Three real numbers $a, b$ and $c$ have a sum of 114 and a product of 46656 . If $b=a r$ and $c=a r^{2}$ for some real number $r$, then the value of $a+c$ is
(A) 78
(B) 76
(C) 24
(D) 54
(E) 36

## Part C: Each correct answer is worth 8.
Since $b=a r, c=a r^{2}$, and the product of $a, b$ and $c$ is 46656 , then $a(a r)\left(a r^{2}\right)=46656$ or $a^{3} r^{3}=46656$ or $(a r)^{3}=46656$ or $a r=\sqrt[3]{46656}=36$.

Therefore, $b=a r=36$.

Since the sum of $a, b$ and $c$ is 114 , then $a+c=114-b=114-36=78$.

ANSWER: (A)