Question

 The three numbers in A.P., where sum is 24 and the product is 440, then the numbers are​

Answer 1

Step-by-step explanation:

Let the three numbers in A.P. be a−d,a, and a+d

According to given information

Sum=(a−d)+(a)+(a+d)=24...(1)

⇒3a=24∴a=8

& Product=(a−d)a(a+d)=440...(2)

⇒(8−d)(8)(8+d)=440

⇒(8−d)(8+d)=55

⇒64−d

2

=55

⇒d

2

=64−55=9

⇒d=±3

Therefore when d=3, the numbers are 5,8,11 and

when d=−3, the numbers are 11,8 and 5.

Thus the three numbers are 5,8 and 11.

Answer 2

Given

Sum is 24 and Product is 440

To find

Three numbers ?

Solution

Let the three numbers are (a-d) (a) (a+d)

Now there sum:-

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

3a = 24

a = 8

We got the value of a is 8

Now there product:-

$\implies$(8-d)(8)(8+d)=440

$\implies$(8-d)(8)(8+d) = 440

$\implies\sf(8^2-d^2) = \cancel\frac{440}{8}$

$\implies 64-d^2=55$

$\implies-d^2 = 55-64$

$\implies \cancel{-}d^2 = \cancel{-}9$

$\implies\:d\:=\:√9$

$\implies\:d\:=\: +_-3$

Now,put the value of a and d,

Case I:- take (+)

Numbers are :- (a-d) = 8-3 = 5

(a) = 8

(a+d)= 8+3=11

Case II:- take (-)

Nubers are:- (a-d) = 8-(-3)= 11

(a) = 8

(a+d) = 8+(-3)= 5

So, in both the cases numbers are 5,8,11

Hence numbers are 5,8,11