# 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 $2 n$ terms of another AP whose first term is $-30$ and the common difference is 8 . Find $n$.

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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 13:27:40