Q13 of 110 Page 41

The sum of three consecutive terms in an A.P. is 6 and their product is –120. Find the three numbers.

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 = 6


⇒ 3a = 6


⇒ a = 6/3 = 2


Also given (a – d) (a) (a + d) = -120


We know that (a – b) (a + b) = a2 – b2


⇒ (a2 – d2) (2) = -120


⇒ (22 – d2) = -120/2


⇒ 4 – d2 = -60


⇒ d2 = 60 + 4 = 64


⇒ d = 8


So, the numbers are


⇒ a – d = 2 – 8 = -6


⇒ a = 2


⇒ a + d = 2 + 8 = 10


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


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