In an exponential sequence, each term can be expressed as:
\[ a_n = a_1 \cdot r^{n-1} \]
where \( a_n \) is the \( n \)-th term, \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
From the information provided:
- The fifth term (\( a_5 \)) is 475.
- The second term (\( a_2 \)) is 35.
Using the definition of an exponential sequence, we have:
\[ a_5 = a_1 \cdot r^{4} = 475 \] \[ a_2 = a_1 \cdot r^{1} = 35 \]
Now we can express \( a_1 \) in terms of \( r \):
\[ a_1 = \frac{35}{r} \]
Next, substitute this expression for \( a_1 \) into the equation for \( a_5 \):
\[ 475 = \left(\frac{35}{r}\right) \cdot r^{4} \]
This simplifies to:
\[ 475 = 35r^{3} \]
Solving for \( r^{3} \):
\[ r^{3} = \frac{475}{35} \]
Calculating the fraction:
\[ r^{3} = 13.57142857 \quad (\text{approx. } 13.57) \]
To find \( r \):
\[ r = \sqrt[3]{13.57142857} \approx 2.4027 \]
Now we can calculate \( a_1 \):
\[ a_1 = \frac{35}{r} \approx \frac{35}{2.4027} \approx 14.591 \]
Finding the third term
The third term (\( a_3 \)) is given by:
\[ a_3 = a_1 \cdot r^{2} \]
Calculating \( a_3 \):
\[ a_3 \approx 14.591 \cdot (2.4027)^2 \approx 14.591 \cdot 5.7757 \approx 84.36 \]
Finding the sixth term
The sixth term (\( a_6 \)) can be calculated using:
\[ a_6 = a_1 \cdot r^{5} \]
Calculating \( a_6 \):
\[ a_6 \approx 14.591 \cdot (2.4027)^5 \approx 14.591 \cdot 61.5081 \approx 897.06 \]
Sum of the first five terms
The sum of the first five terms (\( S_5 \)) in an exponential sequence can be calculated using the formula:
\[ S_5 = a_1 \frac{r^5 - 1}{r - 1} \]
Calculating \( S_5 \):
\[ S_5 \approx 14.591 \cdot \frac{(2.4027)^5 - 1}{2.4027 - 1} \approx 14.591 \cdot \frac{61.5081 - 1}{1.4027} \approx 14.591 \cdot \frac{60.5081}{1.4027} \approx 14.591 \cdot 43.1559 \approx 630.75 \]
Summary of Results
- The third term (\( a_3 \)) is approximately 84.36.
- The sixth term (\( a_6 \)) is approximately 897.06.
- The sum of the first five terms (\( S_5 \)) is approximately 630.75.