Question

In an A.P series,the sum of four numbers is 20 and the sum of the squares the numbers is 120?find the solution.

Answer 1

Let ( a - 3d ),( a - d ) ,( a + d ) ,( a + 3d )

are four terms of AP ,

According to the problem given,

Sum of the four terms = 20

a -3d + a - d + a + d + a + 3d = 20

4a = 20

a = 20 / 4

a = 5 ----( 1 )

Sum of the squares of the numbers

= 120

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

2 ( a² + 3d² ) + 2 ( a² + d² ) = 120

Divide each term with 2 ,

a² + 3d² + a² + d² = 60

2a² + 4d² = 60

2 × 5² + 4d² = 60

50 + 4d² = 60

4d² = 10

d² = 10 / 4

d = √( 10 / 4 ) = √10 / 2

Therefore a = 5 , d = √10 / 2

Required four terms are

a - 3d = 5 - 3√10 /2

a - d = 5 - √10/2

a+ d = 5 + √10 /2

a + 3d = 5 +3 √10 /2

I hope this helps you.

:)