Question

How do you find the sum of the geometric series $\dfrac{1}{9}-\dfrac{1}{3}+1-..........$ to $6$ terms?

Answer 1

From the question given, we have been asked to find the sum of geometric series $\dfrac{1}{9}-\dfrac{1}{3}+1-..........$ to $6$ terms. We can find the sum of the geometric series given in the question by using the sum of the $n$ terms of a geometric progression formula .Formula for the sum of the $n$ terms of a geometric progression is shown below: $S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where $a$ is the first term and $r$ Is the common ratio of the geometric progression.

Complete step by step answer:
From the question given, it has been given that $\dfrac{1}{9}-\dfrac{1}{3}+1-..........$
In this given geometric progression First term $a=\dfrac{1}{9}$ and common ratio $r=\dfrac{{{a}_{2}}}{{{a}_{1}}}\Rightarrow \dfrac{\dfrac{1}{9}}{-\dfrac{1}{3}}=-3$
From the question given, we have been asked to find the sum of the given geometric progression up to $6$ terms.
Therefore $n=6$ for the given geometric progression in the given question.
Now, we have got all the values which we have to substitute in the sum of the $n$ terms of a geometric progression formula.
So, substitute all values we got in the above written sum of $n$ terms of a geometric progression formula.
By substituting all values we get,
$S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
$\Rightarrow S=\dfrac{\dfrac{1}{9}\left( {{\left( -3 \right)}^{6}}-1 \right)}{-3-1}$
$\Rightarrow S=\dfrac{\dfrac{1}{9}\left( 729-1 \right)}{-4}$
$\Rightarrow S=\dfrac{\dfrac{1}{9}\left( 728 \right)}{-4}$
$\Rightarrow S=\dfrac{182}{-9}$
Therefore, we got the sum of $n$ terms of given geometric progression by using the sum of $n$ terms of a geometric progression formula.

Note:
We should be well aware of the progressions concept. Calculation must be done very carefully by the students especially while finding the sum of terms using the formula. We should be well known about the all formulae in progressions concept, because, in progressions concept, we can solve maximum questions by just using the formulae. Similarly the sum of $n$ terms of an arithmetic progression is given as $\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ .