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Question

Find three consecutive terms in an Arithmetic progression whose sum is 18 and sum of their squares is 140.

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Solution

Solution:
Let the terms be ad,a,a+d
Given:
ad+a+a+d=18
or, 3a=18
or, a=6
And
(ad)2+a2+(a+d)2=140
or, (6d)2+62+(6+d)2=140
or, 3612d+d2+36+36+12d+d2=140
or, 108+2d2=140
or, 2d2=140108
or, 2d2=32
or, d2=16
or, d=±4
If d=4, then the terms of progression are
64,6,6+4
2,6,10
If d=4, then the terms of progression are
6(4),6,64
10,6,2

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