Question

find four numbers in ap whose sum is 20 and the sum of whose squaired is 120

Answer 1

(a+3d)+(a--3d) +(a+d)+(a--d) =20
solve and get
4a=20
a=5
(a+3d)^2+(a--3d) ^2+(a+d)^2+(a--d) ^2=120
solve and we get
a^2+5d^2=30
put a=5
25+5d^2=30
therefore
d= 1
now
a+3d=5+3*1=8
a--3d=5--3*1=2
a+d=5+1=6
a--d=5--1=4
there fore number s are 2, 4. 6. 8.