Let the 1st terms of A.P. be ′a′ and common difference ′d′ .
Given a1+a7+a16=40
⇒a+a+6d+a+15d=40
⇒a+7d=340 ….(1)
Sum of first 15 term
S15=215[2a+14d]
=215×2×340=200
If a1,a2,a3,.... are in A.P. such that a1+a7+a16=40, then the sum of the first 15 terms of this A.P is:
Held on 12 Apr 2019 · Verified 6 Jul 2026.
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