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.
