Question 1: Show that each one of the following progressions is a G.P.. Also, find the common ratio in each case:

i) $4, -2, 1, -$ $\frac{1}{2}$ $, \ldots$     ii) $\frac{-2}{3}$ $, -6, -54, \ldots$     iii) $a,$ $\frac{3a^2}{4}$ $,$ $\frac{9a^3}{16}$ $, \ldots$     iv) $\frac{1}{2}$ $,$ $\frac{1}{3}$ $,$ $\frac{2}{9}$ $,$ $\frac{4}{27}$ $, \ldots$

i)      $4, -2, 1, -$ $\frac{1}{2}$ $, \ldots$

Given $a_1 = 4, \ \ \ \ a_2 = -2, \ \ \ \ a_3 = 1, \ \ \ \ a_4 =$ $\frac{-1}{2}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{-2}{4}$ $=$ $\frac{-1}{2}$       and        $\frac{a_3}{a_2}$ $=$ $\frac{-1}{2}$    and             $\frac{a_4}{a_3}$ $=$ $\frac{\frac{-1}{2}}{1}$ $=$ $\frac{-1}{2}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{a_3}{a_2}$ $=$ $\frac{a_4}{a_3}$ $=$ $\frac{-1}{2}$

Therefore $a_1, a_2, a_3$ and $a_4$ are in G.P. where $a = 4$ and $r =$ $\frac{-1}{2}$

ii)    $\frac{-2}{3}$ $, -6, -54, \ldots$

Given $a_1 =$ $\frac{-2}{3}$ $\ \ \ \ a_2 = -6, \ \ \ \ a_3 = -54$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{-6}{\frac{-2}{3}}$ $= 9$               $\frac{a_3}{a_2}$ $=$ $\frac{-54}{-6}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{a_3}{a_2}$ $= 9$

Therefore $a_1, a_2, a_3$  are in G.P. where $a =$ $\frac{-2}{3}$ and $r = 9$

iii)    $a,$ $\frac{3a^2}{4}$ $,$ $\frac{9a^3}{16}$ $, \ldots$

Given $a_1 = a$        $a_2 =$ $\frac{3a^2}{4}$ $, \ \ \ \ a_3 =$ $\frac{9a^3}{16}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{3a^2}{4a}$ $=$ $\frac{3a}{4}$      and         $\frac{a_3}{a_2}$ $=$ $\frac{9a^3}{16}$ $\cdot$ $\frac{4}{3a^2}$ $=$ $\frac{3a}{4}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{a_3}{a_2}$  $=$ $\frac{3a}{4}$

Therefore $a_1, a_2, a_3$  are in G.P. where $a$ is the first term and $r =$ $\frac{3a}{4}$

iv) $\frac{1}{2}$ $,$ $\frac{1}{3}$ $,$ $\frac{2}{9}$ $,$ $\frac{4}{27}$ $, \ldots$

Given $a_1 =$ $\frac{1}{2}$ $, \ \ \ \ a_2 =$ $\frac{1}{3}$ $, \ \ \ \ a_3 =$ $\frac{2}{9}$ $, \ \ \ \ a_4 =$ $\frac{4}{27}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{1/3}{1/2}$ $=$ $\frac{2}{3}$       and        $\frac{a_3}{a_2}$ $=$ $\frac{2/9}{1/3}$ $=$ $\frac{2}{3}$    and             $\frac{a_4}{a_3}$ $=$ $\frac{4/27}{2/9}$ $=$ $\frac{2}{3}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{a_3}{a_2}$ $=$ $\frac{a_4}{a_3}$ $=$ $\frac{2}{3}$

Therefore $a_1, a_2, a_3$ and $a_4$ are in G.P. where $a =$ $\frac{1}{2}$ and $r =$ $\frac{2}{3}$

$\\$

Question 2: Show that the sequence defined by $a_n =$ $\frac{2}{3^n}$ $, n \in N$ is a G.P.

