Question

find 4 number in AP such that their sum is 20 and sum of their square is 120

Answer 1

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Answer 2

Let the terms:- a-3d, a-d, a+d, a+3d

Sum of the terms :-(a-3d)+(a-d)+(a+d)+(a+3d)=20
4a=20
a=5
Sum of the square
(a-3d)2+(a-d)2+(a+d)2+(a+3d)2=20
a2-6d+9d2+a2-2ad+d2+a2+2ad+d2+a2+6ad+9d2=120
4a2+20d2=120
Substitute a=5 into a
4(52)+20d2=120
100+20d2=120
d=1
Thus the four number are
(a-3d)(a-d)(a+d)(a+3d)
(5-3)(5-1)(5+1)(5+3)
=2,4,6,8 is answer