Let a, ar, ar…2….. be terms of G.P.
Given,T2+T3+T4=3⇒ ar (1+r+r2)=3
T6+T7+T8=243⇒
ar5(1+r+r2)=243
by (i) and (ii)
r4=81
⇒r=3
∴a=131
S50=r−1a(r50−1)=26350−1
If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :
Held on 5 Sept 2020 · Verified 6 Jul 2026.
261(349−1)
261(350−1)
132(350−1)
131(350−1)
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