Given $a_n =$ $\frac{2}{3^n}$ $, n \in N$ is a G.P

Given $a_1 =$ $\frac{2}{3}$ $, \ \ \ \ a_2 =$ $\frac{2}{3^2}$ $, \ \ \ \ a_3 =$ $\frac{2}{3^3}$ $, \ \ \ \ a_4 =$ $\frac{2}{3^4}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{2/3^2}{2/3}$ $=$ $\frac{1}{3}$       and        $\frac{a_3}{a_2}$ $=$ $\frac{2/3^3}{2/3^2}$ $=$ $\frac{1}{3}$    and             $\frac{a_4}{a_3}$ $=$ $\frac{2/3^4}{2/3^3}$ $=$ $\frac{1}{3}$

$\therefore$    $\frac{a_2}{a_1}$ $=$ $\frac{a_3}{a_2}$ $=$ $\frac{a_4}{a_3}$ $=$ $\frac{1}{3}$

Therefore $a_n =$ $\frac{2}{3^n}$ $, n \in N$ is a G.P  where $a =$ $\frac{2}{3}$ and $r =$ $\frac{1}{3}$

$\\$

Question 3: Find:

i) the $9^{th}$ term of the G.P.  $1, 4, 16, 64 \ldots$

ii) the ${10}^{th}$ term of the G.P. $\frac{-3}{4}$ $,$ $\frac{1}{2}$ $,$ $\frac{-1}{3}$ $,$ $\frac{2}{9}$ $\ldots$

iii) the $8^{th}$ term of the G.P.  $0.3, 0.06, 0.012, \ldots$

iv) the ${12}^{th}$ term of the G.P. $\frac{1}{a^3x^3}$ $, ax, a^5x^5, \ldots$

v) the $n^{th}$ term of the G.P. $\sqrt{3},$ $\frac{1}{\sqrt{3}}$ $,$ $\frac{1}{3\sqrt{3}}$ $, \ldots$

vi) the ${10} ^{th}$ term of the G.P. $\sqrt{2},$ $\frac{1}{\sqrt{2}}$ $,$ $\frac{1}{2\sqrt{2}}$ $\ldots$

i)       To find the $9^{th}$ term of the G.P.  $1, 4, 16, 64 \ldots$

Here $a = 1, \hspace{1.0cm} r =$ $\frac{a_2}{a_1}$ $=$ $\frac{4}{1}$ $= 4$

We know, $a_n = ar^{n-1}$

$\therefore a_9 = (1)(4)^{9-1} = 4^8 = 65536$

ii)     To find the ${10}^{th}$ term of the G.P.$\frac{-3}{4}$ $,$ $\frac{1}{2}$ $,$ $\frac{-1}{3}$ $,$ $\frac{2}{9}$ $\ldots$

Here $a =$ $\frac{-3}{4}$ $, \hspace{1.0cm} r =$ $\frac{a_2}{a_1}$ $= \Big($ $\frac{1/2}{-3/4}$ $\Big) =$ $\frac{-2}{3}$

We know, $a_n = ar^{n-1}$

$\therefore a_{10} = \Big($ $\frac{-3}{4}$ $\Big) \Big($ $\frac{-2}{3}$ $\Big)^{10-1} =$ $\frac{1}{2}$ $\Big ($ $\frac{2}{3}$ $\Big )^8$

iii)    To find the $8^{th}$ term of the G.P.  $0.3, 0.06, 0.012, \ldots$

Here $a = 0.3, \hspace{1.0cm} r =$ $\frac{a_2}{a_1}$ $=$ $\frac{0.06}{0.3}$ $= 0.2$

We know, $a_n = ar^{n-1}$

$\therefore a_8 = (0.3)(0.2)^{8-1} = (0.3)(0.2)^7$

iv)    To find the ${12}^{th}$ term of the G.P.$\frac{1}{a^3x^3}$ $, ax, a^5x^5, \ldots$

Here $a =$ $\frac{1}{a^3x^3}$ $, \hspace{1.0cm} r =$ $\frac{a_2}{a_1}$ $= \Bigg($ $\frac{ax}{\frac{1}{a^3x^3}}$ $\Bigg) =$ $a^4x^4$

We know, $a_n = ar^{n-1}$

$\therefore a_{12} = \Big($ $\frac{1}{a^3x^3}$ $\Big) (a^4 x^4)^{12-1} = a^{41}x^{41}$

v)     To find the $n^{th}$ term of the G.P.$\sqrt{3},$ $\frac{1}{\sqrt{3}}$ $,$ $\frac{1}{3\sqrt{3}}$ $, \ldots$

Here $a = \sqrt{3}, \hspace{1.0cm} r =$ $\frac{a_2}{a_1}$ $=$ $\frac{1/\sqrt{3}}{\sqrt{3}}$ $=$ $\frac{1}{3}$

