The sum of three consecutive terms of an AP is 21, and the sum of the squares of these terms is 165. Find these terms

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)272(7 + d)2= 165


49 + d214d + 49 + d2 + 14d = 116


2d2 = 18 d2 = 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.


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