Question

find the three positive integers of an ap whose sum is 24 and product is 480 ​

Answer 1

Here is your answer plz mark me the brainliest

Answer 2

Answer:

let the three positive integer be a-d , a , a+ d

a-d + a+ a+d = 24

cancelling -d and +d

3a = 24

a = 24/3

a = 8

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

$a \: ( {a}^{2} - {d}^{2} ) = 480$

sub a=8

$8( {8}^{2} - {d}^{2}) = 480$

$64 - {d}^{2} = \frac{480}{8}$

$64 - {d}^{2} = 60$

$- {d}^{2} = 60 - 64$

$- {d}^{2} = - 4$

cancelling - on bot the sides

${d}^{2} = 4$

d = +- 2

when a =8 and d= 2

6,8,10

when a = 8 and d=-2

10,8,6

hope it helps u mate!!!