Answer :
To solve this problem, we need to follow a series of calculations step-by-step. Let's start with each given part:
### (a) Finding the value of x
We are given three consecutive terms of a geometric progression (GP): [tex]\( x + 1 \)[/tex], [tex]\( x - 3 \)[/tex], and [tex]\( x - 6 \)[/tex]. For any three consecutive terms of a GP, the middle term squared is equal to the product of the first and the third terms. Therefore:
[tex]\[ (x - 3)^2 = (x + 1)(x - 6) \][/tex]
Expand and simplify:
[tex]\[ (x - 3)(x - 3) = (x + 1)(x - 6) \][/tex]
[tex]\[ x^2 - 6x + 9 = x^2 - 5x - 6 \][/tex]
Subtract [tex]\( x^2 \)[/tex] from both sides:
[tex]\[ -6x + 9 = -5x - 6 \][/tex]
Add [tex]\( 5x \)[/tex] to both sides:
[tex]\[ -x + 9 = -6 \][/tex]
Subtract 9 from both sides:
[tex]\[ -x = -15 \][/tex]
So,
[tex]\[ x = 15 \][/tex]
### (b) Finding the common ratio [tex]\( r \)[/tex]
Having found [tex]\( x = 15 \)[/tex]:
The terms are [tex]\( x+1 = 16 \)[/tex], [tex]\( x-3 = 12 \)[/tex], and [tex]\( x-6 = 9 \)[/tex].
The common ratio [tex]\( r \)[/tex] in a GP is the ratio of successive terms. Thus:
[tex]\[ r = \frac{12}{16} = \frac{3}{4} \][/tex]
### (c) Finding the first term given that [tex]\( x + 1 \)[/tex] is the 4th term of the GP
We know that [tex]\( x + 1 = 16 \)[/tex] is the 4th term of the GP. To find the first term [tex]\( a \)[/tex]:
The [tex]\( n \)[/tex]-th term of a GP is given by [tex]\( a \cdot r^{n-1} \)[/tex].
For the 4th term:
[tex]\[ a \cdot r^3 = 16 \][/tex]
Substitute the common ratio [tex]\( r = \frac{3}{4} \)[/tex]:
[tex]\[ a \left( \frac{3}{4} \right)^3 = 16 \][/tex]
[tex]\[ a \cdot \frac{27}{64} = 16 \][/tex]
Multiply both sides by [tex]\( \frac{64}{27} \)[/tex]:
[tex]\[ a = 16 \cdot \frac{64}{27} \][/tex]
[tex]\[ a = \frac{1024}{27} \][/tex]
### (d) Finding the sum of the first 5 terms correct to 3 significant figures
The sum of the first [tex]\( n \)[/tex] terms of a geometric progression is given by:
[tex]\[ S_n = a \cdot \frac{r^n - 1}{r - 1} \][/tex]
We need to find the sum of the first 5 terms ([tex]\( S_5 \)[/tex]):
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{\left( \frac{3}{4} \right)^5 - 1}{\frac{3}{4} - 1} \][/tex]
First, compute [tex]\( \left( \frac{3}{4} \right)^5 \)[/tex]:
[tex]\[ \left( \frac{3}{4} \right)^5 = \frac{243}{1024} \][/tex]
Next, compute the denominator [tex]\( \frac{3}{4} - 1 \)[/tex]:
[tex]\[ \frac{3}{4} - 1 = -\frac{1}{4} \][/tex]
Substitute these values into the sum formula:
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{\frac{243}{1024} - 1}{-\frac{1}{4}} \][/tex]
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{\frac{243 - 1024}{1024}}{-\frac{1}{4}} \][/tex]
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{-781}{1024} \cdot \left(-4\right) \][/tex]
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{3124}{1024} \][/tex]
[tex]\[ S_5 = \frac{3124}{27} \][/tex]
Evaluating this quotient:
[tex]\[ S_5 \approx 115.704 \][/tex]
Thus, the sum of the first 5 terms, correct to three significant figures, is:
[tex]\[ S_5 \approx 115.704 \][/tex]
So, the steps and results are:
- (a) [tex]\( x = 15 \)[/tex]
- (b) The common ratio [tex]\( r = \frac{3}{4} \)[/tex]
- (c) The first term [tex]\( a = \frac{1024}{27} \)[/tex]
