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 arithmetic progression . Find the numbers
Let the three numbers in G.P. be a,ar and ar2
From the given condition
a+ar+ar2=56⇒a(1+r+r2)=56...(1)
Also given a−1,ar−7,ar2−21 forms an A.P.
∴(ar−7)−(a−1)=(ar2−21)−(ar−7)⇒ar−a−6=ar2−ar−14⇒ar2−2ar+a=8⇒a(r2+1−2r)=8⇒a(r−1)2=8...(2)
Using (1) and (2)
⇒7(r2−2r+1)=1+r+r2⇒7r2−14r+7−1−r−r2=0⇒6r2−15r+6=0⇒6r2−12r−3r+6=0⇒6r(r−2)−3(r−2)=0⇒(6r−3)(r−2)=0⇒r=2,12
When r=2, a=8
And three numbers in GP are 8,16 32.
When r=12 , three numbers in G.P. are 32,16 and 8.
Thus in either case, the three required numbers are 8,16 and 32.