Question

The sum of four consecutive numbers in an A.P.
with d > Ois 20. Sum of their square is 120, then
the middle terms are​

Answer 1

Answer:

hey mate your answer

Step-by-step explanation:

Let the terms be a−3d,a−d,a+d,a+3d.

Given, Sum =20

⇒a−3d+a−d+a+d+a+3d=20

⇒4a=20

⇒a=5

Also given, sum of squares =120

⇒(a−3d)2+(a−d)2+(a+d)2+(a+3d)2=120

⇒(5−3d)2+(5−d)2+(5+d)2+(5+3d)2=120

⇒25−30d+9d2+25−10d+d2+25+10d+d2

⇒25+30d+9d2=120

⇒100+20d2=120

⇒20d2=20

⇒d2=1

⇒d=±1

The terms are 

hope it helps you