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:
a−d,a,a+d
a−d+a+a+d=21
3a=21
∴a=7
(a−d)^2+a^2+(a+d)^2=165
a^2+d^2−2ad+a^2+a^2+d^2+2ad=165
3a^2+2d2^=165
2d^2=165−3(7)^2=18
d^2=9
∴d=3
∴ the numbers are
a−d,a,a+d=7−3,7,7+3=4,7,10