The sum of the first n terms of an AP

Solution:

Given that, first term of the first AP (a) = 8

and common difference of the first AP (d) = 20

Let the number of terms in first AP be n.

$\because$ Sum of first $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore$ $S_{n}=\frac{n}{2}[2 \times 8+(n-1) 20]$

$\Rightarrow$ $S_{n}=\frac{n}{2}(16+20 n-20)$

$\Rightarrow$ $S_{n}=\frac{n}{2}(20 n-4)$

$\therefore$ $S_{n}=n(10 n-2)$ ...(i)

Now, first term of the second $\mathrm{AP}\left(a^{\prime}\right)=-30$ and common difference of the second $\operatorname{AP}\left(d^{\prime}\right)=8$

$\therefore$ Sum of first $2 n$ terms of second AP, $S_{2 n}=\frac{2 n}{2}\left[2 a^{\prime}+(2 n-1) d^{\prime}\right]$

$\Rightarrow \quad S_{2 n}=n[2(-30)+(2 n-1)(8)]$

$\left.\Rightarrow \quad S_{2 n}=n[-60+16 n-8)\right]$

$\Rightarrow \quad S_{2 n}=n[16 n-68]$....(ii)

Now, by given condition,

 

Sum of first $n$ terms of the first $A P=$ Sum of first $2 n$ terms of the second $A P$

$\Rightarrow \quad S_{n}=S_{2 n} \quad$ [from Eqs. (i) and (ii)]

$\Rightarrow \quad n(10 n-2)=n(16 n-68)$

$\Rightarrow \quad n[(16 n-68)-(10 n-2)]=0$

$\Rightarrow \quad n(16 n-68-10 n+2)=0$

$\Rightarrow \quad n(6 n-66)=0$

$\therefore \quad n=11 \quad[\because n \neq 0]$

Hence, the required value of n is 11.