given that the sum of the first n terms of a geometric sequence is $4-(2^{(n+2)})/3^{(2n)}$. find the second term and its common ratio.

Question

I tried to find the first term by substituting 1 into n, and I got $28/9$. After that, I subtitute $2$ into $n$ for $S_n$ and dis the $Term_2 = S_2 -S_1$. I wasn't sure whether I am correct or not. Can someone please guide me on how to solve this question?

Answer 1

It seems like you already know how to solve the problem.

Let's denote the actual terms of the sequence by $a_1, a_2, a_3, \dots$. Let $S_n = a_1 + a_2 + \dotsb + a_n$ be the sum of the first $n$ terms.

As you said, you can find the first term by plugging $n = 1$ into the formula for $S_n$: $$a_1 = S_1 = 4 - 2^{1 + 2}/3^{2 \times 1} = \frac{28}{9}.$$

Since $S_1 = a_1$, $S_2 = a_1 + a_2$, you can recover the second term by \begin{align*} a_2 & = S_2 - S_1\\ & = \Big( 4 - \frac{2^{2 + 2}}{3^{2 \times 2}} \Big) - \frac{28}{9}\\ & = \frac{56}{81}. \end{align*}

Since you are told that the sequence is geometric, the common ratio is just $$q = \frac{a_2}{a_1} = \frac{\frac{56}{81}}{\frac{28}{9}} = \frac{2}{9}.$$