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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:


  1. 144

  2. 66

  3. 36

  4. None


Solution

The correct option is B

66


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

∴ a = 4,  (4d)343(4+d)3 = 408  ⇒ d = ± 3

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

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