We know, $a_n = ar^{n-1}$

$\therefore a_n = (\sqrt{3}) \Big($ $\frac{1}{3}$ $\Big)^{n-1}$

vi)     To find the ${10} ^{th}$ term of the G.P.$\sqrt{2},$ $\frac{1}{\sqrt{2}}$ $,$ $\frac{1}{2\sqrt{2}}$ $\ldots$

Here $a = \sqrt{2}, \hspace{1.0cm} r =$ $\frac{a_2}{a_1}$ $=$ $\frac{1/\sqrt{2}}{\sqrt{2}}$ $=$ $\frac{1}{2}$

We know, $a_n = ar^{n-1}$

$\therefore a_{10} = (\sqrt{2}) \Big($ $\frac{1}{2}$ $\Big)^{10-1} = \sqrt{2} \Big($ $\frac{1}{2^9}$ $\Big) =$ $\frac{1}{(\sqrt{2})^{17}}$

$\\$

Question 4: Find the $4^{th}$ term from the end of the G.P. $\frac{2}{27}$ $,$ $\frac{2}{9},$ $\frac{2}{3}$ $, \ldots , 162$

Here $a =$ $\frac{2}{27}$ $, \hspace{1.0cm} r =$ $\frac{a_2}{a_1}$ $= \Big($ $\frac{2/9}{2/27}$ $\Big) =$ $3$

Last term $l = 162$

Now, when we reverse the G.P., we have $a = l$, and $r =$ $\frac{1}{3}$

We know, $a_n = ar^{n-1}$

$\therefore a_{4} = (162) ($ $\frac{1}{3}$ $) =$ $\frac{162}{3^3}$ $= 6$

$\\$

Question 5: Which term of the progression $0.004, 0.02, 0.1, \ldots$ is $12.5$

Given series: $0.004, 0.02, 0.1, \ldots$ is $12.5$

Here $a = 0.004$     $\frac{a_2}{a_1}$ $=$ $\frac{0.02}{0.004}$ $= 5$     $\frac{a_3}{a_2}$ $= \frac{0.1}{0.02}$ $= 5$

$\therefore$ $\frac{a_2}{a_1}$ $=$ $\frac{a_3}{a_2}$ $= 5$

Let $12.5$ be the $n^{th}$ term

$\therefore 12.5 = ar^{n-1}$

$\Rightarrow 12.5 = ( 0.004) ( 5)^{n-1}$

$\Rightarrow 5^{n-1} = 3125$

$\Rightarrow 5^{n-1} = 5^5$

$\therefore n - 1 = 5 \Rightarrow n = 6$

Therefore $12.5$ is the $6^{th}$ term of the given G.P.

$\\$

Question 6: Which term of the G.P.:

i) $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}}, \ldots \text{ is } \frac{1}{512\sqrt{2}} ?$

ii) $2, 2\sqrt{2}, 4, \ldots \text{ is } 128 ?$

iii) $\sqrt{3}, 3, 3\sqrt{3}, \ldots \text{ is } 729?$

iv) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}$ $, \ldots \text{ is }$ $\frac{1}{19683}$ $?$

i)       Given series: $\sqrt{2},$ $\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}}$ $, \ldots$ ?

Here $a = \sqrt{2}$     $\frac{a_2}{a_1} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}$

We know $a_n = ar^{n-1}$

Let $\frac{1}{512\sqrt{2}}$ be the $n^{th}$ term

$\therefore \ \ \$ $\frac{1}{512\sqrt{2}}$ $= ( \sqrt{2}) \Big($ $\frac{1}{2}$ $\Big)^{n-1}$

$\Rightarrow$ $\frac{1}{1024}$ $= \Big($ $\frac{1}{2}$ $\Big)^{n-1}$

$\Rightarrow \Big($ $\frac{1}{2}$ $\Big)^{10} = \Big($ $\frac{1}{2}$ $\Big)^{n-1}$

$\Rightarrow n-1 = 10$

$\Rightarrow n = 11$

Therefore $\frac{1}{512\sqrt{2}}$ is the ${11}^{th}$ term of the given G.P.

ii)      Given series: $2, 2\sqrt{2}, 4, \ldots$

Here $a = 2$     $\frac{a_2}{a_1} = \frac{2\sqrt{2}}{2}$ $= \sqrt{2}$

We know $a_n = ar^{n-1}$

Let $128$ be the $n^{th}$ term

$\therefore 128 = 2 ( \sqrt{2} )^{n-1}$

$\Rightarrow 64 = ( \sqrt{2})^{n-1}$

$\Rightarrow (\sqrt{2})^{12}= ( \sqrt{2})^{n-1}$

$\Rightarrow n - 1 = 12$

$\Rightarrow n = 13$

Therefore $128$ is the ${13}^{th}$ term of the given G.P.

