
How do you find the sum of geometric sequence 2, 4, 8, …. If there are 20 terms.
Answer
493.5k+ views
Hint: Now we are given with a GP. First we will find the first term and the common ratio of GP. To find the common ratio we will take the ratio of any two consecutive terms of GP. Hence we will have the first term and the common ratio. Now we will use the formula to find sum of n terms of GP which is ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where a is the first term and r is the common ratio. Now we will substitute the values of a and r and n = 20 in the formula to find the sum of 20 terms of given GP.
Complete step-by-step solution:
Now let us first understand what a geometric series is.
A geometric series is a series in which the ratio of any two consecutive terms is the same.
Hence a geometric series looks like $a+ar+a{{r}^{2}}+a{{r}^{3}}+...a{{r}^{n}}$ . where a is the first term and r is the common ratio.
Now consider the given geometric series 2, 4, 8, …
Here the first term a = 2.
And the common ratio between two consecutive terms is $\dfrac{8}{4}=\dfrac{4}{2}=2$
Hence we have a = 2 and r = 2.
Now we know that sum of n terms of GP is given by ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ .
Hence substituting a = 2 and r = 2 in the above formula we get,
$\Rightarrow {{S}_{n}}=\dfrac{2\left( {{2}^{n}}-1 \right)}{2-1}$
The above equation represents the sum of n terms of the given GP.
Now since we want to find the sum of 20 terms we will substitute n = 20.
Hence we get,
$\begin{align}
& \Rightarrow {{S}_{20}}=\dfrac{2\left( {{2}^{20}}-1 \right)}{2-1} \\
& \Rightarrow {{S}_{20}}=2\left( {{2}^{20}}-1 \right) \\
\end{align}$
Hence the sum of the first 20 terms of the GP 2, 4, 8, … is $2\left( {{2}^{20}}-1 \right).$
Note: Now note that there are different formulas for the sum of GP. Sum of GP of finite terms is given by ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ when r is not 1.
If r is 1 then the sum of GP is given by an. Also we have a formula for infinite GP which is $\dfrac{a}{1-r}$ .
Also ${{n}^{th}}$ term of GP is given by $a{{r}^{n-1}}$
Complete step-by-step solution:
Now let us first understand what a geometric series is.
A geometric series is a series in which the ratio of any two consecutive terms is the same.
Hence a geometric series looks like $a+ar+a{{r}^{2}}+a{{r}^{3}}+...a{{r}^{n}}$ . where a is the first term and r is the common ratio.
Now consider the given geometric series 2, 4, 8, …
Here the first term a = 2.
And the common ratio between two consecutive terms is $\dfrac{8}{4}=\dfrac{4}{2}=2$
Hence we have a = 2 and r = 2.
Now we know that sum of n terms of GP is given by ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ .
Hence substituting a = 2 and r = 2 in the above formula we get,
$\Rightarrow {{S}_{n}}=\dfrac{2\left( {{2}^{n}}-1 \right)}{2-1}$
The above equation represents the sum of n terms of the given GP.
Now since we want to find the sum of 20 terms we will substitute n = 20.
Hence we get,
$\begin{align}
& \Rightarrow {{S}_{20}}=\dfrac{2\left( {{2}^{20}}-1 \right)}{2-1} \\
& \Rightarrow {{S}_{20}}=2\left( {{2}^{20}}-1 \right) \\
\end{align}$
Hence the sum of the first 20 terms of the GP 2, 4, 8, … is $2\left( {{2}^{20}}-1 \right).$
Note: Now note that there are different formulas for the sum of GP. Sum of GP of finite terms is given by ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ when r is not 1.
If r is 1 then the sum of GP is given by an. Also we have a formula for infinite GP which is $\dfrac{a}{1-r}$ .
Also ${{n}^{th}}$ term of GP is given by $a{{r}^{n-1}}$
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