Question

Find the number of terms of the sequence 32, 24, 16, 8 ... for which the sum of the terms is zero.

• A. 7
• B. 8
• C. 6
• D. 9

Answer 1

The correct option is D 9

$S_n=\left(\frac{n}{2}\right)[2a+(n-1\left)d\right]$
$\Rightarrow 0=\left(\frac{n}{2}\right)[64+(n-1\left)\right(-8\left)\right]$
$\Rightarrow 0=\left(\frac{n}{2}\right)[64-(n-1\left)8\right]$
$\Rightarrow$ n (36 – 4n) = 0
n = 0; n=9

As, n can not be equal to 0, so, the number of terms required is 9.