Question

see the attached file and answer it for me ​

Answer 1

MATE

YOUR SOLUTION IS IN ATTACHED PICTURE ☺️

Answer 2

Let the first term of GP be a and common ratio be r.

Now Given,

$T_{m+n}=p\\\;\\ar^{m+n-1}=p\;\;\;......i)$

And,

$T_{m-n}=q\\\;\\ar^{m-n-1}=q\;\;\;...........ii)$

Dividing Eq. i) by ii)

$\frac{ar^{m+n-1}}{ar^{m-n-1}}=\frac{p}{q}\\\;\\\frac{r^{m+n-1}}{r^{m-n-1}}=\frac{p}{q}\\\;\\r^{m+n-1-(m-n-1)}=\frac{p}{q}\\\;\\r^{m+n-1-m+n+1}=\frac{p}{q}\\\;\\r^{2n}=\frac{p}{q}\\\;\\r=(\frac{p}{q})^\frac{1}{2n}$

Putting this value in eq. i)

$ar^{m+n-1}=p\\\;\\a(\frac{p}{q})^\frac{m+n-1}{2n}\\\;\\a.(\frac{1}{q})^{\frac{m+n-1}{2n}}=\frac{p}{p^\frac{m+n-1}{2n}}\\\;\\a.(\frac{1}{q})^\frac{m+n-1}{2n}=p^{1-(\frac{m+n-1}{2n})}\\\;\\a.(\frac{1}{q})^\frac{m+n-1}{2n}=p^{\frac{2n-m-n+1}{2n}}\\\;\\a.(\frac{1}{q})^\frac{m+n-1}{2n}=p^\frac{n-m+1}{2n}$ $a.(\frac{1}{q})^\frac{m+n-1}{2n}=p^\frac{n-m+1}{2n}\\\;\\a=q^\frac{m+n-1}{2n}.p^\frac{n-m+1}{2n}$

Now,

$T_m=ar^{m-1}\\\;\\T_m=q^\frac{m+n-1}{2n}.p^\frac{n-m+1}{2n}.(\frac{p}{q})^\frac{m-1}{2n}\\\;\\T_m=q^\frac{m+n-1-m+1}{2n}\times p^\frac{n-m+1+m-1}{2n}\\\;\\T_m=q^\frac{n}{2n}\times p^\frac{n}{2n}\\\;\\T_m=q^\frac{1}{2}.\times p^\frac{1}{2}\\\;\\T_n=\sqrt{p}{q}$

And,

$T_n=ar^{n-1}\\\;\\T_n=q^\frac{m+n-1}{2n}.p^\frac{n-m+1}{2n}\times(\frac{p}{q})^\frac{n-1}{2n}\\\;\\T_n=q^\frac{m+n-1-n+1}{2n}\times p^\frac{n-m+1+n-1}{2n}\\\;\\T_n=q^\frac{m}{2n}.p^\frac{2n-m}{2n}$

$T_n=q^\frac{m}{2n}.p^{1+\frac{2n-m}{2n}-1}}\\\;\\T_n=p.q^\frac{m}{2n}\times p^\frac{2n-m-2n}{2n}\\\;\\T_n=p.q^\frac{m}{2n}\times p^{\frac{-m}{2n}}\\\;\\T_n=p.(\frac{p}{q})^\frac{m}{2n}$