2015/052) Find four numbers in A.P. whose sum is 20 and sum of whose squares is 120?

without loss of generality we can choose numbers to be a - 3d, a - d, a + d, a-3d and d > 0

now sum = $(a-3d) + (a-d) + (a+d) + (a+3d) = 4a = 20$ or $a=5$

sum of squares = $(a-3d)^2 + (a- d)^2 + (a+d)^2 + (a+3d)^2 = 4a^2 + 20d^2 = 120$

so putting value of a we get d = 1 so numbers are 2,4,6,8
I found the problem at https://in.answers.yahoo.com/question/index?qid=20150528071206AAbn3Cz