Question

Sum of the three consecutive term of an AP is 12 and sum of their square is 66.Find the number

Answer 1

Answer:

let the three consecutive nos. be a-d,  a  , a+d

ATQ,    a-d+a+a+d=12

3a =12

a=4

(a-d)^2 + a^2 +(a-d)^2=66

after solving and substituting ,

d=√9 =+ or - 3

therefore the nos are 7,4,1

Answer 2

Answer:

Step-by-step explanation:

Let three consecutive no.be (a-d),(a),(a+d).

A=4 by eq given 1.

Asq+dsq+asq+asq+dsq-2ad+2ad=66

3asq+2dsq=66

Put value of a in asq,

66-48=2dsq

9=dsq

+3=d

-

A.p is 4,7,10 or

4,1,-2....