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 arithmetic progression. Find the numbers.

Answer 1

It is given that the sum of three number in G.P. is 56.

Since the numbers are in G.P., so the numbers will be a, ar and a r 2 .

According to the question,

a+ar+a r 2 =56 a( 1+r+ r 2 )=56 (1)

It is given that a−1, ar−7 and a r 2 −21 form an A.P., so,

( ar−7 )−( a−1 )=( a r 2 −21 )−( ar−7 ) ar−a−6=a r 2 −ar−14 a r 2 −ar−ar+a=8 a( r 2 +1−2r )=8 (2)

From equation (1) and (2),

7( r 2 −2r+1 )=( 1+r+ r 2 ) 7 r 2 −14r+7−1−r− r 2 =0 6 r 2 −15r+6=0 6 r 2 −12r−3r+6=0

Simplify further,

6r( r−2 )−3( r−2 )=0 ( 6r−3 )( r−2 )=0 r=2, 1 2

When r=2, then equation (1) becomes,

a( 1+2+ 2 2 )=56 a( 1+2+4 )=56 a= 56 7 a=8

So, the numbers are 8,16 and 32.

When r= 1 2 , then equation (1) becomes,

a( 1+ 1 2 + 1 2 2 )=56 a( 1+ 1 2 + 1 4 )=56 a= 56×4 7 a=32

So, the number are 32,16 and 8.

Therefore, the three numbers are 8, 16 and 32.