Question

Divide $69$ into three parts which are in A.P such that the product of the first two is $483$

Answer 1

Let three parts be $a-d,a,a+d$

Given that $a-d+a+a+d=69$

$\Rightarrow 3a=69$

$\Rightarrow a=\frac{69}{3}=23$

Product of the first two is $=483$

$\Rightarrow a(a-d)=483$

$\Rightarrow 23(23-d)=483$

$\Rightarrow 529-23d=483$

$\Rightarrow 23d=529-483=46$

$\Rightarrow d=\frac{46}{23}=2$

The numbers are $23-2,23,23+2$ or $21,23,25$