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Question

The sum of four consecutive terms of an A.P. is 20 and the sum of their squares is 120. Find those numbers

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Solution

Let the terms be a3d,ad,a+d,a+3d.
Given, Sum =20
a3d+ad+a+d+a+3d=20
4a=20
a=5
Also given, sum of squares =120
(a3d)2+(ad)2+(a+d)2+(a+3d)2=120
(53d)2+(5d)2+(5+d)2+(5+3d)2=120
2530d+9d2+2510d+d2+25+10d+d2
25+30d+9d2=120
100+20d2=120
20d2=20
d2=1
d=±1
The terms are a3d,ad,a+d,a+3d
Thus, 53,51,5+1,5+3
(i.e.) 2,4,6,8 or 8,6,4,2

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