Answer :
Sure, let's solve the problem step-by-step.
Given:
- The first term ([tex]\(t_1\)[/tex]) is 23.
- The third term ([tex]\(t_3\)[/tex]) is 92.
- The sum of all terms of the series is 62813.
### Step 1: Find the Common Ratio
We know that the third term ([tex]\(t_3\)[/tex]) can be expressed in terms of the first term ([tex]\(t_1\)[/tex]) and the common ratio ([tex]\(r\)[/tex]):
[tex]\[ t_3 = t_1 \cdot r^{3-1} \][/tex]
[tex]\[ 92 = 23 \cdot r^2 \][/tex]
From this equation, we can solve for [tex]\(r^2\)[/tex]:
[tex]\[ r^2 = \frac{92}{23} \][/tex]
[tex]\[ r^2 = 4 \][/tex]
Taking the square root of both sides to find [tex]\(r\)[/tex]:
[tex]\[ r = \sqrt{4} \][/tex]
[tex]\[ r = 2 \][/tex]
### Step 2: Use the Sum of the Geometric Series Formula
The sum of a geometric series can be expressed with the formula:
[tex]\[ S_n = t_1 \frac{r^n - 1}{r - 1} \][/tex]
Given that the sum [tex]\(S_n\)[/tex] is 62813, we can substitute the known values into the formula:
[tex]\[ 62813 = 23 \frac{2^n - 1}{2 - 1} \][/tex]
[tex]\[ 62813 = 23 \left(2^n - 1\right) \][/tex]
We can now isolate [tex]\(2^n - 1\)[/tex] by dividing both sides of the equation by 23:
[tex]\[ \frac{62813}{23} = 2^n - 1 \][/tex]
[tex]\[ 2732 = 2^n - 1 \][/tex]
Add 1 to both sides to solve for [tex]\(2^n\)[/tex]:
[tex]\[ 2732 + 1 = 2^n \][/tex]
[tex]\[ 2733 = 2^n \][/tex]
### Step 3: Solve for the Number of Terms ([tex]\(n\)[/tex])
To find [tex]\(n\)[/tex], we use the logarithm:
[tex]\[ 2^n = 2733 \][/tex]
Taking the natural logarithm (or log base 2) on both sides:
[tex]\[ \log_2(2733) = n \][/tex]
Converting to natural logarithm:
[tex]\[ n = \frac{\log(2733)}{\log(2)} \][/tex]
Using the known values for the logs:
[tex]\[ n \approx \frac{\log(2733)}{\log(2)} \approx 11.415741768290092 \][/tex]
Therefore, the number of terms in the series is approximately 11.42, which when rounded down, is 11 full terms in the series, considering that in practical settings, fractional terms may suggest the series is close to but does not exceed a certain number of terms. Thus, the series effectively had 11 terms.
Given:
- The first term ([tex]\(t_1\)[/tex]) is 23.
- The third term ([tex]\(t_3\)[/tex]) is 92.
- The sum of all terms of the series is 62813.
### Step 1: Find the Common Ratio
We know that the third term ([tex]\(t_3\)[/tex]) can be expressed in terms of the first term ([tex]\(t_1\)[/tex]) and the common ratio ([tex]\(r\)[/tex]):
[tex]\[ t_3 = t_1 \cdot r^{3-1} \][/tex]
[tex]\[ 92 = 23 \cdot r^2 \][/tex]
From this equation, we can solve for [tex]\(r^2\)[/tex]:
[tex]\[ r^2 = \frac{92}{23} \][/tex]
[tex]\[ r^2 = 4 \][/tex]
Taking the square root of both sides to find [tex]\(r\)[/tex]:
[tex]\[ r = \sqrt{4} \][/tex]
[tex]\[ r = 2 \][/tex]
### Step 2: Use the Sum of the Geometric Series Formula
The sum of a geometric series can be expressed with the formula:
[tex]\[ S_n = t_1 \frac{r^n - 1}{r - 1} \][/tex]
Given that the sum [tex]\(S_n\)[/tex] is 62813, we can substitute the known values into the formula:
[tex]\[ 62813 = 23 \frac{2^n - 1}{2 - 1} \][/tex]
[tex]\[ 62813 = 23 \left(2^n - 1\right) \][/tex]
We can now isolate [tex]\(2^n - 1\)[/tex] by dividing both sides of the equation by 23:
[tex]\[ \frac{62813}{23} = 2^n - 1 \][/tex]
[tex]\[ 2732 = 2^n - 1 \][/tex]
Add 1 to both sides to solve for [tex]\(2^n\)[/tex]:
[tex]\[ 2732 + 1 = 2^n \][/tex]
[tex]\[ 2733 = 2^n \][/tex]
### Step 3: Solve for the Number of Terms ([tex]\(n\)[/tex])
To find [tex]\(n\)[/tex], we use the logarithm:
[tex]\[ 2^n = 2733 \][/tex]
Taking the natural logarithm (or log base 2) on both sides:
[tex]\[ \log_2(2733) = n \][/tex]
Converting to natural logarithm:
[tex]\[ n = \frac{\log(2733)}{\log(2)} \][/tex]
Using the known values for the logs:
[tex]\[ n \approx \frac{\log(2733)}{\log(2)} \approx 11.415741768290092 \][/tex]
Therefore, the number of terms in the series is approximately 11.42, which when rounded down, is 11 full terms in the series, considering that in practical settings, fractional terms may suggest the series is close to but does not exceed a certain number of terms. Thus, the series effectively had 11 terms.