- (d) The sum of the first 5 terms [tex]\( S_5 \approx 115.704 \)[/tex]
### (a) Finding the value of x
We are given three consecutive terms of a geometric progression (GP): [tex]\( x + 1 \)[/tex], [tex]\( x - 3 \)[/tex], and [tex]\( x - 6 \)[/tex]. For any three consecutive terms of a GP, the middle term squared is equal to the product of the first and the third terms. Therefore:
[tex]\[ (x - 3)^2 = (x + 1)(x - 6) \][/tex]
Expand and simplify:
[tex]\[ (x - 3)(x - 3) = (x + 1)(x - 6) \][/tex]
[tex]\[ x^2 - 6x + 9 = x^2 - 5x - 6 \][/tex]
Subtract [tex]\( x^2 \)[/tex] from both sides:
[tex]\[ -6x + 9 = -5x - 6 \][/tex]
Add [tex]\( 5x \)[/tex] to both sides:
[tex]\[ -x + 9 = -6 \][/tex]
Subtract 9 from both sides:
[tex]\[ -x = -15 \][/tex]
So,
[tex]\[ x = 15 \][/tex]
### (b) Finding the common ratio [tex]\( r \)[/tex]
Having found [tex]\( x = 15 \)[/tex]:
The terms are [tex]\( x+1 = 16 \)[/tex], [tex]\( x-3 = 12 \)[/tex], and [tex]\( x-6 = 9 \)[/tex].
The common ratio [tex]\( r \)[/tex] in a GP is the ratio of successive terms. Thus:
[tex]\[ r = \frac{12}{16} = \frac{3}{4} \][/tex]
### (c) Finding the first term given that [tex]\( x + 1 \)[/tex] is the 4th term of the GP
We know that [tex]\( x + 1 = 16 \)[/tex] is the 4th term of the GP. To find the first term [tex]\( a \)[/tex]:
The [tex]\( n \)[/tex]-th term of a GP is given by [tex]\( a \cdot r^{n-1} \)[/tex].
For the 4th term:
[tex]\[ a \cdot r^3 = 16 \][/tex]
Substitute the common ratio [tex]\( r = \frac{3}{4} \)[/tex]:
[tex]\[ a \left( \frac{3}{4} \right)^3 = 16 \][/tex]
[tex]\[ a \cdot \frac{27}{64} = 16 \][/tex]
Multiply both sides by [tex]\( \frac{64}{27} \)[/tex]:
[tex]\[ a = 16 \cdot \frac{64}{27} \][/tex]
[tex]\[ a = \frac{1024}{27} \][/tex]
### (d) Finding the sum of the first 5 terms correct to 3 significant figures
The sum of the first [tex]\( n \)[/tex] terms of a geometric progression is given by:
[tex]\[ S_n = a \cdot \frac{r^n - 1}{r - 1} \][/tex]
We need to find the sum of the first 5 terms ([tex]\( S_5 \)[/tex]):
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{\left( \frac{3}{4} \right)^5 - 1}{\frac{3}{4} - 1} \][/tex]
First, compute [tex]\( \left( \frac{3}{4} \right)^5 \)[/tex]:
[tex]\[ \left( \frac{3}{4} \right)^5 = \frac{243}{1024} \][/tex]
Next, compute the denominator [tex]\( \frac{3}{4} - 1 \)[/tex]:
[tex]\[ \frac{3}{4} - 1 = -\frac{1}{4} \][/tex]
Substitute these values into the sum formula:
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{\frac{243}{1024} - 1}{-\frac{1}{4}} \][/tex]
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{\frac{243 - 1024}{1024}}{-\frac{1}{4}} \][/tex]
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{-781}{1024} \cdot \left(-4\right) \][/tex]
[tex]\[ S_5 = \frac{1024}{27} \cdot \frac{3124}{1024} \][/tex]
[tex]\[ S_5 = \frac{3124}{27} \][/tex]
Evaluating this quotient:
[tex]\[ S_5 \approx 115.704 \][/tex]
Thus, the sum of the first 5 terms, correct to three significant figures, is:
[tex]\[ S_5 \approx 115.704 \][/tex]
So, the steps and results are:
- (a) [tex]\( x = 15 \)[/tex]
- (b) The common ratio [tex]\( r = \frac{3}{4} \)[/tex]
- (c) The first term [tex]\( a = \frac{1024}{27} \)[/tex]
- (d) The sum of the first 5 terms [tex]\( S_5 \approx 115.704 \)[/tex]
