If the pth and qth terms of a G.P. are q and p, respectively, then show that (p + q)th term is ${(\frac{q^{p}}{p^{q}})}^{\frac{1}{p - q}}$ .

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Solution

$As, a_{p} = q \Rightarrow a r^{(p - 1)} = q . . . . . (i) Also, a_{q} = p \Rightarrow a r^{(q - 1)} = p . . . . . (ii) Dividing (i) by (ii), we get \frac{a r^{(p - 1)}}{a r^{(q - 1)}} = \frac{q}{p} \Rightarrow r^{(p - 1 - q + 1)} = \frac{q}{p} \Rightarrow r^{(p - q)} = \frac{q}{p} \Rightarrow r = {(\frac{q}{p})}^{\frac{1}{(p - q)}} Substituting the value of r in (ii), we get a {[{(\frac{q}{p})}^{\frac{1}{(p - q)}}]}^{(q - 1)} = p \Rightarrow a [{(\frac{q}{p})}^{\frac{(q - 1)}{(p - q)}}] = p \Rightarrow a = p \times {(\frac{p}{q})}^{\frac{(q - 1)}{(p - q)}} \Rightarrow a = p {(\frac{p}{q})}^{\frac{(q - 1)}{(p - q)}} Now, a_{(p + q)} = a r^{(p + q - 1)} = p {(\frac{p}{q})}^{\frac{(q - 1)}{(p - q)}} \times {[{(\frac{q}{p})}^{\frac{1}{(p - q)}}]}^{(p + q - 1)} = p {(\frac{p}{q})}^{\frac{(q - 1)}{(p - q)}} \times {(\frac{q}{p})}^{\frac{(p + q - 1)}{(p - q)}} = p {(\frac{q}{p})}^{\frac{- (q - 1)}{(p - q)}} \times {(\frac{q}{p})}^{\frac{(p + q - 1)}{(p - q)}} = p \times {(\frac{q}{p})}^{\frac{- (q - 1)}{(p - q)} + \frac{(p + q - 1)}{(p - q)}} = p \times {(\frac{q}{p})}^{\frac{- q + 1 + p + q - 1}{(p - q)}} = p \times {(\frac{q}{p})}^{\frac{p}{(p - q)}} = \frac{p \times q^{\frac{p}{(p - q)}}}{p^{\frac{p}{(p - q)}}} = \frac{q^{\frac{p}{(p - q)}}}{p^{\frac{p}{(p - q)} - 1}} = \frac{q^{\frac{p}{(p - q)}}}{p^{\frac{p - p + q}{(p - q)}}} = \frac{q^{\frac{p}{(p - q)}}}{p^{\frac{q}{(p - q)}}} = \frac{q^{p \times \frac{1}{(p - q)}}}{p^{q \times \frac{1}{(p - q)}}} = {(\frac{q^{p}}{p^{q}})}^{\frac{1}{p - q}}$

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