Q. 145.0( 5 Votes )

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

Answer :

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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