4d$

$\therefore$ $d = 4{a_1}$

And $110 < {a_{15}} < 120$

$\therefore$ $110 < {a_1} + 14d < 120 \Rightarrow 110 < 57{a_1} < 120$

$\therefore$ ${a_1} = 2$ ($\because$ ${a_i}\, \in N$)

$d = 8$

$\therefore$ ${S_{10}} = 5[4 + 9 \times 8] = 380$

Given G.P's 2, 2², 2³, .... 60 terms

4, 4², .... n terms

Now, G.M $ = {2^{{{225} \over 8}}}$

${\left( {{{2.2}^2}...\,{{4.4}^2}...} \right)^{{1 \over {60 + n}}}} = {2^{{{225} \over 8}}}$

$\left( {{2^{{{{n^2} + n + 1830} \over {60 + n}}}}} \right) = {2^{{{225} \over 8}}}$

$ \Rightarrow {{{n^2} + n + 1830} \over {60 + n}} = {{225} \over 8}$

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

$n = {{57} \over 8},\,20,\,$ so $n = 20$

$\therefore$ $\sum\limits_{k = 1}^{20} {k(20 - k) = 20 \times {{20 \times 21} \over 2} - {{20 \times 21 \times 41} \over 6}} $

$ = {{20 \times 21} \over 2}\left[ {20 - {{41} \over 3}} \right] = 1330$

$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}} = {3 \over 4}\sum\limits_{n = 1}^{21} {{1 \over {4n - 1}} - {1 \over {4n + 3}}} } $

$ = {3 \over 4}\left[ {\left( {{1 \over 3} - {1 \over 7}} \right) + \left( {{1 \over 7} - {1 \over {11}}} \right) + \,\,....\,\, + \,\,\left( {{1 \over {83}} - {1 \over {87}}} \right)} \right]$

$ = {3 \over 4}\left[ {{1 \over 3} - {1 \over {87}}} \right] = {3 \over 4}{{84} \over {3.87}} = {7 \over {29}}$

Given,

$1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}}\, + \,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,....\,\, + \,11}}$

General term,

${T_n} = {1 \over {1 + 2 + 3\, + \,\,....\,\, + \,\,n}}$

$ = {1 \over {{{n(n + 1)} \over 2}}}$

$ = {2 \over {n(n + 1)}}$

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

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

${t_1} = 2\left[ {{1 \over 1} - {1 \over 2}} \right]$

${t_2} = 2\left[ {{1 \over 2} - {1 \over 3}} \right]$

${t_3} = 2\left[ {{1 \over 3} - {1 \over 4}} \right]$

$ \vdots $

${t_n} = 2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$

$\therefore$ ${S_n} = {t_1} + {t_2} + {t_3}\, + \,\,....\,\, + \,\,{t_n}$

$ = 2\left[ {{1 \over 1} - {1 \over 2} + {1 \over 2} - {1 \over 3} + {1 \over 3} - {1 \over 4}\, + \,\,...\,\, + \,\,{1 \over n} - {1 \over {n + 1}}} \right]$

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

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

$ = {{2n} \over {n + 1}}$

$\therefore$ ${S_{11}} = {{2 \times 11} \over {11 + 1}} = {{22} \over {12}} = {{11} \over 6}$

$S = 1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + $ ...... (1)

${1 \over 6}S = {1 \over 6} + {5 \over {{6^2}}} + {{12} \over {{6^3}}} + $ ...... (2)

$S - {1 \over 6}S = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $ ........

$ \Rightarrow {{5S} \over 6} = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $ ....... (3)

Now, multiplying both sides by ${1 \over 6}$, we get

$ \Rightarrow {{5S} \over {36}} = {1 \over 6} + {4 \over {{6^2}}} + {7 \over {{6^3}}} + {{10} \over {{6^4}}} + $ ...... (4)

Subtract equation (4) from equation (3), we get

${{25} \over {36}}S = 1 + {3 \over 6} + {3 \over {{6^2}}} + {3 \over {{6^3}}} + $ ......

