S2n=22n[2a+(2n−1)d],S4n=24n[2a+(4n−1)d]
⇒S2−S1=24n[2a+(4n−1)d]−22n[2a+(2n−1)d]
=4an+(4n−1)2nd−2na−(2n−1)dn
=2na+nd[8n−2−2n+1]
⇒2na+nd[6n−1]=1000
2a+(6n−1)d=n1000
Now, S6n=26n[2a+(6n−1)d]
=3n⋅n1000=3000
JEE Main 2021 — Mathematics Algebra
Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2−S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to:
Held on 18 Mar 2021 · Verified 6 Jul 2026.
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