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Question

Question: How do you find the sum of the infinite geometric series 30 + 22.5 + 16.875 + ....?...

How do you find the sum of the infinite geometric series 30 + 22.5 + 16.875 + ....?

Explanation

Solution

This question is from the topic of sequence and series. In this question, we will find the sum of given infinite geometric series. In solving this question, we will first know the first term of the given series. After that, we will find out the common ratio of the given geometric series. After that, we will use the formula for infinite sums of geometric series and find the sum. After that, we will get our answer.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find out the sum of the given infinite geometric series.
The given infinite geometric series is given as
30 + 22.5 + 16.875 + .......
Here, we can see that the first term is 30.
So, let us write the first term as
a1=30{{a}_{1}}=30
Now, let us find the common ratio.
The common ratio of geometric series is always equal to next term divided by the previous term.
Let us write the common ration as r.
So, we can write
r=22.530=16.87522.5=34r=\dfrac{22.5}{30}=\dfrac{16.875}{22.5}=\dfrac{3}{4}
So, the formula for infinite sum of geometric series is in the following:
S=a11r{{S}_{\infty }}=\dfrac{{{a}_{1}}}{1-r}; if r<1.
Where, S{{S}_{\infty }} is a sum of infinite series, a1{{a}_{1}} is the first term, and r is the common ratio.
So, for a1=30{{a}_{1}}=30 and r=34r=\dfrac{3}{4}, we can write the sum of infinite geometric series as
S=30134=30434=3014{{S}_{\infty }}=\dfrac{30}{1-\dfrac{3}{4}}=\dfrac{30}{\dfrac{4-3}{4}}=\dfrac{30}{\dfrac{1}{4}}
The above can also be written as
S=30×4=120\Rightarrow {{S}_{\infty }}=30\times 4=120
Hence, we got that the sum of the infinite geometric series 30 + 22.5 + 16.875 + .... will be 120.

Note: We should have a better knowledge in the topic of sequence and series to solve this type of question easily. We should know the formulas of geometric series. Let the series be
a,ar,ar2,ar3,....a,ar,a{{r}^{2}},a{{r}^{3}},.... where, ‘a’ is the first term and ‘r’ is the common ratio.
The nth term of the geometric series will be
an=arn1{{a}_{n}}=a{{r}^{n-1}}
The sum of n terms of geometric series will be
Sn=a(rn1)(r1){{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}
If r>1, then the sum of infinite geometric series is S={{S}_{\infty }}=\infty
If r<1, then the sum of infinite geometric series is S=a1r{{S}_{\infty }}=\dfrac{a}{1-r}