Question

The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. find these terms.

Answer 1

Let the treee consecutive terms in A.P.be a-d, a and a+d.

`∴ (a-d)+a+(a+d)=21`

`⇒ a=7`           ............(1) 

Also , `(a-d)^2+a^2+(a+d)^2=165`

`⇒a^2+d^2-2ad+a^2+a^2+d^2+2ad=165`

`⇒ 3a^2+2d^2=165`

`⇒ 3xx(7)^2+2d^2=165`.............[from (1)]

`⇒ 3xx49+2d^2=165` 

`⇒ 147+2d^2=165` 

`⇒ 2d^2=18`

`⇒d^2=9`

`⇒ d=+-3` 

when `a=7 and d=3`

Required terms =`a-d, a and a+d`

                       `=7-3,7,7+3`

                       `=4,7,10` 

When `a=7 and d=-3`

Required terms=` a-d, a and a+d`

                       ` = 7-(-3),7,7+(-3)`

                       `=10,7,4`