Question

The sum of first four terms of an AP is 26 if the sum of the squares of the first and fourth term is 125 then find the first four terms

Answer 1

The SUM of the first 4 terms of an arithmetic progression is 26.

Letting a1 be the 1st term and d be the common difference:

 

a1+(a1+d)+(a1+2d)+(a1+3d) = 26

4a1 + 6d = 26  

2(2a1+3d)=26

2a1 + 3d = 13   {equation 1}

 

sum of squares of first and fourth term is 125

 

a12 + (a1+3d)2 = 125  

a12 + a12 + 6a1d + 9d2 = 125

2a12 + 6a1d + 9d2 = 125   {equation 2}

 

From equation 1:  a1 = (13-3d)/2

Plug into equation 2

 

2[(13-3d)/2)2] + 6d(13-3d)/2 + 9d2 = 125

2[(169-78d+9d2)/4] + 3d(13-3d) + 9d2= 125

(169-78d+9d2)/2 + 39d - 9d2 + 9d2 = 125

(169-78d+9d2)/2 + 39d = 125

169 - 78d + 9d2 + 78d = 250

9d2 + 169 = 250

9d2 = 81

d2 = 9

d = 3

 

2a1 + 3d = 13

2a1 + 9 = 13

2a1 = 4

a1 = 2

a2 = 2+3 = 5

a3 = 5+3 = 8

a4 = 8+3 = 11

 

Check:  a1+a2+a3+a4 = 26

            2+5+8+11 = 26

            26 = 26

 

a12 + a42 = 125

22 + 112 = 125

4 + 121 = 125

125 = 125