Question

Question 33
The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is –30 and the common difference is 8. Find n.

Answer 1

Given that, first term of the first AP (a) = 8
Common difference of the first AP (d) = 20
Let the number of terms in first AP be n.
Sum of first n terms of an AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\therefore S_n=\frac{n}{2}[2\times 8+(n-1\left)20\right]$
$\Rightarrow S_n=\frac{n}{2}(16+20n-20)$
$\Rightarrow S_n=n(10n-2)$
Now, first term of the second AP (a’) = - 30
Common difference of the second AP (d’) = 8
$\therefore$ Sum of first 2n terms of second AP, $S_{2n}=\frac{2n}{2}[2a^{\prime }+(2n-1\left)d^{\prime }\right]$
$\Rightarrow S_{2n}=n\left[2\right(-30)+(2n-1\left)\right(8\left)\right]$
$\Rightarrow S_{2n}=n[-60+16n-8]$
$\Rightarrow S_{2n}=n[16n-68]$
Now, by given condition,
Sum of first n terms of the first AP = Sum of first 2n terms of the second AP
$\Rightarrow S_n=S_{2n}[fromeqs.(i\left)and\right(ii\left)\right]$
$\Rightarrow n(10n-2)=n(16n-68)$
$\Rightarrow n\left[\right(16n-68)-(10n-2\left)\right]=0$
$\Rightarrow n(16n-68-10n+2)=0$
$\Rightarrow n(6n-66)=0$
$\therefore n=11[\because n\ne 0]$
Hence, the required value of n is 11.