Question

If the sum of three numbers in AP is 9 and their product is 24 then the number is what

Answer 1

Answer:

2,3,4

Step-by-step explanation:

Let the numbers be a-d,a,a+d

Given sum is 9

a-d+a+a+d=9

3a=9

a=9/3

a=3

Given product is 24

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

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

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

(9-d^2)3=24

9-d^2=24/3

9-d^2=8

d^2=9-8

d^2=1

d=+or - 1

if d=1

a-d=3-1=2

a=3

a+d=3+1=4

if d=-1

a-d=3-(-1)=3+1=4

a=3

a+d=3+(-1)=3-1=2

The numbers are 2,3,4

Answer 2

Answer:

As per the data given in the above question.

we have to find the three numbers .

Given ,

Sum of three numbers of an A.P =9

Product of three numbers of an A.P.= 24

NOW,

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

According to given information,

Sum of three numbers are,

(a−d)+(a)+(a+d)=9........(1)

Add the equation(1) ,then we get

$3a = 9 \\ a = \frac{9}{3} \\ a = 3...........(2)$

Product of three numbers are,

(a−d)× a ×(a+d)=24........(3)

Now ,put the value of a=3 in equation (3)..,

$(3−d)\times\: 3 \times(3+d)=24$

Here shift the only 3 toward 24 then,

$(3 - d)(3 + d) = \frac{24}{3}$

$( {3}^{2} - {d}^{2} ) = 8$

$- {d}^{2} = 8 - {3}^{2}$

$- {d}^{2} = 8 - 9$

$- {d}^{2} = - 1$

$d=±1$

Therefore ,

put the values in a−d,a, and a+d

when, d=1, a=3the numbers are 2,3,4

d=1, a=3the numbers are 2,3,4when ,d=−1, a=3the numbers are 4,3,2

Thus the three numbers are 2 ,3 and 4.

Project code #SPJ2