Question

find 4 terms of A.P whose sum is 20 and sum of squares is 120

Answer 1

Let the terms be---> a-3d, a-d, a+d and a+3d

Given, a-3d+a-d+a+d+a+3d=20
            4a = 20
            a=5

It is also given that, 
                              (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120
=> (a^2 + 9(d^2) - 6ad) + (a^2 + d^2 - 2ad) + (a^2 + d^2 + 2ad) + (a^2 + 9(d^2) + 6ad) = 120
=> 4(a^2) + 2*9(d^2) + 2*(d^2) = 120
=> 4(5^2) + 18(d^2) + 2(d^2) = 120
=> 4*25 + 20(d^2) = 120
=> 20(d^2) = 120-100
=>d^2 = 20/20
=>d^2 = 1
=> d= 1

a= 5 and d=1
Thus the terms of the AP are: a-3d, a-d, a+d, a+3d
                                                5-3, 5-1, 5+1, 5+3
                                               
                                  (ANSWER)    2 , 4, 6, 8

Thankyou.
Hope it helps. :)

Answer 2

i hope this will usful to u