Question

The some of three consecutive numbers in an



a.P is 21and their products is 231 find the numbes

Answer 1

Let a be the first term and common difference be d.

A.P. will be a, a+d, a+2d......

Then , a(a+d)(a+2d)=231 .........(1)

a+a+d+a+2d=21

3a+3d=21

a+d=7 .......... (2)

By eq. (1) a=7-d

Putting this value in eq. (1),

(7-d)(7-d+d)(7-d+2d)=231

(49-7d)(7+d)=231

343-49d+49d-7y^2=231

7d^2=343-231

d^2=112/7

d=√16=4

Put d=4 in eq.(2)

a=7-4=3.

A.P. 3, 3+4, 3+2×4

A.P. 3,7, 11 are the numbers.