Question

Find three positive integers in AP such that their sum is 24 and their product is 480

Answer 1

hey dear!
let the three numbers in AP be a-d, a, a+d where a is first term and d is the common difference.
hope it will help✌

Answer 2

Answer:

Let the three terms in AP be  a-d , a , a+d

Now we are given that The sum of three numbers in A.P is 24

So, $a-d+a+a+d= 24$

$3a= 24$

$a=8$ --- 1

We are also given that their product is 480

$(a-d)(a)(a+d)= 480$

$(a^3-ad^2)= 480$

Substitute the value of d from 1

$(8^3-8d^2)= 480$

d=-2, 2

At a = 8 and d = -2

a+d=8-2=6

a=8

a-d=8+2=10

At a = 8 and d = 2

a+d=8+2=10

a=8

a-d=8-2=6