The sum of first n terms of an A.P. whose first terms 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 the common difference is 8. Find n.

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Solution

For $1^{s t}$ AP
$a = 8, d = 20$
For $2^{n d}$ AP
$a^{'} = - 30, d^{'} = 8$
According to the question, $S_{n} = S_{2 n}^{'}$
$\Rightarrow \frac{n}{2} [2 a + (n - 1) d] = \frac{2 n}{2} [2 a^{'} + (2 n - 1) d^{'}]$
$\Rightarrow [2 (8) + (n - 1) 20] = 2 [2 (- 30) + (2 n - 1) 8]$
$\Rightarrow 2 \times 8 + n \times 20 - 20 = 2 [- 60 + 16 n - 8]$
$\Rightarrow 20 n - 4 = - 136 + 32 n$
$\Rightarrow - 32 n + 20 n = - 136 + 4$
$\Rightarrow n = 11$
Hence, the required value of n is 11.

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The sum of first n terms of an A.P. whose first terms 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 the common difference is 8. Find n.

For 1st APa=8,d=20For 2nd APa′=−30,d′=8According to the question, Sn=S′2n⇒n2[2a+(n−1)d]=2n2[2a′+(2n−1)d′]⇒[2(8)+(n−1)20]=2[2(−30)+(2n−1)8]⇒2×8+n×20−20=2[−60+16n−8]⇒20n−4=−136+32n⇒−32n+20n=−136+4⇒n=11Hence, the required value of n is 11.