To Find: The three numbers which are in AP.
Given: Sum and sum of the squares of three numbers are 21 and 165 respectively.
Let required number be (a – d), (a), (a + d). Then,
(a – d) + a + (a + d) = 21 ⇒ 3a = 21 ⇒ a = 7
Thus, the numbers are (7 – d), 7 and (7 + d).
But their sum of the squares of three numbers is 165.
(7 – d)2 + 7

2 + (7 + d)2= 165
⇒ 49 + d2 –14d + 49 + d2 + 14d = 116
⇒ 2d2 = 18 ⇒ d

2 = 9 ⇒ d = ± 3
When d=3 numbers are 4, 7, 10
When d= (– 3) numbers are 10, 7, 4
So, Numbers are 4, 7, 10 or 10, 7, 4.