Answer :
We are given a geometric series with first term
[tex]$$
a_1 = 3125,
$$[/tex]
last term
[tex]$$
a_n = 1,
$$[/tex]
and common ratio
[tex]$$
r = \frac{1}{5}.
$$[/tex]
### Step 1. Determine the Number of Terms
The formula for the [tex]\( n \)[/tex]th term of a geometric series is
[tex]$$
a_n = a_1 \cdot r^{n-1}.
$$[/tex]
Substitute the given values:
[tex]$$
1 = 3125 \cdot \left(\frac{1}{5}\right)^{n-1}.
$$[/tex]
Notice that [tex]\(3125\)[/tex] can be written as [tex]\(5^5\)[/tex] because
[tex]$$
3125 = 5^5.
$$[/tex]
Substitute this into the equation:
[tex]$$
1 = 5^5 \cdot \left(\frac{1}{5}\right)^{n-1}.
$$[/tex]
Recall that [tex]\(\left(\frac{1}{5}\right)^{n-1} = 5^{-(n-1)}\)[/tex]. Therefore, the equation becomes:
[tex]$$
1 = 5^5 \cdot 5^{-(n-1)} = 5^{5 - (n-1)} = 5^{6-n}.
$$[/tex]
Since any power of 5 equals 1 only if the exponent is 0, we have
[tex]$$
6 - n = 0.
$$[/tex]
Thus,
[tex]$$
n = 6.
$$[/tex]
### Step 2. Calculate the Sum of the Series
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a geometric series is given by
[tex]$$
S_n = a_1 \cdot \frac{1 - r^n}{1 - r}, \quad \text{for } r \neq 1.
$$[/tex]
Substitute [tex]\( a_1 = 3125 \)[/tex], [tex]\( r = \frac{1}{5} \)[/tex], and [tex]\( n = 6 \)[/tex] into the formula:
[tex]$$
S_6 = 3125 \cdot \frac{1 - \left(\frac{1}{5}\right)^6}{1 - \frac{1}{5}}.
$$[/tex]
#### Compute the Denominator:
[tex]$$
1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5}.
$$[/tex]
#### Compute [tex]\( r^n \)[/tex]:
[tex]$$
\left(\frac{1}{5}\right)^6 = \frac{1}{5^6} = \frac{1}{15625}.
$$[/tex]
#### Substitute Back into the Sum Formula:
[tex]$$
S_6 = 3125 \cdot \frac{1 - \frac{1}{15625}}{\frac{4}{5}}.
$$[/tex]
Simplify the numerator:
[tex]$$
1 - \frac{1}{15625} = \frac{15625 - 1}{15625} = \frac{15624}{15625}.
$$[/tex]
The expression for [tex]\( S_6 \)[/tex] is now:
[tex]$$
S_6 = 3125 \cdot \frac{\frac{15624}{15625}}{\frac{4}{5}}.
$$[/tex]
Dividing by a fraction is equivalent to multiplying by its reciprocal:
[tex]$$
S_6 = 3125 \cdot \frac{15624}{15625} \cdot \frac{5}{4}.
$$[/tex]
### Step 3. Write the Final Answer
After performing the arithmetic calculations, we obtain:
[tex]$$
S_6 = 3906.
$$[/tex]
Thus, the sum of the series is
[tex]$$
\boxed{3906}.
$$[/tex]
[tex]$$
a_1 = 3125,
$$[/tex]
last term
[tex]$$
a_n = 1,
$$[/tex]
and common ratio
[tex]$$
r = \frac{1}{5}.
$$[/tex]
### Step 1. Determine the Number of Terms
The formula for the [tex]\( n \)[/tex]th term of a geometric series is
[tex]$$
a_n = a_1 \cdot r^{n-1}.
$$[/tex]
Substitute the given values:
[tex]$$
1 = 3125 \cdot \left(\frac{1}{5}\right)^{n-1}.
$$[/tex]
Notice that [tex]\(3125\)[/tex] can be written as [tex]\(5^5\)[/tex] because
[tex]$$
3125 = 5^5.
$$[/tex]
Substitute this into the equation:
[tex]$$
1 = 5^5 \cdot \left(\frac{1}{5}\right)^{n-1}.
$$[/tex]
Recall that [tex]\(\left(\frac{1}{5}\right)^{n-1} = 5^{-(n-1)}\)[/tex]. Therefore, the equation becomes:
[tex]$$
1 = 5^5 \cdot 5^{-(n-1)} = 5^{5 - (n-1)} = 5^{6-n}.
$$[/tex]
Since any power of 5 equals 1 only if the exponent is 0, we have
[tex]$$
6 - n = 0.
$$[/tex]
Thus,
[tex]$$
n = 6.
$$[/tex]
### Step 2. Calculate the Sum of the Series
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a geometric series is given by
[tex]$$
S_n = a_1 \cdot \frac{1 - r^n}{1 - r}, \quad \text{for } r \neq 1.
$$[/tex]
Substitute [tex]\( a_1 = 3125 \)[/tex], [tex]\( r = \frac{1}{5} \)[/tex], and [tex]\( n = 6 \)[/tex] into the formula:
[tex]$$
S_6 = 3125 \cdot \frac{1 - \left(\frac{1}{5}\right)^6}{1 - \frac{1}{5}}.
$$[/tex]
#### Compute the Denominator:
[tex]$$
1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5}.
$$[/tex]
#### Compute [tex]\( r^n \)[/tex]:
[tex]$$
\left(\frac{1}{5}\right)^6 = \frac{1}{5^6} = \frac{1}{15625}.
$$[/tex]
#### Substitute Back into the Sum Formula:
[tex]$$
S_6 = 3125 \cdot \frac{1 - \frac{1}{15625}}{\frac{4}{5}}.
$$[/tex]
Simplify the numerator:
[tex]$$
1 - \frac{1}{15625} = \frac{15625 - 1}{15625} = \frac{15624}{15625}.
$$[/tex]
The expression for [tex]\( S_6 \)[/tex] is now:
[tex]$$
S_6 = 3125 \cdot \frac{\frac{15624}{15625}}{\frac{4}{5}}.
$$[/tex]
Dividing by a fraction is equivalent to multiplying by its reciprocal:
[tex]$$
S_6 = 3125 \cdot \frac{15624}{15625} \cdot \frac{5}{4}.
$$[/tex]
### Step 3. Write the Final Answer
After performing the arithmetic calculations, we obtain:
[tex]$$
S_6 = 3906.
$$[/tex]
Thus, the sum of the series is
[tex]$$
\boxed{3906}.
$$[/tex]
