In an A.P. the sum of three consecutive terms is 27 and their product is 504. Find the terms. ( Consider the terms to be in ascending order. )

Answer

The three consecutive terms are 4, 9, 14, or 14 , 9, 4.

Explanation

let The three consecutive terms in an A.P. be `a-d`, `a` and `a+d`.
From the first condition,
`(a-d)+a(a+d)=27`
`thereforea-d+a+a=d=27" " therefore3a=27`
`thereforea=9`
From the second condition,
`(a-d)xxaxx(a+d)=504`
`therefore(9-d)xx9xx(9+d)=504` ...(Substituting `a = 90`)
`therefore(9-d)(9+d)=(504)/(9)=56`
`therefore81-d^(2)=56" " therefore 81-56=d^(2)`
`therefored^(2)=25thereforepm5`
The three consecutive terms are `a-d=9-5=4`,
`a=9,a+d=9+5=14 or`
`a-d=9-(-5)=9+5=14. a=9`
`a+d=9+(-5)=9-5=4`