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Question

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than by  4th by a18.


Solution

Let the four numbers in G.P. be  a,ar,ar2,ar3

Thus, ar2=a+9 and ar=ar3+18

ar2a=9 and arar3=18

Now~ ar2a=9a2r1=9....(1)

and~ arar3=18ar(1r2)18

ar(r21)=18      ....(2)

Dividing (2) by (1), we have :

ar(r2)1a(r2)1=189r=2

Now, substituting r in (1), we have :

a(41)=9a=3

ar=3×(2)=6

ar2=3×(2)2=12

ar3=3×(2)3=24

Thus, required numbers are 3, -6, 12, -24.

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