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Sum
In an AP, if Sn = 3n2 + 5n and ak = 164, find the value of k.
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Solution
∵ nth term of an AP
an = Sn – Sn – 1
= 3n2 + 5n – 3(n – 1)2 – 5(w – 1)
∵ Sn = 3n2 + 5n ......(Given)
= 3n2 + 511 – 3n2 – 3 + 6n – 5n + 5
⇒ an = 6n + 2 ........(i)
or ak = 6k + 2 = 164 .......[∵ ak = 164 (Given)]
⇒ 6k = 164 – 2 = 162
∴ k = 27
Concept: Sum of First n Terms of an A.P.
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