Question

sum of the three consecutive term of an AP is 9 and product is 24. find the number​

Answer 1

Answer

Let the terms be a-d, a and a+d

Now,

a-d+a+a+d=9

3a =9

a=3

Now,

(a-d) (a) (a+d) =24

replacing a

(3-d) (3+d) 3 =24

Using identity

( 3^2-d^2) 3=24

( 9-d^2)3 =24

(9-d^2) =24/3

(9-d^2) =8

-d^2=8-9

-d^2 =-1

d^2 =1

d=√1

d = +1, -1

Answer 2

Step-by-step explanation:

Let the terms be a-d, a and a+d

Now,

a-d+a+a+d=9

3a =9

a=3

Vo) 4G LTE1 l LTE 2.l

Now,

(a-d) (a) (a+d) =24

replacing a

(3-d) (3+d) 3 =24

Using identity

( 3^2-d^2) 3=24

(9-d^2)3 =24

(9-d^2) =24/3

(9-d^2) =8

-d^2=8-9

-d^2 =-1

d^2 =1

d=

$\sqrt{1}$

d = +1, -1