Question

The sum of three consecutive numbers in an A. P. is 24 and their product is 440. Find the numbers​

Answer 1

Answer:

let the 3 consecutive number be a-d,a,a+d

Step-by-step explanation:

according to given condition

sum .. a-d+a+a+d=24

therefore 3a=24

a=8

now product ,

a(a2-d2)=440

8(64-d2)=440

512-440=d2

72=d2

d= root72

number are

a-d=8-root72

a=8

a+d=8+ root72

Answer 2

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

A/q

a-d+a+a+d=24

3a=24

a=8........[1]

Now,

(a-d)*a*(a+d)=440

a³-a²d+a²d-ad²=440

a³-ad²=440...........[2]

Now substituting the value of 'a' from equation 1 to equation 2

we get,

8³-8d²=440

512-440=8d²

72=8d²

d²=9

d=3

Hence the numbers are 5, 8 and 11