Question

If the $l^{th}$ ,$m^{th}$ and $n^{th}$ term of G.P. are a, b and c respectively then show that $a^{m-n}.b^{n-l},c^{l-m}=1$.

Answer 1

Let $A$ and $R$ be the first term and common ratio of Geometric Progression respectively.

We have $n^{th}$ term of GP, $A{R}^{n-1}$

Given, $l^{th}$ term of GP, $a$

$\therefore A{R}^{l-1}=a$ ......... $\left(1\right)$

Given, $m^{th}$ term of GP, $b$

$\therefore A{R}^{m-1}=b$ ....... $\left(2\right)$

Given, $n^{th}$ term of GP, $c$

$\therefore A{R}^{n-1}=c$ ...... $\left(3\right)$

Consider, $a^{m-n}.b^{n-l}.c^{l-m}$

Substitute from $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get,

$(A{R}^{l-1}{)}^{m-n}.(A{R}^{m-1}{)}^{n-l}.(A{R}^{n-1}{)}^{l-m}$

$=\left(A{)}^{m-n}\right(R^{(l-1)(m-n)}).(A{)}^{n-l}(R^{(n-l)(m-1)}.(A^{l-m}\left)\right(\left(R^{(n-1)(l-m)}\right)$

$=A^{m-n+n-l+l-m}{R}^{lm-ln-m+n+nm-n-lm-l+ln-l-mn+m}$

$=A^0{R}^0$

$=1$