Question

Find four nos in AP whose sum is 20 and sum of squares is 120.

Answer 1

We can take the no. As (a-2d),(a-d),(a+d),(a+2d)

As their sum is 20

(a-3d)+(a-d)+(a+d)+(a+3d)=20

4a=20

a=5

Then its mentioned that the sum of squares is 120

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

after squaring

a^2+9d^2–6ad+a^2+d^2–2ad+a^2+d^2+2ad+a^2+9d^2+6ad=120

2a^2+10d^2–8ad+2a^2+10^2+8ad=120

2a^2+10d^2+2a^2+10d^2=120

2(2a^2+10d^2)=120

As a=5

2(2*5*5+10d^2)=120

2(50+10d^2)=120

50+10d^2=120/2

10d^2=60–50

10d^2=10

d^2=10/10

d^2=1

d=1

So by puttimg the values of a and d

In (a-3d),(a-d),(a+d),(a+3d)

(5–3),(5–1),(5+1),(5+3)

=2,4,6,8

So the four numbers are 2,4,6,8