Question

The sum of 3 positive numbers in AP is 189. The sum of their squares is 11915. Find their product.

• A. 7930
• B. 8970
• C. 9703
• D. 7960
• E. None of these

Answer 1

The correct option is A 7930

Let the numbers in AP series be $a-d,a,a+d$
So $a-d+a+a+d=189$ or $3a=189$ or $a=63$
As per second part of the problem $(a-d{)}^2+(a{)}^2+(a+d{)}^2=4023$ or $3a^2+3d^2=4023$
or $3\times (63{)}^2+2d^2=4023$
or $2d^2=11915-3\times 63\times 63$
$=11915-11907$
$=08$
or $d^2=4$ or $d=2$
So their product is $(a-(d)\times a\times (a+\left(d\right)$
$=(63-2)\times 2\times (63+2)$
$=61\times 2\times 65$
$=130\times 61$
$=7930$