Q14 of 110 Page 41

Find the three consecutive terms in an A. P. whose sum is 18 and the sum of their squares is 140.

We know that the three consecutive terms of an A.P. may be taken as a – d, a, a + d.


Given, a – d + a + a + d = 18


⇒ 3a = 18


⇒ a = 18/3 = 6


Also given (a – d)2 + a2 + (a + d)2 = 140


We know that (a – b)2 = a2 – 2ab + b2


And (a + b)2 = a2 + 2ab + b2


⇒ a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 140


⇒ 3a2 + 2d2 = 140


⇒ 3(6)2 + 2d2 = 140


⇒ 108 + 2d2 = 140


⇒ 2d2 = 140 – 108 = 32


⇒ d2 = 32/2 = 16


⇒ d = 4


So, the numbers are


⇒ a – d = 6 – 4 = 2


⇒ a = 6


⇒ a + d = 6 + 4 = 10


∴ The three consecutive numbers are 2, 6, 10.


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