For the infinite G.P.
S∞=1−ra=5
Also a+ar+ar2+ar3+ar4=5
⇒a(1−r1−r5)=2598
i.e. 1−r5=12598⇒r5=5333⇒r=(53)53
Now S21=221[20ar+(21−1)10ar2]
=21[10ar+100ar2]
=21⋅a11
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5. Let the sum of its first five terms be 2598. Then the sum of the first 21 terms of an AP, whose first term is 10ar,nth term is an and the common difference is 10ar2, is equal to
Held on 27 Jul 2022 · Verified 6 Jul 2026.
21a11
22a11
15a16
14a16
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