Question

The three numbers are in A.P and their sum is 21. If the first and second are decrease by 1 each and third is increased by 7, they form a G.P Find the numbers of A.P.

Answer 1

Correct Answer - `(12,7,2) or (3,7,11)`
Let the required numbers be `(a-d), a, (a+d)`. Then, `3a=21 rArr a=7`.
So, these numbers are `(7-d), 7, (7+d)`.
`:. (7-d), 6` and `(8+d)` are in GP.
So, `6/((7-d))=((8+d))/6 rArr (7-d) (8+d)=36 rArr d^(2)+d-20=0`
`rArr (d+5) (d-4)=0 rArr d= -5 or d=4`.
Hence, the required numbers are `(12, 7, 2) or (3, 7, 11)`.