Question

The (p + q)th and (p − q)th terms of a G.P. are m and n, respectively. Prove that its pth term is `sqrt(mn)` and gth term is `m(n/m)^(p/(2q))`.

Answer 1

`T_(p + q) ⇒ ar^(p + q - 1) = m`   ....(1)

`T_(p - q) ⇒ ar^(p - q - 1) = 2`   ....(2)

Multiply Equation 1 and Equation 2:

`(ar^(p + q - 1)) xx (ar^(p - q - 1)) = m xx n`

`a^2r^((p + q - 1) + (p - q - 1)) = mn`

`a^2r^(2p-2) = mn`

`(ar^(p - 1 ))^2 = mn`

Taking the square root of both sides:

`ar^(p - 1) = sqrt(mn)`

Since `ar^(p - 1)` is the formula for the p^th term(T_p) we have:

`T_p = sqrt(mn)` Proved.

First, find r by dividing Equation 1 by Equation 2:

`(ar^(p + q - 1))/(ar^(p - q - 1)) = m/n`

`r^((p + q - 1) + (p - q - 1)) = m/n`

`r^(2q) = m/n`

r = `(m/n)^(1/(2q))`

`a = mr^(-(p + q - 1))`

`T_q = mr^(-(p + q - 1)) = r^(q - 1)`

`T_q = mr^(-p - q + 1 + q - 1)`

`T_q = mr^(-p)`

`T_q = m[(m/n)^(1/(2q))]^-p`

`T_q = m(m/n)^((-p)/(2q))`