Find n and S, if a = 5, d = 3, a_ {n} = 50a_ {11.
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Answer:
As we know nth term, a
n
=a+(n−1)d
& Sum of first n terms, S
n
=
2
n
(2a+(n−1)d), where a & d are the first term amd common difference of an AP.
Given, a=5,d=3,a
n
=50
⇒a+(n−1)d=50
⇒5+(n−1)3=50
⇒5+3n−3=50⇒3n=48⇒n=16
∴S
16
=
2
16
[2a+(16−1)d]=8[2×5+15×3]=440
Hence, n=16,S
16
=440
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