Sequences and Series

${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$

${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$

$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$

$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$

From $\left( 1 \right)$ $a = {1 \over {mn}} \Rightarrow a - d = 0$

If $n$ is odd, the required sum is

${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$

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

[ As $\left( {n - 1} \right)$ is even

$\therefore$ using given formula for the sum of $\left( {n - 1} \right)$ terms.]

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

${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $

$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$

$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $

$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$

$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$

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

$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$

The product is $p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$

$ = {2^{1/4 + 2/8 + 3/16 + .......\infty }}$

Now let

$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$

${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)$

Subtracting $(2)$ from $(1)$

$ \Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty $

or ${1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1$

$\therefore$ $P = {2^S} = 2$

${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$

$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$

$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$

$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$

$ = {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$

$ = 2025 - 1600 = 425$

$l = A{R^{p - 1}}$

$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$

$m = A{R^{q - 1}}$

$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$

$n = A{R^{r - 1}}$

$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$

Now, $\left| {\matrix{ {\log l} & p & 1 \cr {\log m} & q & 1 \cr {\log n} & r & 1 \cr } } \right|$

$ = \left| {\matrix{ {\log A + \left( {p - 1} \right)\log R} & p & 1 \cr {\log A + \left( {q - 1} \right)\log R} & q & 1 \cr {\log A + \left( {r - 1} \right)\log R} & r & 1 \cr } } \right|$

Operating ${C_1} - \left( {\log R} \right){C_2} + \left( {\log R - \log A} \right){C_3}$

$ = \left| {\matrix{ 0 & p & 1 \cr 0 & q & 1 \cr 0 & r & 1 \cr } } \right| = 0$

Let $a=$ first team of $G.P.$ and $r=$ common ratio of $G.P.;$

Then $G.P.$ is $a,$ $ar,$ $a{r^2}$

Given ${S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20$

$ \Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)$

Also ${a^2} + {a^2}{r^2} + {a^2}{r^4} + ...$ to $\infty = 100$

$ \Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100$

$ \Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)$

From $(i),$ ${a^2} = 400{\left( {1 - r} \right)^2};$

From $(ii),$ we get $100\left( {1 - r} \right)\left( {1 + r} \right)$

$\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}$

$ \Rightarrow 1 + r = 4 - 4r$

$ \Rightarrow 5r = 3$

$ \Rightarrow r = 3/5.$

$1,\,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log _3}\left( {{{4.3}^x} - 1} \right)$ are in $A.P.$

$ \Rightarrow 2{\log _9}\left( {{3^{1 - x}} + 2} \right)$

$\,\,\,\,\,\,\,\,\,$ $ = 1 + {\log _3}\left( {{{4.3}^x} - 1} \right)$

$ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$

$\,\,\,\,\,\,\,\,\,$ $ = {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right)$

$ \Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$

$\,\,\,\,\,\,\,\,\,$ $ = {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right]$

$ \Rightarrow {3^{1 - x}} + 2 = 3\,\left( {{{4.3}^x} - 1} \right)$

$ \Rightarrow {3.3^{ - x}} + 2 = {12.3^x} - 3.$

Put ${3^x} = t$

$ \Rightarrow {3 \over t} + 2 = 12t - 3$

or $12{t^2} - 5t - 3 = 0;$

Hence $t = - {1 \over 3},{3 \over 4} \Rightarrow {3^x} = {3 \over 4}$

(as ${3^x}\,\, \ne \,\, - ve$ )

$ \Rightarrow x = {\log _3}\left( {{3 \over 4}} \right)$

or $x = {\log _3}3 - {\log _3}4$

$ \Rightarrow x = 1 - {\log _3}4$

$a{r^4} = 2$

$a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}$

$ = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512$