$ \Rightarrow {{25S} \over {36}} = 1 + {{{3 \over 5}} \over {1 - {1 \over 6}}}$

$ = 1 + {3 \over 6} \times {6 \over 5}$

$ = 1 + {3 \over 5} = {8 \over 5}$

$ \Rightarrow S = {8 \over 5} \times {{36} \over {25}} = {{288} \over {125}}$

${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$ & ${a_0} = {a_1} = 0$

${a_2} = 2{a_1} - {a_0} + 1 = 1$

${a_3} = 2{a_2} - {a_1} + 1 = 3$

${a_4} = 2{a_3} - {a_2} + 1 = 6$

${a_5} = 2{a_4} - {a_3} + 1 = 10$

$\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}} = {{{a_2}} \over {{7^2}}} + {{{a_3}} \over {{7^3}}} + {{{a_4}} \over {{7^4}}} + \,\,...} $

$s = {1 \over {{7^2}}} + {3 \over {{7^3}}} + {6 \over {{7^4}}} + {{10} \over {{7^5}}} + \,\,...$

${{{1 \over 7}s = {1 \over {{7^3}}} + {3 \over {{7^4}}} + {6 \over {{7^5}}} + \,\,...} \over {{{6s} \over 7} = {1 \over {{7^2}}} + {2 \over {{7^3}}} + {3 \over {{7^4}}} + \,\,...}}$

${{{{6s} \over {49}} = \,\,\,\,\,\,\,\,\,{1 \over {{7^3}}} + {2 \over {{7^4}}} + \,\,...} \over {{{36s} \over {49}} = {1 \over {{7^2}}} + {1 \over {{7^3}}} + {1 \over {{7^4}}} + \,\,...}}$

${{36s} \over {49}} = {{{1 \over {{7^2}}}} \over {1 - {1 \over 7}}}$

${{36s} \over {49}} = {7 \over {49 \times 6}}$

$s = {7 \over {216}}$

a, A₁, A₂ ........... A_n, 100

Let d be the common difference of above A.P. then

${{a + d} \over {100 - d}} = {1 \over 7}$

$ \Rightarrow 7a + 8d = 100$ ...... (i)

and $a + n = 33$ ..... (ii)

and $100 = a + (n + 1)d$

$ \Rightarrow 100 = a + (34 - a){{(100 - 7a)} \over 8}$

$ \Rightarrow 800 = 8a + 7{a^2} - 338a + 3400$

$ \Rightarrow 7{a^2} - 330a + 2600 = 0$

$ \Rightarrow a = 10,\,{{260} \over 7},$ but $a \ne {{260} \over 7}$

$\therefore$ $n = 23$

${{{A_4}} \over {{r^3}}}.\,{{{A_4}} \over r}.\,{A_4}r\,.\,{A_4}{r^3} = {1 \over {1296}}$

${A_4} = {1 \over 6}$

${A_2} = {7 \over {36}} - {1 \over 6} = {1 \over {36}}$

So ${A_6} + {A_8} + {A_{10}} = 1 + 6 + 36 = 43$

$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + $ ..... ...... (i)

${1 \over 7}S = {2 \over 7} + {6 \over {{7^2}}} + {{12} \over {{7^3}}} + {{20} \over {{7^4}}} + $ .... ....... (ii)

(i) - (ii)

${6 \over 7}S = 2 + {4 \over 7} + {6 \over {{7^2}}} + {8 \over {{7^3}}} + $ ...... ....... (iii)

${6 \over {{7^2}}}S = {2 \over 7} + {4 \over {{7^2}}} + {6 \over {{7^3}}} + $ ..... ......... (iv)

(iii) - (iv)

${\left( {{6 \over 7}} \right)^2}S = 2 + {2 \over 7} + {2 \over {{7^2}}} + {2 \over {{7^3}}} + $ ......

$ = 2\left[ {{1 \over {1 - {1 \over 7}}}} \right] = 2\left( {{7 \over 6}} \right)$

$\therefore$ $4S = 8{\left( {{7 \over 6}} \right)^3} = {\left( {{7 \over 3}} \right)^3}$

a₁, a₂, a₃ .... are in A.P. (Let common difference is d₁)

b₁, b₂, b₃ .... are in A.P. (Let common difference is d₂)

and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$\because$ a₁b₁ = 1

$\therefore$ b₁ = ${1 \over 2}$

a₁₀b₁₀ = 1

$\therefore$ b₁₀ = ${1 \over 3}$

Now, a₁₀ = a₁ + 9d₁ $\Rightarrow$ d₁ = ${1 \over 9}$

b₁₀ = b₁ + 9d₂ $\Rightarrow$ d₂ = ${1 \over 9}$$\left[ {{1 \over 3} - {1 \over 2}} \right]$ = $-$ ${{1 \over {54}}}$

