Find the four terms in an A.P. whose sum is 20 and the sum of whose sqare is 120

let the four terms be a-3d,a-d,a+d,a+3d

ATQ 

a-3d+a-d+a+d+a+3d = 20

4a = 20

a = 5...(i)

also (a-3d)² + (a-d)² + (a+d)² + (a+3d)² = 120

4a² + 20d² = 120

a² + 5d² = 30

25 + 5d² = 30

d² = 1 

therefore d =  1, - 1

for a = 5 and d = 1

a - 3d = 5 - 3 = 2

a - d = 5 -1 = 4

a + d = 5 + 1 = 6

a + 3d = 5+3 = 8 

so AP: 2,4,6,8..........

for a = 5 and d = -1

a - 3d = 5+3 = 8

a - d = 5+1 = 6

a + d = 5 - 1 = 4

a + 3d = 5 - 3 = 2

so AP: 8,6,4,2........