Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
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4
Let the numbers be (a−3d),(a−d),(a+d),(a+3d).
Then, Sum of numbers =20
⟹(a−3d)+(a−d)+(a+d)+(a+3d)=20⟹4a=20⟹a=5
It is given that, sum of the squares =120
⟹(a−3d)
2
+(a−d)
2
+(a+d)
2
+(a+3d)
2
=120
⟹4a
2
+20d
2
=120
⟹a
2
+5d
2
=30
⟹25+5d
2
=30
⟹5d
2
=5⟹d=±1
If d=1, the, the numbers are 2,4,6,8.
If d=−1, then the numbers are 8,6,4,2.
Thus, the numbers are 2,4,6,8 or 8,6,4,2.
Hope it helps you
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