Question

Find the ${n^{{\text{th}}}}$ term of the Geometric Progression $100,{\text{ }} - 110,{\text{ }}121,.....$

Answer 1

In Geometric Progression, the ratio of any two consecutive terms must be the same and this ratio is called the common ratio of Geometric Progression. Take the ratio of any two consecutive terms to identify the common ratio. Then use the general term of G.P. i.e. $a{r^{n - 1}}$ to find out the ${n^{{\text{th}}}}$ term.
Here $a$ is the first term and $r$ is the common ratio of G.P.

Complete step-by-step answer:
According to the question, the given Geometric Progression is $100,{\text{ }} - 110,{\text{ }}121,.....$
We know that in Geometric Progression, the ratio of any two consecutive terms must be the same and this ratio is called the common ratio of G.P.
Let the common ratio of above G.P. is denoted as $r$. Then dividing any two consecutive terms will give us the value of $r$. So we have:
$\Rightarrow r = \dfrac{{ - 110}}{{100}} = \dfrac{{121}}{{ - 110}} \\\ \Rightarrow r = - \dfrac{{11}}{{10}} = - \dfrac{{11}}{{10}} \\\$
Thus the common ratio of above G.P. is $- \dfrac{{11}}{{10}}$.
Further, we know that the general term of G.P. is given as:
$\Rightarrow {T_n} = a{r^{n - 1}}$, where $a$ is the first term and $r$ is the common ratio of G.P.
The first term of above G.P. is $100$.
$\Rightarrow a = 100$
So, putting the values of $a$ and $r$ in general terms, we’ll get the ${n^{{\text{th}}}}$ term. Doing this, we have:
$\Rightarrow {T_n} = 100 \times {\left( { - \dfrac{{11}}{{10}}} \right)^{n - 1}}$

Thus the ${n^{{\text{th}}}}$ term of the above Geometric Progression is $100 \times {\left( { - \dfrac{{11}}{{10}}} \right)^{n - 1}}$.

Note: The sum of first $n$ terms of G.P. can be calculated by using the formula:
$\Rightarrow {S_n} = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right),{\text{ }}\left| r \right| > 1$
If the G.P. is having infinite number of terms, then the sum of all of its terms is calculated using the formula:
$\Rightarrow {S_n} = \dfrac{a}{{1 - r}},{\text{ }}0 < r < 1$