Question

If the (m+n)th term of a gp is p and (m-n)th terma is q, show that mth term and nth term are √pq and p(q/p)^m/2n

Answer 1

Let the given Geometrical progression be : 

      a, a r , a r²,  a r³ , .....

$m+n^{th} \: term: a \: r^{m+n-1} = p\\m-n^{th} \: term: a \: r^{m-n-1} = q\\\\=\ \textgreater \ r^{2n} = p/q\\=\ \textgreater \ r = (p/q)^{\frac{1}{2n}}\\\\Also, \: a^2 \: r^{2m-2}=p q\\ \implies a \: r^{m-1}=\sqrt{pq}=m^{th} \: term \: of \: GP\\\\n^{th} \: term =a \: r^{n-1}=\frac{p}{r^{m+n-1}}*r^{n-1}=\frac{p}{r^m}=p*r^{-m}\\\\=p*(\frac{p}{q})^{\frac{-m}{2n}}=p*(\frac{q}{p})^{\frac{m}{2n}}$

So its is proved.

Answer 2

Step-by-step explanation:

see the attachment

hope this will help you

thank you