Uncover valuable information and solutions with IDNLearn.com's extensive Q&A platform. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.
Sagot :
Sure, let's solve each part of these problems step-by-step.
### Problem b
Let's start with the first part of problem b.
#### Given:
- GP Series: [tex]\(5 + 10 + 20 + \ldots\)[/tex]
- Sum of terms to be equal to 315.
This is a geometric progression (GP) with:
- First term, [tex]\( a = 5 \)[/tex]
- Common ratio, [tex]\( r = \frac{10}{5} = 2 \)[/tex]
- Sum of the first [tex]\( n \)[/tex] terms, [tex]\( S_n = 315 \)[/tex]
The formula for the sum of the first [tex]\( n \)[/tex] terms of a geometric progression is:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
Plugging in the values:
[tex]\[ 315 = 5 \cdot \frac{2^n - 1}{2 - 1} \][/tex]
This simplifies to:
[tex]\[ 315 = 5 \cdot (2^n - 1) \][/tex]
[tex]\[ 63 = 2^n - 1 \][/tex]
[tex]\[ 2^n = 64 \][/tex]
Recognize that [tex]\( 64 = 2^6 \)[/tex], so:
[tex]\[ n = 6 \][/tex]
Thus, 6 terms of the series [tex]\(5 + 10 + 20 + \ldots\)[/tex] must be taken so that the sum becomes 315.
### Problem a
Next part (a) of problem b:
#### Given:
- 3rd term, [tex]\( a_3 = 8 \)[/tex]
- 7th term, [tex]\( a_7 = 128 \)[/tex]
In a geometric progression, the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
So, for the 3rd term:
[tex]\[ a \cdot r^2 = 8 \quad \text{(1)} \][/tex]
And for the 7th term:
[tex]\[ a \cdot r^6 = 128 \quad \text{(2)} \][/tex]
Dividing equation (2) by equation (1):
[tex]\[ \frac{a \cdot r^6}{a \cdot r^2} = \frac{128}{8} \][/tex]
[tex]\[ r^4 = 16 \][/tex]
[tex]\[ r = \sqrt[4]{16} = 2 \][/tex]
Substitute [tex]\(r = 2\)[/tex] back into equation (1):
[tex]\[ a \cdot 2^2 = 8 \][/tex]
[tex]\[ a \cdot 4 = 8 \][/tex]
[tex]\[ a = 2 \][/tex]
Now, to find the sum of the first 8 terms, we use the sum formula for the first [tex]\(n\)[/tex] terms of a GP:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For the first 8 terms:
[tex]\[ S_8 = 2 \cdot \frac{2^8 - 1}{2 - 1} \][/tex]
[tex]\[ S_8 = 2 \cdot (256 - 1) \][/tex]
[tex]\[ S_8 = 2 \cdot 255 \][/tex]
[tex]\[ S_8 = 510 \][/tex]
Thus, the sum of the first 8 terms of this GP is 510.
### Problem b
#### Given:
- 4th term, [tex]\( a_4 = 27 \)[/tex]
- 7th term, [tex]\( a_7 = 729 \)[/tex]
Again, in a geometric progression, the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
So, for the 4th term:
[tex]\[ a \cdot r^3 = 27 \quad \text{(3)} \][/tex]
And for the 7th term:
[tex]\[ a \cdot r^6 = 729 \quad \text{(4)} \][/tex]
Dividing equation (4) by equation (3):
[tex]\[ \frac{a \cdot r^6}{a \cdot r^3} = \frac{729}{27} \][/tex]
[tex]\[ r^3 = 27 \][/tex]
[tex]\[ r = \sqrt[3]{27} = 3 \][/tex]
Substitute [tex]\(r = 3\)[/tex] back into equation (3):
[tex]\[ a \cdot 3^3 = 27 \][/tex]
[tex]\[ a \cdot 27 = 27 \][/tex]
[tex]\[ a = 1 \][/tex]
Hence, [tex]\( a = 1 \)[/tex] and [tex]\( r = 3 \)[/tex].
