**Question:**

If the nth terms of the two AP’s 9,7,5,……….. and 24,21,18,…. are the same, then find the value of n.Also,that term.

**Solution:**

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.