# 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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