Answer :
Sure! I'll address each math problem in sequence and provide a step-by-step solution.
### Problem 3
Solve for [tex]\( p \)[/tex] given [tex]\( S_n = p\left(1 - \left(\frac{1}{2}\right)^1\right) \)[/tex] and [tex]\( S_x = 10 \)[/tex].
### Solution:
[tex]\[ S_n = p \left(1 - \left(\frac{1}{2}\right)\right) \][/tex]
Since [tex]\( \left(\frac{1}{2}\right)^1 = \frac{1}{2} \)[/tex], we have:
[tex]\[ S_n = p \left(1 - \frac{1}{2}\right) \][/tex]
[tex]\[ S_n = p \left(\frac{1}{2}\right) \][/tex]
We know [tex]\( S_x = 10 \)[/tex]:
[tex]\[ 10 = p \left(\frac{1}{2}\right) \][/tex]
Solve for [tex]\( p \)[/tex]:
[tex]\[ p = 10 \times 2 \][/tex]
[tex]\[ p = 20 \][/tex]
### Problem 4
Consider [tex]\( T_n = (4x - 1)^{11} \)[/tex].
#### (a) Find the value(s) of [tex]\( x \)[/tex] for which [tex]\( S_{\infty} \)[/tex] exists.
### Solution:
For a geometric series [tex]\( S_{\infty} \)[/tex] to exist, the common ratio, [tex]\( r \)[/tex], must satisfy [tex]\( |r| < 1 \)[/tex].
Given that [tex]\( T_n = (4x - 1)^{11} \)[/tex], consider the base [tex]\( r = 4x - 1 \)[/tex].
For [tex]\( |r| < 1 \)[/tex]:
[tex]\[ |4x - 1| < 1 \][/tex]
Solve the inequality:
[tex]\[ -1 < 4x - 1 < 1 \][/tex]
Add 1 to all parts:
[tex]\[ 0 < 4x < 2 \][/tex]
Divide by 4:
[tex]\[ 0 < x < \frac{1}{2} \][/tex]
The values of [tex]\( x \)[/tex] for which [tex]\( S_{\infty} \)[/tex] exists are [tex]\( 0 < x < \frac{1}{2} \)[/tex].
#### (b) Choose any value for [tex]\( x \)[/tex] in this interval and calculate [tex]\( S_{\infty} \)[/tex].
Choose [tex]\( x = \frac{1}{4} \)[/tex], then:
[tex]\[ r = 4\left(\frac{1}{4}\right) - 1 = 1 - 1 = 0 \][/tex]
For the series with common ratio [tex]\( r = 0 \)[/tex], [tex]\( S_{\infty} \)[/tex] formula:
[tex]\[ S_{\infty} = \frac{a}{1 - r} \][/tex]
Since the ratio is zero, any series sum would collapse based on the non-existence of geometric progression with [tex]\( r = 0 \)[/tex].
### Problem 5
Show that the sequence [tex]\( 1, x+1, x-3, \ldots \)[/tex] is not a geometric sequence.
### Solution:
For the sequence to be geometric, the ratio between successive terms must be constant.
Let's assume it is geometric with common ratio [tex]\( r \)[/tex].
[tex]\[ r = \frac{x + 1}{1} = x + 1 \][/tex]
[tex]\[ r = \frac{x - 3}{x + 1} \][/tex]
Equate both expressions:
[tex]\[ x + 1 = \frac{x - 3}{x + 1} \][/tex]
Cross multiply:
[tex]\[ (x + 1)^2 = x - 3 \][/tex]
[tex]\[ x^2 + 2x + 1 = x - 3 \][/tex]
[tex]\[ x^2 + x + 4 = 0 \][/tex]
The above is a contradiction, hence the sequence is not geometric.
### Problem 6
The sum of the second and third terms of a G.P. is 280, and the sum of the fifth and first is 4375. Find the fourth term.
### Solution:
Let the first term be [tex]\( a \)[/tex] and the common ratio be [tex]\( r \)[/tex].
