Let the required numbers be `(a-d)`, a and `(a+d)`. Then,
`(a-d)+a+(a+d)=15 rArr 3a=15 rArr a=5`.
So, the numbers are `(5-d), 5` and `(5+d)`.
Adding 1, 3, 9 respectively to these numbers, we the numbers
`(6-d), 8` and `(14+d)`
These numbers are in GP.
`:. 8^(2)=(6-d) (14+d) rArr 84-8d-d^(2)=64`
`rArr d^(2)+8d-20=0`
`rArr d^(2)+10 d-2d-20=0`
`rArr d(d+10)-2(d+10)=0`
`rArr (d+10)(d-2)=0`
`rArr d=-10 or d=2`.
So, the required numbers are (3, 5, 7) or (15, 5, -5).
