Question

yeh sum arithmetic progression chapter se related hai plz jldi solve krna

Answer 1

Let the four terms be (a-3d),(a-d),(a+d),(a+3d)
Therefore,
Now,
(a-3d)+(a-d)+(a+d)+(a+3d)=20
a-3d+a-d+a+d+a+3d=20
4a=20
a=5
Also,
(a-3d)2+(a-d)2+(a+d)2+(a+3d)2=120
a2-6ad+9d2+a2-2ad+d2+a2+2ad+d2+a2+6ad+9d2=120
a2+9d2+a2+d2+a2+d2+a2+9d2=120
4a2+20d2=120
4(a2+5d2)=120
a2+5d2=30
25+5d2=30
5d2=30-25
5d2=5
d2=1
d=+-1
Therefore, when a=5, d=1
The four numbers are=2,4,6,8
But, a=5, d=-1
The four numbers are=8,6,4,2