Now, a₄ = 2 + ${{3 \over {9}}}$ = ${{7 \over {3}}}$

b₄ = ${{1 \over {2}}}$ $-%$ ${{3 \over {54}}}$ = ${{4 \over {9}}}$

$\therefore$ a₄b₄ = ${{28 \over {27}}}$

$x = \sum\limits_{n = 0}^\infty {{a^n} = {1 \over {1 - a}};\,y = \sum\limits_{n = 0}^\infty {{b^n} = {1 \over {1 - b}};\,z = \sum\limits_{n = 0}^\infty {{c^n} = {1 \over {1 - c}}} } } $

Now,

a, b, c $\to$ AP

1 $-$ a, 1 $-$ b, 1 $-$ c $\to$ AP

${1 \over {1 - a}},\,{1 \over {1 - b}},\,{1 \over {1 - c}} \to HP$

x, y, z $\to$ HP

$\therefore$ ${1 \over x},{1 \over y},{1 \over z} \to AP$

$A = \sum\limits_{n = 1}^\infty {{1 \over {{{(3 + {{( - 1)}^n})}^n}}}} $ and $B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{(3 + {{( - 1)}^n})}^n}}}} $

$A = {1 \over 2} + {1 \over {{4^2}}} + {1 \over {{2^3}}} + {1 \over {{4^4}}} + $ ........

$B = {{ - 1} \over 2} + {1 \over {{4^2}}} - {1 \over {{2^3}}} + {1 \over {{4^4}}} + $ ......

$A = {{{1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$, $B = {{ - {1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$

$A = {{11} \over {15}}$, $B = {{ - 9} \over {15}}$

$\therefore$ ${A \over B} = {{ - 11} \over 9}$

Let $S = 1\,.\,{3^0} + 2\,.\,{3^1} + 3\,.\,{3^2} + \,\,......\,\, + \,\,10\,.\,{3^9}$

$3S = 1\,.\,{3^1} + 2\,.\,{3^2} + \,\,..........\,\, + \,\,10\,.\,{3^{10}}$

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$ - 2S = (1\,.\,{3^0} + 1\,.\,{3^1} + 1\,.\,{3^2} + \,\,........\,\, + \,\,1\,.\,{3^9}) - 10\,.\,{3^{10}}$

$ \Rightarrow S = {1 \over 2}\left[ {10\,.\,{3^{10}} - {{{3^{10}} - 1} \over { - 3 - 1}}} \right]$

$ \Rightarrow S = {{19\,.\,{3^{10}} + 1} \over 4}$

x, y > 0 and x³y² = 2¹⁵

Now, 3x + 2y = (x + x + x) + (y + y)

So, by A.M $\ge$ G.M inequality

${{3x + 2y} \over 5} \ge \root 5 \of {{x^3}\,.\,{y^2}} $

$\therefore$ $3x + 2y \ge 5\root 5 \of {{2^{15}}} \ge 40$

$\therefore$ Least value of $3x + 4y = 40$

$\sum\limits_{i = 1}^n {{a_i} = 192} $

$\Rightarrow$ a₁ + a₂ + a₃ + ...... + a_n = 192

$ \Rightarrow {n \over 2}[{a_1} + {a_n}] = 192$

$ \Rightarrow {a_1} + {a_n} = {{384} \over n}$ ..... (1)

Now, $\sum\limits_{i = 1}^{{n \over 2}} {{a_{2i}} = 120} $

$\Rightarrow$ a₂ + a₄ + a₆ + ...... + a_n = 120

Here total ${n \over 2}$ terms present.

$\therefore$ ${{{n \over 2}} \over 2}[{a_2} + {a_n}] = 120$

$ \Rightarrow {n \over 4}[{a_1} + 1 + {a_n}] = 120$

$ \Rightarrow {a_1} + {a_n} + 1 = {{480} \over n}$ ..... (2)

Subtracting (1) from (2), we get

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

$ \Rightarrow 1 = {{96} \over n}$

$\Rightarrow$ n = 96