9if the pth and qth terms of a gp are q and p respectively show that p+q th term is (q^pp^q )^1/p-q

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Solution

$Let the first term is a and common ratio is r so, according to the condition, q = {ar}^{p - 1} . . . . . . . . . (1) p = {ar}^{q - 1} dividing we get, \frac{r^{p - 1}}{r^{q - 1}} = \frac{q}{p} r^{p - q} = \frac{q}{p} r = {(\frac{q}{p})}^{\frac{1}{p - q}} Putting this in equation 1 we get, q = a {(\frac{q}{p})}^{\frac{p - 1}{p - q}} a = q^{1 - \frac{p - 1}{p - q}} \times p^{\frac{p - 1}{p - q}} = q^{\frac{1 - q}{p - q}} p^{\frac{p - 1}{p - q}} Now p + q th term, T_{p + q} = {ar}^{p + q - 1} = q^{\frac{1 - q}{p - q}} p^{\frac{p - 1}{p - q}} \times {(\frac{q}{p})}^{\frac{p + q - 1}{p - q}} = {[q^{1 - q} \times p^{p - 1} \times \frac{q^{p + q - 1}}{p^{p + q - 1}}]}^{\frac{1}{p - q}} = {[q^{1 - q + p + q - 1} \times p^{p - 1 - p - q + 1}]}^{\frac{1}{p - q}} = {(q^{p} \times p^{- q})}^{\frac{1}{p - q}} = {(\frac{q^{p}}{p^{q}})}^{\frac{1}{p - q}}$

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