Question

If the sum of three numbers in A.P. is 12 and sum of their cubes is 408, then sum of their squares is:

• A. 144
• B. 66
• C. 36
• D. None

Answer 1

The correct option is B

66

a - d, a , a+4 are the terms

∴ a = 4, ${(4-d)}^3$ + $4^3$ + ${(4+d)}^3$ = 408 ⇒ d = ± 3

∴ sum of their squares = 1 + 16 + 49 = 66