Sum of the Terms of a Geometric Sequence-Geometric Series ... - Free Printable

Let's solve each problem step by step.

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Problem 3:

In a geometric sequence, the first term is \(100\) and the common ratio is \(\frac{1}{4}\). Find the sum of the first 5 terms.

#### Solution:
The formula for the sum of the first \(n\) terms of a geometric sequence is:
\[
S_n = a \frac{1 - r^n}{1 - r}
\]
where:
- \(a\) is the first term,
- \(r\) is the common ratio,
- \(n\) is the number of terms.

Here:
- \(a = 100\),
- \(r = \frac{1}{4}\),
- \(n = 5\).

Substitute these values into the formula:
\[
S_5 = 100 \frac{1 - \left(\frac{1}{4}\right)^5}{1 - \frac{1}{4}}
\]

First, calculate \( \left(\frac{1}{4}\right)^5 \):
\[
\left(\frac{1}{4}\right)^5 = \frac{1}{4^5} = \frac{1}{1024}
\]

Next, calculate the denominator \(1 - \frac{1}{4}\):
\[
1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}
\]

Now substitute back into the formula:
\[
S_5 = 100 \frac{1 - \frac{1}{1024}}{\frac{3}{4}}
\]

Simplify the numerator \(1 - \frac{1}{1024}\):
\[
1 - \frac{1}{1024} = \frac{1024}{1024} - \frac{1}{1024} = \frac{1023}{1024}
\]

So the expression becomes:
\[
S_5 = 100 \cdot \frac{\frac{1023}{1024}}{\frac{3}{4}}
\]

Dividing by a fraction is the same as multiplying by its reciprocal:
\[
S_5 = 100 \cdot \frac{1023}{1024} \cdot \frac{4}{3}
\]

Simplify the multiplication:
\[
S_5 = 100 \cdot \frac{1023 \cdot 4}{1024 \cdot 3} = 100 \cdot \frac{4092}{3072}
\]

Simplify the fraction \(\frac{4092}{3072}\):
\[
\frac{4092}{3072} = \frac{4092 \div 12}{3072 \div 12} = \frac{341}{256}
\]

So:
\[
S_5 = 100 \cdot \frac{341}{256} = \frac{34100}{256}
\]

Simplify \(\frac{34100}{256}\):
\[
\frac{34100}{256} = \frac{34100 \div 4}{256 \div 4} = \frac{8525}{64}
\]

Thus, the sum of the first 5 terms is:
\[
\boxed{\frac{8525}{64}}
\]

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Problem 6:

Find the sum \(2 + 4 + 8 + 16 + \cdots + 512\).

#### Solution:
This is a geometric sequence where:
- The first term \(a = 2\),
- The common ratio \(r = 2\),
- The last term is \(512\).

First, determine the number of terms \(n\). The \(n\)-th term of a geometric sequence is given by:
\[
a_n = a \cdot r^{n-1}
\]
Set \(a_n = 512\):
\[
512 = 2 \cdot 2^{n-1}
\]
\[
512 = 2^{n}
\]
Since \(512 = 2^9\), we have:
\[
2^n = 2^9 \implies n = 9
\]

Now, use the sum formula for the first \(n\) terms of a geometric sequence:
\[
S_n = a \frac{r^n - 1}{r - 1}
\]
Here:
- \(a = 2\),
- \(r = 2\),
- \(n = 9\).

Substitute these values:
\[
S_9 = 2 \frac{2^9 - 1}{2 - 1}
\]
\[
S_9 = 2 \frac{512 - 1}{1}
\]
\[
S_9 = 2 \cdot 511 = 1022
\]

Thus, the sum is:
\[
\boxed{1022}
\]

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Problem 10:

In a geometric sequence, \(a_1 + a_2 = 8\) and \(a_3 + a_4 = 72\). Find the sum of the first 8 terms of the sequence.

#### Solution:
Let the first term be \(a\) and the common ratio be \(r\). Then:
\[
a_1 = a, \quad a_2 = ar, \quad a_3 = ar^2, \quad a_4 = ar^3
\]

From the given conditions:
1. \(a_1 + a_2 = 8\):
\[
a + ar = 8 \implies a(1 + r) = 8 \quad \text{(Equation 1)}
\]

2. \(a_3 + a_4 = 72\):
\[
ar^2 + ar^3 = 72 \implies ar^2(1 + r) = 72 \quad \text{(Equation 2)}
\]

Divide Equation 2 by Equation 1:
\[
\frac{ar^2(1 + r)}{a(1 + r)} = \frac{72}{8}
\]
\[
r^2 = 9 \implies r = 3 \quad (\text{since } r > 0)
\]

Substitute \(r = 3\) into Equation 1:
\[
a(1 + 3) = 8 \implies a \cdot 4 = 8 \implies a = 2
\]

Now, find the sum of the first 8 terms using the formula:
\[
S_n = a \frac{r^n - 1}{r - 1}
\]
Here:
- \(a = 2\),
- \(r = 3\),
- \(n = 8\).

Substitute these values:
\[
S_8 = 2 \frac{3^8 - 1}{3 - 1}
\]
\[
S_8 = 2 \frac{6561 - 1}{2}
\]
\[
S_8 = 2 \cdot 3280 = 6560
\]

Thus, the sum of the first 8 terms is:
\[
\boxed{6560}
\]

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Problem 16:

Find the sum to infinity of the following geometric sequences:
a. \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots\)
b. \(1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \cdots\)

#### Solution:
The sum to infinity of a geometric sequence is given by:
\[
S_\infty = \frac{a}{1 - r} \quad \text{if } |r| < 1
\]

##### Part (a):
The sequence is:
\[
\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots
\]
Here:
- \(a = \frac{1}{2}\),
- \(r = \frac{1}{2}\).

Since \(|r| = \frac{1}{2} < 1\), the sum to infinity is:
\[
S_\infty = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1
\]

##### Part (b):
The sequence is:
\[
1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \cdots
\]
Here:
- \(a = 1\),
- \(r = \frac{2}{3}\).

Since \(|r| = \frac{2}{3} < 1\), the sum to infinity is:
\[
S_\infty = \frac{1}{1 - \frac{2}{3}} = \frac{1}{\frac{1}{3}} = 3
\]

Thus, the sums to infinity are:
\[
\boxed{1 \text{ and } 3}
\]

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Final Answers:

1. Problem 3: \(\boxed{\frac{8525}{64}}\)
2. Problem 6: \(\boxed{1022}\)
3. Problem 10: \(\boxed{6560}\)
4. Problem 16: \(\boxed{1 \text{ and } 3}\)