Chapter12: Sequences, Series And Binomial Theorem
Section: Chapter Questions
Problem 327PT: Find the twenty-third term of an arithmetic sequence whose seventh term is 11 and common difference...
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Question
Geo
![### Progression of the Geometric Sequence
**Task:**
Find the sum of the 10th term of the geometric progression given below:
\[
-5, 10, -20, \ldots
\]
**Details:**
- The geometric sequence presented starts with the terms -5, 10, -20, and continues. Find the sum of the first 10 terms.
**Explanation:**
Given that we have a geometric progression:
\[ a_1 = -5 \]
\[ a_2 = 10 \]
\[ a_3 = -20 \]
we can find the common ratio (r) by dividing the second term by the first term ( \( r = \frac{a_2}{a_1} = \frac{10}{-5} = -2 \) ).
Taking these values, we can use the sum of the first n terms (S_n) of a geometric series formula for n terms (if \(|r| > 1\)):
\[ S_n = a \left( \frac{r^n - 1}{r - 1} \right) \]
In this particular case:
\[ S_{10} = -5 \left( \frac{(-2)^{10} - 1}{-2 - 1} \right) \]
Calculation:
\[
(-2)^{10} = 1024
\]
\[
S_{10} = -5 \left( \frac{1024 - 1}{-3} \right) = -5 \left( \frac{1023}{-3} \right) = -5 (-341) = 1705
\]
Thus, the sum of the first 10 terms is 1705.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66c7b497-dac7-4855-b923-2e60bbc73063%2Fa13a8d38-9d82-449b-bb5e-5c5559431ed7%2Fwguu5ox_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Progression of the Geometric Sequence
**Task:**
Find the sum of the 10th term of the geometric progression given below:
\[
-5, 10, -20, \ldots
\]
**Details:**
- The geometric sequence presented starts with the terms -5, 10, -20, and continues. Find the sum of the first 10 terms.
**Explanation:**
Given that we have a geometric progression:
\[ a_1 = -5 \]
\[ a_2 = 10 \]
\[ a_3 = -20 \]
we can find the common ratio (r) by dividing the second term by the first term ( \( r = \frac{a_2}{a_1} = \frac{10}{-5} = -2 \) ).
Taking these values, we can use the sum of the first n terms (S_n) of a geometric series formula for n terms (if \(|r| > 1\)):
\[ S_n = a \left( \frac{r^n - 1}{r - 1} \right) \]
In this particular case:
\[ S_{10} = -5 \left( \frac{(-2)^{10} - 1}{-2 - 1} \right) \]
Calculation:
\[
(-2)^{10} = 1024
\]
\[
S_{10} = -5 \left( \frac{1024 - 1}{-3} \right) = -5 \left( \frac{1023}{-3} \right) = -5 (-341) = 1705
\]
Thus, the sum of the first 10 terms is 1705.
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