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Question Number 123869 by benjo_mathlover last updated on 28/Nov/20

Answered by liberty last updated on 29/Nov/20

let :a ,ar,ar^2 ,ar^3 the four consecutive terms in GP given the condition Σ_(i=1) ^4 u_i = a+ar+ar^2 +ar^3 =80 a(1+r+r^2 (1+r))=80 ; a(1+r)(1+r^2 )=80 and ((ar+ar^3 )/2)=30 ; ar(1+r^2 )= 60 ⇒ ((a(1+r)(1+r^2 ))/(ar(1+r^2 ))) = ((80)/(60))=(4/3) , we get 3+3r=4r r = 3 then a = ((60)/(3×10)) = 2. thus the terms is 2, 6, 18, 54.

let:a,ar,ar2,ar3thefourconsecutivetermsinGPgiventhecondition∑4i=1ui=a+ar+ar2+ar3=80a(1+r+r2(1+r))=80;a(1+r)(1+r2)=80andar+ar32=30;ar(1+r2)=60⇒a(1+r)(1+r2)ar(1+r2)=8060=43,weget3+3r=4rr=3thena=603×10=2.thusthetermsis2,6,18,54.

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