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Question

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.

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Solution

Let the first term of a G.P be a and its common ratio be r.
a1+a2+a3=56a+ar+ar2 = 56 a 1+r+r2 = 56 a=561+r+r2 .......i Now, according to the question:a-1, ar-7 and ar2-21 are in A.P. 2ar-7 = a-1 + ar2 - 212ar - 14 = ar2+a-22ar2-2ar+a-8 = 0a1-r2 = 8a = 81-r2 .......iiEquating (i) and (ii):81-r2=561+r+r281+r+r2 = 561+r2-2r 1+r+r2 =7 1+r2-2r1+r+r2 =7+7r2-14r6r2-15r+6=0 32r2-5r+2 = 02r2-4r-r+2=02r(r-2)-1(r-2)=0(r-2)(2r-1)=0r=2, 12When r=2, a=8. [Using (ii)]And, the required numbers are 8, 16 and 32.When r=12, a=32. [Using (ii)]And, the required numbers are 32, 16 and 8.

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