Question

The sum of three num of an ap is 87 and their products is 4005 find the numbers

Answer 1

Let the numbers be (a-2d),(a) and (a+2d).

A/c,

=> (a-2d)+(a)+(a+2d). = 87

=> 3a =87

=> a = 87/3

=> a = 29.

Again,

=>(a-2d)(a)(a+2d) = 4005

$= > ( {a}^{2} - {4d}^{2} )(a) = 4005$

Putting a= 29 in the given equation; we get

$= > ({29}^{2} - {4d}^{2} )29 = 4005$

$= > {29}^{2} - {4d}^{2} = \frac{4005}{29}$