iii)     Given series: $\sqrt{3}, 3, 3\sqrt{3}, \ldots$

Here $a = \sqrt{3}$     $\frac{a_2}{a_1}$ $=$ $\frac{3}{\sqrt{3}}$ $= \sqrt{3}$

We know $a_n = ar^{n-1}$

Let $729$ be the $n^{th}$ term

$\therefore 729 = (\sqrt{3}) ( \sqrt{3} )^{n-1}$

$\Rightarrow (\sqrt{3})^{n-1} =$ $\frac{( \sqrt{2})^{12}}{\sqrt{3}}$

$\Rightarrow (\sqrt{3})^{n-1} = ( \sqrt{3})^{11}$

$\Rightarrow n - 1 = 11$

$\Rightarrow n = 12$

Therefore $729$ is the ${12}^{th}$ term of the given G.P.

iv)     Given series: $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}$ $, \ldots$

Here $a =$ $\frac{1}{3}$     $\frac{a_2}{a_1}$ $=$ $\frac{1/9}{1/3}$ $=$ $\frac{1}{3}$

We know $a_n = ar^{n-1}$

Let $\frac{1}{19683}$ be the $n^{th}$ term

$\therefore$ $\frac{1}{19683}$ $= \Big($ $\frac{1}{3}$ $\Big) \Big($ $\frac{1}{3}$ $\Big )^{n-1}$

$\Rightarrow (\sqrt{3})^{n-1} =$ $\frac{( \sqrt{2})^{12}}{\sqrt{3}}$

$\Rightarrow (\frac{1}{3})^{9} =$ $\frac{( \sqrt{2})^{12}}{\sqrt{3}}$

$\Rightarrow n - 1 + 1 = 9$

$\Rightarrow n = 9$

Therefore $\frac{1}{19683}$ is the ${9}^{th}$ term of the given G.P.

$\\$

Question 7: Which term of the progression $18, -12, 8, \ldots$ is $\frac{512}{729}$?

Given series $18, -12, 8, \ldots$

Here $a = 18$     $\frac{a_2}{a_1}$ $=$ $\frac{-12}{18}$ $=$ $\frac{-2}{3}$

We know $a_n = ar^{n-1}$

Let $\frac{512}{729}$ be the $n^{th}$ term

$\Rightarrow$ $\frac{512}{729}$ $= ( 18) \Big($ $\frac{-2}{3}$ $\Big)^{n-1}$

$\Rightarrow \Big($ $\frac{-2}{3}$ $\Big)^{n-1} =$ $\frac{512}{729}$ $\times$ $\frac{1}{18}$

$\Rightarrow \Big($ $\frac{-2}{3}$ $\Big)^{n-1} =$ $\frac{256}{6561}$

$\Rightarrow \Big($ $\frac{-2}{3}$ $\Big)^{n-1} = \Big($ $\frac{-2}{3}$ $\Big)^{8}$

$\therefore n - 1 = 8 \Rightarrow n = 9$

Therefore $\frac{512}{729}$ is the $9^{th}$ term of the given G.P.

$\\$

Question 8: Find the $4^{th}$ terms from the end of the G.P. $\frac{1}{2}, \frac{1}{6}, \frac{1}{8}, \frac{1}{54}$ $, \ldots$ $\frac{1}{4374}$

Given series: $\frac{1}{2}, \frac{1}{6}, \frac{1}{8}, \frac{1}{54}$ $, \ldots$

Reversing the G.P. we get $a =$ $\frac{1}{4374}$ $, \hspace{1.0cm} r = 3$

We know $a_n = ar^{n-1}$

$\therefore a_4 =$ $\Big( \frac{1}{4374}$ $\Big) (3)^{4-1}$

$\Rightarrow a_4 = \Big($ $\frac{1}{4374}$ $\Big) (3)^{3}$

$\Rightarrow a_4 =$ $\frac{1}{162}$

$\\$

Question 9: The $4^{th}$ term of the G.P. is $27$ and the $7^{th}$ term is $729$, find the G.P.