i) Common ratio and first term:
- Common ratio [tex]\( r = 3 \)[/tex]
- First term [tex]\( a = 1 \)[/tex]
ii) Geometric Series:
First 8 terms of the series:
[tex]\[ a, ar, ar^2, ar^3, ar^4, ar^5, ar^6, ar^7 \][/tex]
Calculating:
[tex]\[ 1, 1 \cdot 3, 1 \cdot 3^2, 1 \cdot 3^3, 1 \cdot 3^4, 1 \cdot 3^5, 1 \cdot 3^6, 1 \cdot 3^7 \][/tex]
[tex]\[ 1, 3, 9, 27, 81, 243, 729, 2187 \][/tex]
So, the series is [tex]\( 1, 3, 9, 27, 81, 243, 729, 2187 \)[/tex].
iii) Sum of first 8 terms:
Using the sum formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For the first 8 terms:
[tex]\[ S_8 = 1 \cdot \frac{3^8 - 1}{3 - 1} \][/tex]
[tex]\[ S_8 = \frac{6561 - 1}{2} \][/tex]
[tex]\[ S_8 = \frac{6560}{2} \][/tex]
[tex]\[ S_8 = 3280 \][/tex]
Thus, the sum of the first 8 terms of this GP is 3280.
### Summarized Answer
1. Number of terms for the sum to be 315:
- [tex]\( n = 6 \)[/tex]
2. Sum of the first 8 terms of GP where [tex]\( a_3 = 8 \)[/tex] and [tex]\( a_7 = 128 \)[/tex]:
- [tex]\( S_8 = 510 \)[/tex]
3. For GP where [tex]\( a_4 = 27 \)[/tex] and [tex]\( a_7 = 729 \)[/tex]:
- i) Common ratio [tex]\( r = 3 \)[/tex] and first term [tex]\( a = 1 \)[/tex]
- ii) Geometric series: [tex]\( 1, 3, 9, 27, 81, 243, 729, 2187 \)[/tex]
- iii) Sum of first 8 terms: [tex]\( S_8 = 3280 \)[/tex]
### Problem b
Let's start with the first part of problem b.
#### Given:
- GP Series: [tex]\(5 + 10 + 20 + \ldots\)[/tex]
- Sum of terms to be equal to 315.
This is a geometric progression (GP) with:
- First term, [tex]\( a = 5 \)[/tex]
- Common ratio, [tex]\( r = \frac{10}{5} = 2 \)[/tex]
- Sum of the first [tex]\( n \)[/tex] terms, [tex]\( S_n = 315 \)[/tex]
The formula for the sum of the first [tex]\( n \)[/tex] terms of a geometric progression is:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
Plugging in the values:
[tex]\[ 315 = 5 \cdot \frac{2^n - 1}{2 - 1} \][/tex]
This simplifies to:
[tex]\[ 315 = 5 \cdot (2^n - 1) \][/tex]
[tex]\[ 63 = 2^n - 1 \][/tex]
[tex]\[ 2^n = 64 \][/tex]
Recognize that [tex]\( 64 = 2^6 \)[/tex], so:
[tex]\[ n = 6 \][/tex]
Thus, 6 terms of the series [tex]\(5 + 10 + 20 + \ldots\)[/tex] must be taken so that the sum becomes 315.
