Find the sum of the first $3n$ terms of a geometric series given the sum of the first $n$ terms is $48$ and the sum of first $2n$ terms is $60$

Question

In a certain geometric series, the sum of the first $n$ terms is $48$ and the sum of the first $2n$ terms is $60$. Find the sum of the first $3n$ terms.

$$48= a_1\frac{1-r^n}{1-r}$$

What do I do after this?

Answer 1

The general expression for the sum of the $n$ first terms of a geometric series is:

$$S_n=a_1{1-q^n\over 1-q}$$

So dividing $S_{2n}$ by $S_n$ one gets:

$${1-q^{2n}\over 1-q^n}={5\over 4}$$

And this is rearranged as $4q^{2n}-5q^n+1=0$. The trivial root $q^n=1$ is to be eliminated and we are left with $q^n=1/4$. We then rewrite the summation formula as ${a_1\over 1-q}={S_n\over 1-q^n}={4\over 3}\cdot 48=64$. So we have

$$S_{3n}=a_1{1-\left(q^n\right)^3\over 1-q}=64\cdot\left(1-{1\over 64}\right)=63$$

Answer 2

Let $a$ be first term of the geometric series. $r$ be the common ratio.

We know that for a geometric series, $S_n=\frac {a(1-r^n)}{1-r}$.Where $S_n$ is the sum of first $n$ terms.

So, we have given $S_n=48$ and $S_{2n}=60$. Let $S_{3n}=x$.

Now, $\frac {S_n}{S_{2n}}=\frac {48}{60}=\frac {1-r^n}{1-r^{2n}}$.

Solving this we get $r^n=\frac 14$.$ \Rightarrow$ $r^{2n}=\frac {1}{16}$ and $r^{3n}=\frac {1}{64}$. (Note: $r^n=1$ is also a root. but we discard it since if it happens, then $S_n=0$ which is a contradiction to the given $S_n=48$).

Now $\frac {S_{2n}}{S_{3n}}=\frac {60}{x}=\frac {1-r^{2n}}{1-r^{3n}}$.

Putting values of $r^{2n}$ and $r^{3n}$, we get $x=63$.

Answer 3

Define the sequence $(S_k)$ as follows: $S_1 = $ the sum of the first $n$ terms, $S_2 =$ the sum of terms $n+1$ to $2n$, and in general $S_r =$ the sum of terms $(r-1)n+1$ to $rn$.

Then $S_1,S_2,\ldots$ is a geometric progression (you should check this for yourself), with $S_1=48,S_2=12$. It is now immediate that $S_3=3$.

Answer 4

$$a_1+a_2+\cdots+a_n=48;$$ $$a_{n+1}+a_{n+2}+\cdots+a_{2n}=12;$$ $$\frac{a_{2n+1}+a_{2n+2}+\cdots+a_{3n}}{a_{n+1}+a_{n+2}+\cdots+a_{2n}}=\frac{a_{n+1}+a_{n+2}+_\cdots+a_{2n}}{a_1+a_2+\cdots+a_n}=\frac14=r^n,$$ where $r=\frac{a_{k+1}}{a_k}$ is the common ratio of the geometric progression; so $$a_{2n+1}+a_{2n+2}+\cdots+a_{3n}=3$$ and $$a_1+a_2+\cdots+a_{3n}=48+12+3=63.$$