Question

The sum of three numbers in AP is 21 and their product is 231. Find the numbers.

Answer 1

Let the required numbers be $(a-d),a,(a+d)$.

Then,

$a-d+a+a+d=21$

$3a=21$

$a=7$

Also,

$(a-d)a(a+d)=231$

$a(a^2-d^2)=231$

$7(49-d^2)=231$

$7d^2=112$

$d^2=16$

$d=\pm 4$

Hence, the required numbers are $(3,7,11)$ or $(11,7,3)$.