Question

How do you find the sum of the terms of the geometric $1,-2,4,-8......-8192$?

Answer 1

Hint: To find the sum of the terms of the given geometric progression, we will first have to find the common ratio of the progression. Then we will substitute the first term, common ratio, and the number of terms of the progression in the formula to find the sum of $n$ terms of the G.P, which is given by, ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ .

Complete step by step solution:
Given the geometric progression:
$1,-2,4,-8......-8192$.
In this progression, we don’t know the total number of terms and therefore we will first have to find the value of $n$ which denotes the total number of terms in the general expression of the geometric progression, which can be given as $a+ar+a{{r}^{2}}+a{{r}^{3}}+...a{{r}^{n}}$, where $a$ is the first term and $r$ is the common ratio.
Here, the first term is $a=1$ and the common ratio $r$ can be calculated by taking the ratio between any two consecutive terms, therefore common ratio can be given by,
$r=\dfrac{4}{-2}$
On further simplifying the above expression we get the value of common ratio as,
$\Rightarrow r=\dfrac{4}{-2}=-2$
Now, to calculate the total number of terms in the given geometric progression, we will use the formula
${{T}_{n}}=a{{r}^{n-1}}$, where ${{T}_{n}}$ is the ${{n}^{th}}$term of the G.P.
Substituting the values of $a$, $r$ and ${{T}_{n}}$ in the above formula we get,
$\Rightarrow -8192=1\times {{\left( -2 \right)}^{n-1}}$
Now, simplifying both the sides and writing both the terms in powers of $2$, we get,
$\Rightarrow {{\left( -2 \right)}^{13}}={{\left( -2 \right)}^{n-1}}$
Now, compare the powers on both the side, since the bases are the same, we can equate the powers, therefore, we get,
$\Rightarrow 13=n-1$
Solving the above equation for $n$, we get,
$\Rightarrow n=14$
Now, finally substituting the values of $a$, $r$ and $n$ in the formula to find the sum of $n$ terms of the G.P., we get,
$\Rightarrow {{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\dfrac{1\left( {{\left( -2 \right)}^{14}}-1 \right)}{\left( -2 \right)-1}$
Taking the powers and simplifying the fraction, we get,
$\Rightarrow {{S}_{n}}=\dfrac{1\left( 16,384-1 \right)}{-3}$
$\Rightarrow {{S}_{n}}=\dfrac{-\left( 16,383 \right)}{3}$
Therefore, the sum of the given G.P. $1,-2,4,-8......-8192$ is $\dfrac{-\left( 16,383 \right)}{3}$

Note: Remember that this is one of the formulas for calculating the sum of a given G.P. The above formula can be used when the terms are finite. If there are infinite terms in a G.P, then we can use the formula $\dfrac{a}{1-r}$ to calculate the sum of the terms.