Find four consecutive terms in an A.P. whose sum is 46 and the product of the 1st and the 3rd is 56. The terms are in ascending order.

Explanation

Let four consecutive termsin an A.P. be `a-3d, a-d, a+d` and `a+3d`.
From the first condition.
`(a-3d)+(a-d)+(a+d)+(a+3d)=46`
`:.4a=46:.a=46/4:.a=23/2`………1
From th second condition
`(a-3d)xx(a+d)=56`
`:.(23/2-3d)(23/2+d)=56` .....(Substituting the value of `a`)
`:.((23-6d)/2)((23+2d)/2)=56`
`:.(23-6d)(23+2d)=224`
`:.529-92d-12d^(2)=224`
`:.12d^(2)+92d-529=-224`....(Multiplying both the sides by `-1`)
`:.12d^(2)+92d-529+224=0`
`:.12d^(2)+92d-305=0`
`:.12d^(2)-30d+122d-305=0`
`:.6d(2d-5)+51(2d-5)=0`

`:.(2d-5)(6d+61)=0` brgt `:.2d-5=0` or `6d+61=0`
`:.2d=5` or `6d=-61`
`:.d=5/2` or `d=-61/6`
The terms are in ascending order `:,d=-61/6` is unacceptable.
`:.d=5/2`
`a-3d=23/2-3xx5/2=23/2-15/2=(23-15)/2=8/2=4`
`a-d=23/2-5/2=(23-5)/2=15/2=9`
`a+d=23/2+5/2=(23+5)/2=28/2=14`
`a+3d=3/2+3xx5/2=23/2+15/2=(23+15)/2=38/2=19`
The required terms are 4,9,14, and 19.