Given $a_4 = 27 \Rightarrow ar^3 = 27$     … … … … … i)

Given $a_7 = 729 \Rightarrow ar^6 = 729$     … … … … … ii)

Dividing ii) by i) we get

$\frac{ar^6}{ar^3}$ $=$ $\frac{729}{27}$

$\Rightarrow r^3 = 27$

$\Rightarrow r^3 = 3^3$

$\Rightarrow r = 3$

Substituting in i) we get

$a(3)^3 = 27 \Rightarrow a = 1$

Therefore the G.P. is $1, 3, 9, 27, \ldots$

$\\$

Question 10:The $7^{th}$ term of the G.P. is $8$ times the $4^{th}$ term and $5^{th}$ term is $48$. Find the G.P.

Given $a_7 = a_4$

$\Rightarrow ar^6 = 8 ar^3$

$\Rightarrow r^3 = 8$

$\Rightarrow r = 2$

Substituting we get $a(2)^4 = 48 \Rightarrow a =$ $\frac{48}{4 \times 14}$ $=3$

Therefore the G.P. is $3, 6, 12, 24, \ldots$

$\\$

Question 11: If the G.P.’s $5, 10, 20, \ldots$ and $1280, 640, 320, \ldots$ have their $n^{th}$ terms equal, find the value of $n$.

For G.P. $5, 10, 20, \ldots$

Therefore $a = 5$          $\frac{a_2}{a_1}$ $=$ $\frac{10}{5}$ $= 2$

We know $a_n = ar^{n-1}$

$\Rightarrow a_n = (5)( 2)^{n-1}$    … … … … … i)

For G.P. $1280, 640, 320, \ldots$

Therefore $a = 1280$           $\frac{a_2}{a_1}$ $=$ $\frac{640}{1280}$ $=$ $\frac{1}{2}$

We know $a_n = ar^{n-1}$

$\Rightarrow a_n = (1280) \Big($ $\frac{1}{2}$ $\Big)^{n-1}$    … … … … … ii)

From i) and ii) we get

$(5)( 2)^{n-1} = (1280) \Big($ $\frac{1}{2}$ $\Big)^{n-1}$

$\Rightarrow 2^{2n - 2} = 256$

$\Rightarrow 2^{2n - 2} = 2^8$

$\Rightarrow 2n - 2 = 8$

$\Rightarrow n = 5$

$\\$

Question 12: The $5^{th}, 8^{th}$ and ${11}^{th}$ term of a G.P. are $p, q$, and $s$ respectively, prove tht $q^2 = ps$.

Given $a_5 = ar^4 = p$     … … … … … i)

$a_8 = ar^7 = q$     … … … … … ii)

$a_{11} = ar^{10} = s$     … … … … … iii)

Therefore $ps = (ar^4) \cdot ( ar^{10} ) = ( ar^7)^{2} = q^2$. Hence proved.

$\\$

Question 13: The $4^{th}$ term of a G.P. is square of its $2^{nd}$ term and the first term is $-3$. Find its $7^{th}$ term.

Given $a_4 = (a_2)^2$   and $a = -3$

$\Rightarrow ar^3 = ( ar)^2$

$\Rightarrow ar^3 = a^2 r^2$

$\Rightarrow r = a$

$\therefore a_7 = ar^6 = ( -3)( -3)^6 = (-3)^7 = - 2187$

$\\$

Question 14: In a G.P. the $3^{rd}$ term is $24$ and the $6^{th}$ term is $192$. Find the ${10}^{th}$ term.

$ar^2 = 24$      … … … … … i)

$ar^5 = 192$      … … … … … ii)

Dividing ii) and i)  we get

$r^3 =$ $\frac{192}{24}$ $= 8$

$r^3 = 2^3 \Rightarrow r = 2$

Substituting in i) we get

$a(2)^2 = 24 \Rightarrow a = 6$

$\therefore a_{10} = ar^9 = 6(2)^9 = 3072$

$\\$

Question 15: If $a, b, c$ and $p$ are different real numbers such that

$(a^2 + b^2 + c^2 )p^2 - 2 ( ab + bc + cd) p + ( b^2 + c^2 +d^2) \leq 0$, then show that $a, b, c$ and $d$ are in G.P.

