Question

if in a gp the (p+q)th term is a and (p-q)th term is b , prove that the pth term is sqrt (ab).

Answer 1

Let m be the first term and r be the common ratio of GP. Given,
$m \times {r}^{p + q - 1} = a - - - (1) \\ m \times {r}^{p - q - 1} = b - - - (2) \\ multiplying \: (1) \: \: and \: \: (2)$
${m}^{2} \times {r}^{2p - 2} = a \times b \\ {(m \times {r}^{p - 1} )}^{2} = ab \\ m \times {r}^{p - 1} = \sqrt{ab} \\ i.e. \: \: pth \: term \: is \: \: \sqrt{ab}$

Answer 2

Step-by-step explanation:

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