Question
Find the number of terms of the sequence 32, 24, 16, 8 ... for which the sum of the terms is zero.
Answer: Option D
:
D
Sn=(n2)[2a+(n−1)d]
⇒0=(n2)[64+(n−1)(−8)]
⇒0=(n2)[64−(n−1)8]
⇒n (36– 4n) = 0
n = 0; n=9
As, n can not be equal to 0, so, the number of terms required is 9.
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:
D
Sn=(n2)[2a+(n−1)d]
⇒0=(n2)[64+(n−1)(−8)]
⇒0=(n2)[64−(n−1)8]
⇒n (36– 4n) = 0
n = 0; n=9
As, n can not be equal to 0, so, the number of terms required is 9.
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