### Problem a
Next part (a) of problem b:
#### Given:
- 3rd term, [tex]\( a_3 = 8 \)[/tex]
- 7th term, [tex]\( a_7 = 128 \)[/tex]
In a geometric progression, the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
So, for the 3rd term:
[tex]\[ a \cdot r^2 = 8 \quad \text{(1)} \][/tex]
And for the 7th term:
[tex]\[ a \cdot r^6 = 128 \quad \text{(2)} \][/tex]
Dividing equation (2) by equation (1):
[tex]\[ \frac{a \cdot r^6}{a \cdot r^2} = \frac{128}{8} \][/tex]
[tex]\[ r^4 = 16 \][/tex]
[tex]\[ r = \sqrt[4]{16} = 2 \][/tex]
Substitute [tex]\(r = 2\)[/tex] back into equation (1):
[tex]\[ a \cdot 2^2 = 8 \][/tex]
[tex]\[ a \cdot 4 = 8 \][/tex]
[tex]\[ a = 2 \][/tex]
Now, to find the sum of the first 8 terms, we use the sum formula for the first [tex]\(n\)[/tex] terms of a GP:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For the first 8 terms:
[tex]\[ S_8 = 2 \cdot \frac{2^8 - 1}{2 - 1} \][/tex]
[tex]\[ S_8 = 2 \cdot (256 - 1) \][/tex]
[tex]\[ S_8 = 2 \cdot 255 \][/tex]
[tex]\[ S_8 = 510 \][/tex]
Thus, the sum of the first 8 terms of this GP is 510.
### Problem b
#### Given:
- 4th term, [tex]\( a_4 = 27 \)[/tex]
- 7th term, [tex]\( a_7 = 729 \)[/tex]
Again, in a geometric progression, the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
So, for the 4th term:
[tex]\[ a \cdot r^3 = 27 \quad \text{(3)} \][/tex]
And for the 7th term:
[tex]\[ a \cdot r^6 = 729 \quad \text{(4)} \][/tex]
Dividing equation (4) by equation (3):
[tex]\[ \frac{a \cdot r^6}{a \cdot r^3} = \frac{729}{27} \][/tex]
[tex]\[ r^3 = 27 \][/tex]
[tex]\[ r = \sqrt[3]{27} = 3 \][/tex]
Substitute [tex]\(r = 3\)[/tex] back into equation (3):
[tex]\[ a \cdot 3^3 = 27 \][/tex]
[tex]\[ a \cdot 27 = 27 \][/tex]
[tex]\[ a = 1 \][/tex]
Hence, [tex]\( a = 1 \)[/tex] and [tex]\( r = 3 \)[/tex].
i) Common ratio and first term:
- Common ratio [tex]\( r = 3 \)[/tex]
- First term [tex]\( a = 1 \)[/tex]
ii) Geometric Series:
First 8 terms of the series:
[tex]\[ a, ar, ar^2, ar^3, ar^4, ar^5, ar^6, ar^7 \][/tex]
Calculating:
[tex]\[ 1, 1 \cdot 3, 1 \cdot 3^2, 1 \cdot 3^3, 1 \cdot 3^4, 1 \cdot 3^5, 1 \cdot 3^6, 1 \cdot 3^7 \][/tex]
[tex]\[ 1, 3, 9, 27, 81, 243, 729, 2187 \][/tex]
So, the series is [tex]\( 1, 3, 9, 27, 81, 243, 729, 2187 \)[/tex].
iii) Sum of first 8 terms:
Using the sum formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
For the first 8 terms:
[tex]\[ S_8 = 1 \cdot \frac{3^8 - 1}{3 - 1} \][/tex]
[tex]\[ S_8 = \frac{6561 - 1}{2} \][/tex]
[tex]\[ S_8 = \frac{6560}{2} \][/tex]
[tex]\[ S_8 = 3280 \][/tex]
Thus, the sum of the first 8 terms of this GP is 3280.
### Summarized Answer
1. Number of terms for the sum to be 315:
- [tex]\( n = 6 \)[/tex]
2. Sum of the first 8 terms of GP where [tex]\( a_3 = 8 \)[/tex] and [tex]\( a_7 = 128 \)[/tex]:
- [tex]\( S_8 = 510 \)[/tex]
3. For GP where [tex]\( a_4 = 27 \)[/tex] and [tex]\( a_7 = 729 \)[/tex]:
- i) Common ratio [tex]\( r = 3 \)[/tex] and first term [tex]\( a = 1 \)[/tex]
- ii) Geometric series: [tex]\( 1, 3, 9, 27, 81, 243, 729, 2187 \)[/tex]
- iii) Sum of first 8 terms: [tex]\( S_8 = 3280 \)[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.
