Let a be first term and d be common difference of this A.P.
Given S3n=3S2n
⇒23n[2a+(3n−1)d]=3×22n[2a+(2n−1)d]
⇒2a+(3n−1)d=4a+(4n−2)d
⇒2a+(n−1)d=0
Now S2nS4n=22n[2a+(2n−1)d]24n[2a+(4n−1)d]=[2a+(n−1)d+nd]2[2a+(n−1)d+3nd]
=nd6nd=6
JEE Main 2021 — Mathematics Algebra
Let Sn be the sum of the first n terms of an arithmetic progression. If S3n=3S2n, then the value of S2nS4n is :
Held on 25 Jul 2021 · Verified 6 Jul 2026.
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