Question

The sum of three numbers in A.P. is $27$, and their product is $504$; find them.

Answer 1

let $3$ numbers in A.P. be $(a-d),a,(a+d)$

$\Rightarrow$ Sum of $3$ numbers $=27$

$\Rightarrow$$a-d+a+a+d=27$

$\Rightarrow$$3a=27$

$a=9$

$\Rightarrow$Product of numbers $=504$

$\Rightarrow$$(a-d)a(a+d)=504$

$\Rightarrow$$(a^2-d^2)a=504$

$\Rightarrow$$[9^2-d^2]9=504$

$\Rightarrow$$81-d^2=56$

$\Rightarrow$$d^2=81-56$

$\Rightarrow$$d^2=25$

$\Rightarrow$$d=\pm 5$

If $d=5,$ A.P. $=4,9,14$
If $d=-5,$ A.P. $=14,9,4$

$\therefore$ These numbers are $4,9,14$