Question

The sum of 3 consecutive terms of an A.P. is 36 and their product is 1140.Find the terms.​

Answer 1

Let the 3 consecutive terms be
(a-d) , (a) , (a+d) .
Now acc. to question ,
a - d + a + a + d= 36
3a=36
a=12
Now ,
(a - d) (a) (a + d)=1140
(a^2 - d^2) (a)=1140
(12^2 - d^2) (12)=1140
(144 - d^2) (12)=1140
1728 - 12d^2 = 1140
588 = 12d^2
49 = d^2
d = +-7 ( here +7 and -7 both will be considered)

Terms when d = 7
a = 12
(a - d) = 5
(a) = 12
(a + d) = 19

Terms when d = -7
a = 12
(a - d) = 19
(a)= 12
(a + d) = 5

Answer 2

Answer:

let three terms are

a-d,a,a+d

as per 1st condition,

a-d+a+a+d=36

3a=36

a=36/3

[a=12]