Find the four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Explanation

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`
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