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Question

The sum of three consecutive terms in increasing order of a G.P. is 56. Subtracting 1, 7 and 21 respectively from them, an A.P. is formed. Find those numbers.

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Solution

Let the GP be ar,a,ar
thereforear,a,ar=56a(1r+1+r)=56a(r2+r+1r)=56a=56rr2+r+1
Given: ar1,a7,ar21 is an A.P.
(a7)(ar1)=(ar21)(a7)a7ar+1=ar21a+7a(11r)6=a(r1)14a(11rr+1)=8a(2r1r2r)=856rr2+r+1(r2+2r1r)=8
7r2+r+1(r2+2r1)=17r2+14r7=r2r114r+r=7r2r2+7115r=6r2+66r215r+6=06r23r12r+6=03r(2r1)6(2r1)=0(3r6)(2r1)=0r=63=2 r=12a=56rr2+r+1 a=56rr2+r+1=16 =16
The GP's are
ar,a,ar162,16,16×2=8,16,32 and 1612,16,16×12=32,16,8

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