If the nth term of the A.P. 9, 7, 5, ... is

Here, we are given two A.P. sequences whose nth terms are equal. We need to find n. 

So let us first find the nth term for both the A.P.

First A.P. is 9, 7, 5 …

Here,

First term (a) = 9

Common difference of the A.P. $(d)=7-9$

$=-2$

Now, as we know,

$a_{n}=a+(n-1) d$

So, for $n^{\text {th }}$ term,

$a_{n}=9+(n-1)(-2)$

$=9-2 n+2$

$=11-2 n$......(1)

Second A.P. is 15, 12, 9 …

Here,

First term (a) = 15

Common difference of the A.P. $(d)=12-15$

$=-3$

Now, as we know,

$a_{n}=a+(n-1) d$

So, for nth term,

$a_{n}=15+(n-1)(-3)$

$=15-3 n+3$

$=18-3 n$.......(2)

Now, we are given that the nth terms for both the A.P. sequences are equal, we equate (1) and (2),

$11-2 n=18-3 n$

$3 n-2 n=18-11$

$n=7$

Therefore, $n=7$