Find G.P. such that

3r=1tr=42

3r=1(tr)2=1092

Using summation formula of G.P. having 1st term as a and common ratio as r we have,

a(r31)(r1)=42 ……..(A)

Squaring the terms of the G.P. yields another G.P. having 1st term as a2 and common ratio as r2. Thus,

a2(r61)(r21)=1092

a2(r31)(r3+1)(r1)(r+1)=1092

Using (A), 42×a(r3+1)(r+1)=1092

a(r2r+1)=26

Given, a(r2+r+1)=42

r2+r+1r2r+1=4226=2113

8r234r+8=0 or 4r217r+4=0

r=4,1/4

Using, a=26r2r+1;a=2,32

The 3 terms of G.P. are either 2, 8, 32 or 32, 8, 2

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