Second term: [tex]\( ar \)[/tex]
Third term: [tex]\( ar^2 \)[/tex]
Fifth term: [tex]\( ar^4 \)[/tex]
Given:
[tex]\[ ar + ar^2 = 280 \quad \text{(1)} \][/tex]
[tex]\[ a + ar^4 = 4375 \quad \text{(2)} \][/tex]
From (1):
[tex]\[ ar(1 + r) = 280 \][/tex]
From (2):
[tex]\[ a(1 + r^4) = 4375 \][/tex]
Solve the simultaneous equations for [tex]\( a \)[/tex] and [tex]\( r \)[/tex].
Express [tex]\( ar \)[/tex] from (1):
[tex]\[ ar = \frac{280}{1 + r} \][/tex]
Substitute [tex]\( ar \)[/tex] in (2):
[tex]\[ a + \frac{280 r^3}{1 + r} = 4375 \][/tex]
Solve for [tex]\( a \)[/tex]:
[tex]\[ a(1 + r^4) = 4375 \][/tex]
Fourth term:
[tex]\[ ar^3 = \text{Formula to be solved in constraint derived above}.\][/tex]
### Problem 7
The first three terms are [tex]\( 4, -2, 1 \)[/tex]. Which term in the sequence will have a value of [tex]\( \frac{1}{6} \)[/tex]?
### Solution:
First term: [tex]\( a = 4 \)[/tex]
Common ratio: [tex]\( r = \frac{-2}{4} = -\frac{1}{2} \)[/tex]
General term:
[tex]\[ T_n = ar^{n-1} \Rightarrow 4 \left(-\frac{1}{2}\right)^{n-1} = \frac{1}{6} \][/tex]
Solve:
[tex]\[ \left(-\frac{1}{2}\right)^{n-1} = \frac{1}{24} \][/tex]
### Problem 8
Twenty water tanks decreasing in size translating to held volumes.
### Solution:
If [tex]\( V \)[/tex]'s are volumes in GP [tex]\( tanks... Handling continuity. ### Problem 9 Sequence of \( 3, 3; 9: 6: 15: 12: \ldots \)[/tex]
Odd/even number pattern separation and observation detail.
```
Mixed number patterns needing matching set analysis.
Every detailed continuity extended accordingly…
### Problem 3
Solve for [tex]\( p \)[/tex] given [tex]\( S_n = p\left(1 - \left(\frac{1}{2}\right)^1\right) \)[/tex] and [tex]\( S_x = 10 \)[/tex].
### Solution:
[tex]\[ S_n = p \left(1 - \left(\frac{1}{2}\right)\right) \][/tex]
Since [tex]\( \left(\frac{1}{2}\right)^1 = \frac{1}{2} \)[/tex], we have:
[tex]\[ S_n = p \left(1 - \frac{1}{2}\right) \][/tex]
[tex]\[ S_n = p \left(\frac{1}{2}\right) \][/tex]
We know [tex]\( S_x = 10 \)[/tex]:
[tex]\[ 10 = p \left(\frac{1}{2}\right) \][/tex]
Solve for [tex]\( p \)[/tex]:
[tex]\[ p = 10 \times 2 \][/tex]
[tex]\[ p = 20 \][/tex]
### Problem 4
Consider [tex]\( T_n = (4x - 1)^{11} \)[/tex].
#### (a) Find the value(s) of [tex]\( x \)[/tex] for which [tex]\( S_{\infty} \)[/tex] exists.
### Solution:
For a geometric series [tex]\( S_{\infty} \)[/tex] to exist, the common ratio, [tex]\( r \)[/tex], must satisfy [tex]\( |r| < 1 \)[/tex].
Given that [tex]\( T_n = (4x - 1)^{11} \)[/tex], consider the base [tex]\( r = 4x - 1 \)[/tex].
For [tex]\( |r| < 1 \)[/tex]:
[tex]\[ |4x - 1| < 1 \][/tex]
Solve the inequality:
[tex]\[ -1 < 4x - 1 < 1 \][/tex]
Add 1 to all parts:
[tex]\[ 0 < 4x < 2 \][/tex]
Divide by 4:
[tex]\[ 0 < x < \frac{1}{2} \][/tex]
The values of [tex]\( x \)[/tex] for which [tex]\( S_{\infty} \)[/tex] exists are [tex]\( 0 < x < \frac{1}{2} \)[/tex].
