The $(m + n)^{t h}$ and $(m - n)^{t h}$ terms of a G.P. are p and q respectively. Show that $m^{t h}$ and $n^{t h}$ terms are $\sqrt{p q}$ and $p {(\frac{q}{p})}^{\frac{m}{2 n}}$ respectively.

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Solution

$m + n^{t h}$ term of a G.P.
${a r}^{m + n - 1} = p ({m - n)}^{t h}$ term of G.P.=2
${a r}^{m - n - 1} = 2 p q = a^{2} r^{2 m - 2} {a r}^{m - 1} = \sqrt{p q}$ $m^{t h} t e r m = \sqrt{p q} \frac{p}{q} = r^{2 n} r = (\frac{p}{q})^{\frac{1}{2 n}} a (\frac{p}{q})^{\frac{m - 1}{2 n}} = \sqrt{p q} a = (\sqrt{p q}) (\frac{q}{p})^{\frac{m - 1}{2 n}} n^{t h} t e r m = a r^{n - 1} = \sqrt{p q} (\frac{q}{p})^{\frac{m - 1}{2 n}} (\frac{p}{q})^{\frac{n - 1}{2 n}} = \sqrt{p q} (\frac{p}{q})^{\frac{1 - m}{2 n} + \frac{n - 1}{2 n}} = \sqrt{p q} (\frac{p}{q})^{\frac{1}{2}} (\frac{q}{p})^{\frac{m}{2 n}} = p (\frac{q}{p})^{\frac{m}{2 n}}$

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The (m + n)th and the (m - n)th terms of a GP are p and q respectively. Show that the mth and the nth terms of the GP are $\sqrt{p q}$ and ${(\frac{q}{p})}^{(\frac{m}{2 n})}$

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The (m+n)th and (m−n)th terms of a G.P. are p and q respectively. Show that mth and nth terms are √pq and p(qp)m2n respectively.

m+nth term of a G.P.arm+n−1=p(m−n)thterm of G.P.=2arm−n−1=2pq=a2r2m−2arm−1=√pqmthterm=√pqpq=r2nr=(pq)12na(pq)m−12n=√pqa=(√pq)(qp)m−12nnthterm=arn−1=√pq(qp)m−12n(pq)n−12n=√pq(pq)1−m2n+n−12n=√pq(pq)12(qp)m2n=p(qp)m2n

The $(m + n)^{t h}$ and $(m - n)^{t h}$ terms of a G.P. are p and q respectively. Show that $m^{t h}$ and $n^{t h}$ terms are $\sqrt{p q}$ and $p {(\frac{q}{p})}^{\frac{m}{2 n}}$ respectively.