A basic question from arithmetic progression chapter in which we the sum of four numbers is 28 and their sum of squares is 216, we need to calculate those numbers and those numbers must be in the form of AP.
Book – RS Aggarwal, Class 10, chapter 5B, question no 11
Let the numbers be, a−3d,a−d,a+3d,a+d
Given,
a−3d+a−d+a+3d+a+d=28
⇒4a=28,
∴a=7
(a−3d)^2+(a−d)^2+(a+3d)^2+(a+d)^2=216
2(a^2+9d^2)+2(a^2+d^2)=216
4a^2+20d^2=216
4(7^2)+20d^2=216
⇒d=±1
for d=1
the series is 7−3(−1),7−(−1),7−1,7−3⇒10,8,6,4
For d=1
the series is 7−3,7−1,7+1,7+3⇒4,6,8,10
Therefore the numbers are, 4,6,8,10