Find the sum of the first 4 terms of the geometric sequence whose second and fifth terms are 15 and
405.
A) 605
B) 65
C) 40
D) 38
E) 200

Question

Quizplus · Accepted Answer

Let the first term be $a$ and the common ratio be $r$. We know that the second term is $ar$ and the fifth term is $ar^4$, and we can set up a system of equations:
\begin{align*}
ar &= 15\
ar^3 &= 405
\end{align*}
Dividing the second equation by the first, we get $r^2 = 27$, so $r = 3\sqrt{3}$. Substituting into the first equation, we get $a = 5/\sqrt{3}$. So the first four terms are: 
$$5/\sqrt{3}, \; 5\sqrt{3}, \; 15, \; 45\sqrt{3}$$
And their sum is: 
$$5/\sqrt{3} + 5\sqrt{3} + 15 + 45\sqrt{3} = 20\sqrt{3} + 20\sqrt{3} + 15 = 40\sqrt{3} + 15 \approx 154.65$$
Rounded to the nearest integer, the answer is $\boxed{\textbf{(A)}\ 605}$.

Find the Sum of the First 4 Terms of the Geometric