Question

four umbers are in AP & their sum is 20 if sum of their squares is 120 the find the middle two numbers??

Answer 1

the numbers are 2 , 4 , 6 ,8 .
the middle two numbers are 4 & 6

(since you asked for direct answer , I haven't solved)

Answer 2

___heya mate ___

☆☆here is your answer ☆☆

here is the full method

a, a+d, a+2d, a+3d 

or for convenience we can take 

a-3d, a-d, a+d, a+3d (considering increase of "2d") 

now addition of 4 numbers is 20 

[a-3d] + [a-d] + [a+d] + [a+3d] = 20 

4a = 20 (-3d, -d, +d, +3d cancels each other) 

a = 20/4 

a= 5 (A) 

now, sum of squares is 120 

(a-3d)**2 + (a-d)**2 + (a+d)**2 + (a+3d)**2 = 120 (I used ** for squaring or exponentiating) 

[a**2 + 9d**2 - 6ad] + [a**2 + d**2 - 2ad] + [a**2 + d**2 + 2ad] + [a**2 + 9d**2 + 6ad] = 120 

4a**2 + 20d**2 = 120 (-6ad -2ad +2ad + 6ad cancels each other) 
4[5]**2 + 20d**2 = 120 use (A) 
4[25] + 20d**2 = 120 
100 + 20d**2 = 120 
20d**2 = 120-100 
20d**2 = 20 
d**2 = 20/20 
d**2 = 1 
d=+1 or -1 
in AP the difference cannot be negative. 
hence d = 1 

now the numbers are 

a-3d, a-d, a+d, a+3d 

5-3, 5-1, 5+1, 5+3 

2,4,6,8 

ans is 4 and 6

hope the answer will help you!!!