Answer :
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]
