Question

find four number in Ap whose sum is 28 and the sum of whose square is 216 ​

Answer 1

Answer:

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(a2+9d2)+2(a2+d2)=216

4a2+20d2=216

4(72)+20d2=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