Question

find four numbers in AP whose sum is 20 and the sum of whose square is 120.

Answer 1

Heya !!



Let the required numbers are ( a - 3d ) , ( a - d ) , ( a + d ) and ( a + 3d ).




Then, ( a - 3d ) + ( a - d ) + ( a + d ) + ( a + 3d ) = 20



=> 4a = 20


=> a = 5


And,



( a - 3d)² + ( a - d )² + ( a + d )² + ( a + 3d )² = 120




=> 4 ( a² + 5d² ) = 120




=> (a² + 5d² ) = 30




=> 25 + d² = 30




=> 5d² = 5


=> d = √1 = 1




First term ( a ) = 5

And,


Common difference ( d ) = 1



Hence,


The required numbers are (2,4,6,8) .

Answer 2

arithmatic progression is 

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") 

TIPS: if you have to find count of odd numbers, u take the increase of "d" and for even numbers, u take the increase of "2d" 
it gives us, an easier calculation, as shown in the addition and squarring steps of this problem. 

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