# The sum of first $n$ terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first $2n$ terms of another A.P. whose first term is $-30$ and common difference is 8. Find $n$.

Given:

The sum of first $n$ terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first $2n$ terms of another A.P. whose first term is $-30$ and common difference is 8.
To do:

We have to find the value of $n$.

Solution:

Let the first A.P. be $A_1$ and the second A.P. be $A_2$.

First term of the first A.P. $a = 8$
Common difference of the first A.P. $d = 20$
Let the number of terms in first A.P. be $n$
Sum of first $n$ terms of an A.P., $S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[2\times8+(n-1)20]$
$=\frac{n}{2}(16+20n-20)$

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

$=n(10 n-2)$......(i)

First term of the second A.P. $\left(a^{\prime}\right)=-30$

Common difference of the second A.P. $\left(d^{\prime}\right)=8$
$\therefore$ Sum of first $2 n$ terms of second A.P.,

$\mathrm{S}_{2 n}=\frac{2 n}{2}\left[2 a^{\prime}+(2 n-1) d^{\prime}\right]$
$=n[2(-30)+(2 n-1) 8]$
$=n[-60+16 n-8]$
$=n[16 n-68]$......(ii)
According to the question,

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

$\Rightarrow \mathrm{S}_{n}=\mathrm{S}_{2 n}$
$\Rightarrow n(10 n-2)=n(16 n-68)$
$\Rightarrow n[(16 n-68)-(10 n-2)]=0$
$\Rightarrow n(16 n-68-10 n+2)=0$
$\Rightarrow n(6 n-66)=0$

$6n=66$ or $n=0$ which is not possible
$\therefore n=11$
Hence, the required value of $n$ is 11.

Updated on: 10-Oct-2022

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