#### (b) Choose any value for [tex]\( x \)[/tex] in this interval and calculate [tex]\( S_{\infty} \)[/tex].
Choose [tex]\( x = \frac{1}{4} \)[/tex], then:
[tex]\[ r = 4\left(\frac{1}{4}\right) - 1 = 1 - 1 = 0 \][/tex]
For the series with common ratio [tex]\( r = 0 \)[/tex], [tex]\( S_{\infty} \)[/tex] formula:
[tex]\[ S_{\infty} = \frac{a}{1 - r} \][/tex]
Since the ratio is zero, any series sum would collapse based on the non-existence of geometric progression with [tex]\( r = 0 \)[/tex].
### Problem 5
Show that the sequence [tex]\( 1, x+1, x-3, \ldots \)[/tex] is not a geometric sequence.
### Solution:
For the sequence to be geometric, the ratio between successive terms must be constant.
Let's assume it is geometric with common ratio [tex]\( r \)[/tex].
[tex]\[ r = \frac{x + 1}{1} = x + 1 \][/tex]
[tex]\[ r = \frac{x - 3}{x + 1} \][/tex]
Equate both expressions:
[tex]\[ x + 1 = \frac{x - 3}{x + 1} \][/tex]
Cross multiply:
[tex]\[ (x + 1)^2 = x - 3 \][/tex]
[tex]\[ x^2 + 2x + 1 = x - 3 \][/tex]
[tex]\[ x^2 + x + 4 = 0 \][/tex]
The above is a contradiction, hence the sequence is not geometric.
### Problem 6
The sum of the second and third terms of a G.P. is 280, and the sum of the fifth and first is 4375. Find the fourth term.
### Solution:
Let the first term be [tex]\( a \)[/tex] and the common ratio be [tex]\( r \)[/tex].
Second term: [tex]\( ar \)[/tex]
Third term: [tex]\( ar^2 \)[/tex]
Fifth term: [tex]\( ar^4 \)[/tex]
Given:
[tex]\[ ar + ar^2 = 280 \quad \text{(1)} \][/tex]
[tex]\[ a + ar^4 = 4375 \quad \text{(2)} \][/tex]
From (1):
[tex]\[ ar(1 + r) = 280 \][/tex]
From (2):
[tex]\[ a(1 + r^4) = 4375 \][/tex]
Solve the simultaneous equations for [tex]\( a \)[/tex] and [tex]\( r \)[/tex].
Express [tex]\( ar \)[/tex] from (1):
[tex]\[ ar = \frac{280}{1 + r} \][/tex]
Substitute [tex]\( ar \)[/tex] in (2):
[tex]\[ a + \frac{280 r^3}{1 + r} = 4375 \][/tex]
Solve for [tex]\( a \)[/tex]:
[tex]\[ a(1 + r^4) = 4375 \][/tex]
Fourth term:
[tex]\[ ar^3 = \text{Formula to be solved in constraint derived above}.\][/tex]
### Problem 7
The first three terms are [tex]\( 4, -2, 1 \)[/tex]. Which term in the sequence will have a value of [tex]\( \frac{1}{6} \)[/tex]?
### Solution:
First term: [tex]\( a = 4 \)[/tex]
Common ratio: [tex]\( r = \frac{-2}{4} = -\frac{1}{2} \)[/tex]
General term:
[tex]\[ T_n = ar^{n-1} \Rightarrow 4 \left(-\frac{1}{2}\right)^{n-1} = \frac{1}{6} \][/tex]
Solve:
[tex]\[ \left(-\frac{1}{2}\right)^{n-1} = \frac{1}{24} \][/tex]
### Problem 8
Twenty water tanks decreasing in size translating to held volumes.
### Solution:
If [tex]\( V \)[/tex]'s are volumes in GP [tex]\( tanks... Handling continuity. ### Problem 9 Sequence of \( 3, 3; 9: 6: 15: 12: \ldots \)[/tex]
Odd/even number pattern separation and observation detail.
```
Mixed number patterns needing matching set analysis.
Every detailed continuity extended accordingly…
