If the nth terms of the two AP’s 9,7,5,

Let the first term, common difference and number of terms of the AP 9, 7, 5,…are a1,d1 and n1 respectively.

i.e., first term (a1) = 9and common difference (d1)= 7 – 9 = – 2.

$\therefore$ Its nth term, $\quad T_{n_{1}}^{\prime}=a_{1}+\left(n_{1}-1\right) d_{1}$

$\Rightarrow \quad T_{n_{1}}^{\prime}=9+\left(n_{1}-1\right)(-2)$

$\Rightarrow \quad T_{n_{1}}^{\prime}=9-2 n_{1}+2$

$\Rightarrow \quad T_{n_{1}}^{\prime}=11-2 n_{1} \quad\left[\because n\right.$th term of an AP, $\left.I_{n}=a+(n-1) d\right] \ldots$ (I)

Let the first term, common difference and the number of terms of the AP $24,21,18, \ldots$ are $a_{2}, d_{2}$ and $n_{2}$, respectively.

i.e., first term, $\left(a_{2}\right)=24$ and common difference $\left(d_{2}\right)=21-24=-3$.

$\therefore$ Its $n$th term, $\quad T_{n_{2}}^{\prime \prime}=a_{2}+\left(n_{2}-1\right) d_{2}$

$\Rightarrow \quad T_{n_{2}}^{n \prime}=24+\left(n_{2}-1\right)(-3)$

$\Rightarrow \quad T_{n_{2}}^{n^{\prime \prime}}=24-3 n_{2}+3$

$\Rightarrow \quad T_{n_{2}^{\prime \prime}}^{\prime 2}=27-3 n_{2}$ ...(ii)

Now, by given condition,

nth terms of the both APs are same, i.e., $T_{n_{1}}^{\prime}=T_{n_{2}}^{\prime \prime}$

$11-2 n_{1}=27-3 n_{2} \quad$ [from Eqs. (i) and (ii) $]$

$\Rightarrow \quad n=16$

$\therefore$ nth term of first AP, $T^{\prime} n_{1}=11-2 n_{1}=11-2(16)$

$=11-32=-21$

and $n$th term of second $A P, T^{\prime \prime} n_{2}=27-3 n_{2}=27-3(16)$

$=27-48=-21$

Hence, the value of n is 16 and that term i.e., nth term is -21.