Question

The sum of three numbers in ap is 27 and the sum of their squares is293 then find the ap?

Answer 1

Step-by-step explanation:

let a-d,a and a+d be the three terms of that AP

a-d+a+a+d=27

3a=27

a=9

(a-d)²+a²+(a+d)²=293

(9-d)²+81+(9+d)²=293

(9-d)²+(9+d)²=212

81-18d+d²+81+18d+d²=212

2d²=50

d²=25

d=±5

but the terms with negative D will not satisfy the first condition of sum of three terms being 27

thus we will take d as+5

AP is 4,9,14,......

Answer 2

Answer: 4 , 9 , 14

Step-by-step explanation:Let the numbers be a−d,a,a+d

so 3a=27

⇒ a=9

Also (a−d)  

2

+a  

2

+(a+d)  

2

=293.

3a  

2

+2d  

2

=293

d  

2

=25

⇒ d=±5

therefore numbers are 4,9,14.