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Question

The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165: Find these terms.

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Solution

Let the numbers be:

ad,a,a+d

ad+a+a+d=21

3a=21

a=7

(ad)2+a2+(a+d)2=165

a2+d22ad+a2+a2+d2+2ad=165

3a2+2d2=165

2d2=1653(7)2=18

d2=9

d=3

the numbers are

ad,a,a+d=73,7,7+3=4,7,10

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