Q. 135.0( 6 Votes )
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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