If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.

Solution:

Given:

nth term of the A.P. 9, 7, 5... is the same as the nth term of the A.P. 15, 12, 9...

Considering $9,7,5 \ldots$

$a=9, d=(7-9)=-2$

$\mathrm{n}^{\text {th }}$ term $=9+(n-1)(-2) \quad\left[a_{n}=a+(n-1) d\right]$

$=9-2 n+2$

$=11-2 n \quad \ldots(\mathrm{i})$

Considering $15,12,9, \ldots$

$a=15, d=(12-15)=-3$

$n^{\text {th }}$ term $=15+(n-1)(-3) \quad\left[a_{n}=a+(n-1) d\right]$

$=15-3 n+3$

$=18-3 n \ldots$ (ii)

Equating (i) and (ii), we get:

$11-2 n=18-3 n$

$\Rightarrow n=7$

Thus, 7th terms of both the A.P.s are the same.