Given $(a^2 + b^2 + c^2 )p^2 - 2 ( ab + bc + cd) p + ( b^2 + c^2 +d^2) \leq 0$

$\Rightarrow (a^2p^2 + b^2p^2 + c^2p^2 ) - 2 ( abp + bcp + cdp)+ ( b^2 + c^2 +d^2) \leq 0$

$\Rightarrow (a^2p^2 - 2abp+b^2) + ( b^2p^2 - 2bcp +c^2) + ( c^2 p^2 - 2cdp + d^2) \leq 0$

$\Rightarrow (ap-b)^2 + ( bp-c)^2 + ( cp-d)^2 \leq 0$

Since all the terms are square, therefore they cannot be less than zero. Hence,

$\Rightarrow (ap-b)^2 + ( bp-c)^2 + ( cp-d)^2 = 0$

$\Rightarrow (ap-b)^2 = 0 \hspace{1.5cm} ( bp-c)^2=0 \hspace{1.5cm} ( cp-d)^2 = 0$

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

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

$\Rightarrow a, b, c \text{ and } d$ are in G.P.

$\\$

Question 16:  If  $\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} =\frac{c+dx}{c-dx}$ $(x \neq 0)$, then show that $a, b, c$ and $d$ are in G.P.

Given $\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} =\frac{c+dx}{c-dx}$

Take the first two terms

$\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx}$

Apply componendo and dividendo we get

$\frac{(a+bx)+ (a-bx)}{(a+bx) - (a-bx)} = \frac{(b+cx) + (b-cx)}{(b+cx)+ (b-cx)}$

$\frac{2a}{2bx} = \frac{2b}{2cx}$

$\frac{a}{b} = \frac{b}{c}$     … … … … … i)

Similarly,  Take the last two terms

$\frac{b+cx}{b-cx} =\frac{c+dx}{c-dx}$

Apply componendo and dividendo we get

$\frac{(b+cx) + (b-cx)}{(b+cx)+ (b-cx)} = \frac{(c+dx)+ (c-dx)}{(c+dx) - (c-dx)}$

$\frac{2b}{2cx} = \frac{2c}{2dx}$

$\frac{b}{c} = \frac{c}{d}$     … … … … … ii)

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

Therefore $a, b, c$ and $d$ are in G.P.

$\\$

Question 17: If the $p^{th}$ and the $q^{th}$ terms of a G.P. are $q$ and $p$ respectively, show that $(p+q)^{th}$ term is $\Big( \frac{q^p}{p^q} \Big)^{\frac{1}{p-q}}$

Given $a_p = ar^{p-1} = q$    … … … … … i)

$a_q = ar^{q-1} = p$     … … … … … ii)

Dividing i) by ii)

$\frac{ar^{p-1}}{ar^{q-1}} = \frac{q}{p}$

$\frac{r^{p-1}}{r^{q-1}} = \frac{q}{p}$

$r^{p-1-q+1} = \frac{q}{p}$

$r^{p-q} = \frac{q}{p}$

$r = \Big( \frac{q}{p} \Big)^{\frac{1}{p-q}}$     … … … … … i)

Substituting the value of r in ii)  we get

$a \Big[ \Big( \frac{q}{p} \Big)^{\frac{1}{p-q}} \Big] ^{q-1} = p$

$a \Big[ \Big( \frac{q}{p} \Big)^{\frac{q-1}{p-q}} \Big] = p$

$a = p \Big( \frac{p}{q} \Big)^{\frac{q-1}{p-q}}$     … … … … … ii)

Therefore $a_{p+q} = a r^{p+q-1}$     … … … … … iii)

Substituting i) and ii) in iii) we get

$a_{p+q} = \Big[ p \Big( \frac{p}{q} \Big)^{\frac{q-1}{p-q}} \Big] \cdot \Big[ \Big( \frac{q}{p} \Big)^{\frac{1}{p-q}} \Big]^{p+q-1}$

$= \Big[ p \Big( \frac{p}{q} \Big)^{\frac{q-1}{p-q}} \Big] \cdot \Big[ \Big( \frac{q}{p} \Big)^{\frac{p+q-1}{p-q}} \Big]$

$= \Big[ p \Big( \frac{q}{p} \Big)^{\frac{-(q-1)}{p-q}} \Big] \cdot \Big[ \Big( \frac{q}{p} \Big)^{\frac{p+q-1}{p-q}} \Big]$

$= p \Big( \frac{q}{p} \Big)^{ \frac{p+q-1}{p-q} - \frac{q-1}{p-q} }$

$= p \Big( \frac{q}{p} \Big)^{ \frac{p}{p-q} }$

$= \frac{ q^{\frac{p}{p-q}} }{ p^{\frac{q}{p-q} } }$

$= \Big( \frac{q^p}{p^q} \Big)^{\frac{1}{p-q}}$