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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.

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3 Answers 3

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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 $r=\left(\dfrac nm\right)^{\frac1{m-n}}$

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  • $\begingroup$ You keep beating me! $\endgroup$ Aug 24, 2014 at 15:51
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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} $.

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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.

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