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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.
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