If $(p + q)$ th term and $(p - q)$ th terms of G.P are $a$ and $b$ respectively. prove that $p^{t h}$ term is $\sqrt{a b}$ .

Open in App

Solution

$n^{t h}$ term of an GP is $a_{1} r^{n - 1}$

where $a_{1}$ is the first term and $r$ is common ratio

$a_{p + q} = a_{1} r^{(p + q - 1)} a = a_{1} r^{(p + q - 1)} (1)$

$a_{p - q} = a_{1} r^{(p - q - 1)} b = a_{1} r^{(p - q - 1)} (2)$

Multiply $(1)$ and $(2)$ , we get

$a b = {a_{1}}^{2} r^{(p + q - 1 + p - q - 1)} a b = {a_{1}}^{2} r^{(2 p - 2)} a b = (a_{1} r^{p - 1})^{2} \sqrt{a b} = a_{1} r^{p - 1} = a_{p}$

Suggest Corrections

Similar questions

{(\frac{q^{p}}{p^{q}})}^{\frac{1}{p - q}}

If $(p + q)^{t h}$ and (p - q)th terms of a G.P. are m and n respectively, then write its pth term.

If 5th, 8th and 11th terms of a G.P. are p, q and s respectively, prove that $q^{2} = p s$ .

p^{t h}, q^{t h}

r^{t h}

a^{q - r} b^{r - p} c^{p - q} = 1

p^{t h}, q^{t h}

r^{t h}

a^{q - r} b^{r - p} c^{p - q} = 1

Join BYJU'S Learning Program