Question

How do you find the sum of a geometric series for which $ {{a}_{1}}=48 $ , $ {{a}_{n}}=3 $ and $ r=-\dfrac{1}{2} $ ?

Answer 1

Hint: From the given terms of geometric sequence, we find the general term of the series. We find the formula for $ {{a}_{n}} $ , the $ {{n}^{th}} $ term of the series. From the given data we find the series of the G.P. We put the values for the formula of summation.

Complete step-by-step answer:
We have been given a series of geometric series whose $ {{a}_{1}}=48 $ , $ {{a}_{n}}=3 $ and $ r=-\dfrac{1}{2} $ .
We express the geometric sequence in its general form.
We express the terms as $ {{a}_{n}} $ , the $ {{n}^{th}} $ term of the series.
The first term be $ {{a}_{1}} $ and the common ratio be $ r $ where
$ r=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}} $ .
We can express the general term $ {{a}_{n}} $ based on the first term and the common ratio.
The formula being $ {{a}_{n}}={{a}_{1}}{{r}^{n-1}} $ .
The first term is 48. So, $ {{a}_{1}}=48 $ . The common ratio is $ r=-\dfrac{1}{2} $ .
We put the values of $ {{t}_{1}} $ and $ r $ to find the general form.
We express general term $ {{a}_{n}} $ as $ {{a}_{n}}={{a}_{1}}{{r}^{n-1}}=48\times {{\left( -\dfrac{1}{2} \right)}^{n-1}} $ .
So, $ 48\times {{\left( -\dfrac{1}{2} \right)}^{n-1}}=3 $ which gives \[{{\left( -\dfrac{1}{2} \right)}^{n-1}}=\dfrac{3}{48}=\dfrac{1}{16}={{\left( -\dfrac{1}{2} \right)}^{4}}\].
The simplification gives $ n-1=4\Rightarrow n=5 $ .
The value of $ \left| r \right|<1 $ for which the sum of the first n terms of an G.P. is
$ {{S}_{n}}={{t}_{1}}\dfrac{1-{{r}^{n}}}{1-r} $ .
Putting the values, we get $ {{S}_{5}}=48\times \dfrac{1-{{\left( -\dfrac{1}{2} \right)}^{5}}}{1-\left( -\dfrac{1}{2} \right)}=32\times \left( 1+\dfrac{1}{32} \right)=33 $ .
Therefore, the sum of a geometric series for which $ {{a}_{1}}=48 $ , $ {{a}_{n}}=3 $ and $ r=-\dfrac{1}{2} $ is 33.
So, the correct answer is “33”.

Note: The sequence is an increasing sequence where the common ratio is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number. The ratio formula should always be according $ r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}} $ .