Question

The sum of three numbers m GP is 56. If we subtract 1.7,21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Answer 1

Let the required numbers be a, ar and `ar^(2)`. Then,
`a+ar+ar^(2)=56` ...(i)
Also, `(a-1), (ar-7)` and `(ar^(2)-21)` are in AP.
`:. 2(ar-7)=(a-1)+(ar^(2)-21)rArr a+ar^(2)=2ar+8`. ...(ii)
Using (ii) in (i), we get
`ar+(2ar+8)=56 rArr 3ar=48 rArr ar=16 rArr r=16/a`.
Putting `r=16/a` in (i), we get
`a+16+256/a=56 rArr a^(2)+16a+256=56a`
`rArr a^(2)-40a+256=0`
`rArr a^(2)-8a-32a+256=0`
`rArr a(a-8)-32(a-8)=0`
`rArr (a-8) (a-32)=0`
`rArr a=8 or a=32`.
Now, `a=8 rArr r=16/8 rArr r=2`.
And, `a=32 rArr r=16/32 rArr r=1/2`.
Hence, the required numbers are (8, 16, 32) or (32, 16, 8).