Question

Find $S_n$ for each of the geometirc series described below.
(i) $a=3,t_8=384,n=8$
(ii) $a=5,r=3,n=12$

Answer 1

(i) It is given that the first term of the geometric series is $a=3$, the $8$th term is $T_8=384$ and the number of terms are $n=8$.

We know that the general term of an geometric progression with first term $a$ and common ratio $r$ is $T_n=a{r}^{n-1}$, therefore,

$T_8=3r^{8-1}\Rightarrow 384=3r^7\Rightarrow r^7=\frac{384}{3}\Rightarrow r^7=128\Rightarrow r^7=2^7\Rightarrow r=2>1$

We also know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(r^n-1)}{r-1}$ if $r>1$

Now, substitute $a=3,r=2$ and $n=8$ in $S_n=\frac{a(r^n-1)}{r-1}$ as follows:

$S_8=\frac{3[{\left(2\right)}^8-1]}{2-1}=\frac{3[{\left(2\right)}^8-1]}{1}=3(256-1)=3\times 255=765$

Hence the sum is $765$.

(ii) It is given that the first term of the geometric series is $a=5$, the common ratio is $r=3>1$ and the number of terms are $n=12$.

We know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(r^n-1)}{r-1}$ if $r>1$

Now, substitute $a=5,r=3$ and $n=12$ in $S_n=\frac{a(r^n-1)}{r-1}$ as follows:

$S_{12}=\frac{5[{\left(3\right)}^{12}-1]}{3-1}=\frac{5[{\left(3\right)}^{12}-1]}{2}=\frac{5}{2}[{\left(3\right)}^{12}-1]$

Hence the sum is $\frac{5}{2}[{\left(3\right)}^{12}-1]$.