An important question from arithmetic progression chapter in which we have been asked to find the terms, if the sum of three consecutive terms of an AP is 21 and the sum of squares of these terms is 165.

RS Aggarwal, Class 10, chapter 5B, question no 9

Let the numbers be:

aâˆ’d,a,a+d

aâˆ’d+a+a+d=21

3a=21

âˆ´a=7

(aâˆ’d)^2+a^2+(a+d)^2=165

a^2+d^2âˆ’2ad+a^2+a^2+d^2+2ad=165

3a^2+2d2^=165

2d^2=165âˆ’3(7)^2=18

d^2=9

âˆ´d=3

âˆ´Â the numbers are

aâˆ’d,a,a+d=7âˆ’3,7,7+3=4,7,10