Question

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

Answer 1

Sum of first n terms Sn=n/2[2a+(n-1)*d]
n/2[2×8+(n-1)×20]=2n/2[2×-30+(2n-1)×8]
16+20n-20=2(-60+16n-8)
-4+20n=2(-68+16n)
-2+10n=-68+16n
66=6n
n=11

Answer 2

Answer:

The value of n is 11.

Step-by-step explanation:

The sum of first n terms of an AP is

$S_n=\frac{n}{2}[2a+(n-1)d]$

Where, a is first term and d is common difference.

It is given that the first term is 8 and the common difference is 20. The sum of n terms is

$S_n=\frac{n}{2}[2(8)+(n-1)20]$

$S_n=\frac{n}{2}[16+20n-20]$

$S_n=\frac{n}{2}[20n-4]$

$S_n=n(10n-2)$

The first term of another AP is -30 and common difference is 8. The sum of 2n terms is

$S_2n=\frac{2n}{2}[2(-30)+(2n-1)8]$

$S_2n=n(-60+16n-8)$

$S_2n=n(16n-68)$

It is given that Sum of n terms of an AP is equal to the sum of 2n terms of another AP.

$n(10n-2)=n(16n-68)$

$10n-2=16n-68$

$-2+68=16n-10n$

$66=6n$

$11=n$

Therefore the value of n is 11.