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 $.

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.