The mth term of a Geometrical Progression is n and nth term is m. Find (m+n)th term The mth term of a Geometrical Progression is n and nth term is m. Find (m+n)th term.
I've tried this:
Tm = arm-1 = n (Eq 1)
Tn = arn-1 = m (Eq 2)
Subracting 2 from 1
rm - r - rn + r = n-m
rm - rn = n-m
rm + m  = rn + n
I don't know how to proceed. I don't even know if I have done this correctly until this point.
 A: HINT:
Eliminating $r$ $$\left(\frac na\right)^{n-1}=\left(\frac ma\right)^{m-1}$$
$$a^{m-n}=\frac{m^{m-1}}{n^{n-1}}\implies  a=\left(\frac{m^{m-1}}{n^{n-1}}\right)^{\frac1{m-n}}$$
Divide the given relations to find $r=\left(\dfrac nm\right)^{\frac1{m-n}}$ 
A: Solution. Let the first term be $a$ and the common ratio be $r$. Then, the $m$th term is $ar^{m-1}$ and the $n$th term is $ar^{n-1}$, so we get $$ \begin {eqnarray*} ar^{m-1} &=& n, \\ ar^{n-1} &=& m. \end {eqnarray*} $$We seek the $(m+n)$th term, which is $ar^{m+n-1}$. Dividing the two equations, we get $$ r^{m-n} = \frac{n}{m} \implies r^{m+n} = \frac {n}{m} \cdot r^{2n} \implies r^{m+n-1} = \frac {n \, r^{2n-1}}{m}. $$Now, multiply by $a$ to get $\frac{an}{m} \cdot r^{2n-1} $. 
A: A "symmetrical" approach:
For a GP, the ratio of two terms is the common ratio raised to the difference in the number of terms. Hence, 
$$\begin{align}r=\left(\frac{T_{m+n}}{T_n}\right)^{\frac 1m}&=\left(\frac {T_m}{T_n}\right)^\frac 1{m-n}\\
T_{m+n}&=T_n\left(\frac {T_m}{T_n}\right)^\frac m{m-n}\\
&=m\left(\frac nm\right)^{\frac m{n-m}}\qquad \blacksquare\end{align}$$
An interesting question, nevertheless.
