Answer :
Let's go through each part of the question step-by-step:
### Part (a)
#### (1) Determine the value of [tex]\( x \)[/tex].
Given the arithmetic sequence:
[tex]\[ (x + 5),\ (37 - x),\ (x + 13) \][/tex]
For the sequence to be arithmetic, the difference between consecutive terms must be the same.
The common difference between the first and second terms:
[tex]\[ (37 - x) - (x + 5) = 37 - x - x - 5 = 32 - 2x \][/tex]
The common difference between the second and third terms:
[tex]\[ (x + 13) - (37 - x) = x + 13 - 37 + x = 2x - 24 \][/tex]
Set these two differences equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ 32 - 2x = 2x - 24 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 32 + 24 = 2x + 2x \][/tex]
[tex]\[ 56 = 4x \][/tex]
[tex]\[ x = 14 \][/tex]
So, the value of [tex]\( x \)[/tex] is [tex]\( 14 \)[/tex].
#### (2) Determine the general term of the sequence in the form [tex]\( T_n = \ldots \)[/tex].
The first term [tex]\( T_1 \)[/tex] is:
[tex]\[ T_1 = (x + 5) = 14 + 5 = 19 \][/tex]
The common difference [tex]\( d \)[/tex] is:
[tex]\[ d = (37 - x) - (x + 5) = 32 - 2x = 32 - 2 \cdot 14 = 32 - 28 = 4 \][/tex]
The general term of an arithmetic sequence is given by:
[tex]\[ T_n = a + (n - 1)d \][/tex]
Here, [tex]\( a = 19 \)[/tex] and [tex]\( d = 4 \)[/tex], so:
[tex]\[ T_n = 19 + (n - 1) \cdot 4 = 19 + 4n - 4 = 4n + 15 \][/tex]
Thus, the general term is:
[tex]\[ T_n = 4n + 15 \][/tex]
### Part (b)
The sum of the first three terms of a geometric sequence is 91 and its common ratio is 3. Let [tex]\( a \)[/tex] be the first term.
The terms are:
[tex]\[ a, 3a, 9a \][/tex]
The sum of the first three terms is:
[tex]\[ a + 3a + 9a = 13a \][/tex]
Given that the sum is 91:
[tex]\[ 13a = 91 \][/tex]
Solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{91}{13} = 7 \][/tex]
So, the first term [tex]\( a \)[/tex] is [tex]\( 7 \)[/tex].
### Part (c)
Given that [tex]\( S_2 = 90 \)[/tex] and [tex]\( S_{\infty} = \frac{375}{4} \)[/tex], we need to find the first term [tex]\( a \)[/tex] and the common ratio [tex]\( r \)[/tex].
The sum of the first two terms in a geometric series is:
[tex]\[ S_2 = a + ar = a(1 + r) \][/tex]
Given:
[tex]\[ a(1 + r) = 90 \quad \text{(1)} \][/tex]
The sum to infinity of a convergent geometric series is:
[tex]\[ S_{\infty} = \frac{a}{1 - r} \][/tex]
Given:
[tex]\[ \frac{a}{1 - r} = \frac{375}{4} \quad \text{(2)} \][/tex]
We now have two equations:
[tex]\[ a(1 + r) = 90 \][/tex]
[tex]\[ \frac{a}{1 - r} = \frac{375}{4} \][/tex]
Let's solve these equations simultaneously.
From equation (1):
[tex]\[ a = \frac{90}{1 + r} \][/tex]
Substitute [tex]\( a \)[/tex] into equation (2):
[tex]\[ \frac{\frac{90}{1 + r}}{1 - r} = \frac{375}{4} \][/tex]
[tex]\[ \frac{90}{(1 + r)(1 - r)} = \frac{375}{4} \][/tex]
[tex]\[ \frac{90}{1 - r^2} = \frac{375}{4} \][/tex]
[tex]\[ 360 = 375(1 - r^2) \][/tex]
[tex]\[ 360 = 375 - 375r^2 \][/tex]
[tex]\[ 375r^2 = 15 \][/tex]
[tex]\[ r^2 = \frac{15}{375} = \frac{1}{25} \][/tex]
[tex]\[ r = \frac{1}{5} \quad \text{(since \( r \) must be positive for the series to be convergent)} \][/tex]
Substituting [tex]\( r = \frac{1}{5} \)[/tex] back into equation (1) to find [tex]\( a \)[/tex]:
[tex]\[ a(1 + \frac{1}{5}) = 90 \][/tex]
[tex]\[ a \cdot \frac{6}{5} = 90 \][/tex]
[tex]\[ a = 90 \cdot \frac{5}{6} = 75 \][/tex]
Thus, the first term [tex]\( a \)[/tex] is [tex]\( 75 \)[/tex] and the common ratio [tex]\( r \)[/tex] is [tex]\( \frac{1}{5} \)[/tex].
### Part (a)
#### (1) Determine the value of [tex]\( x \)[/tex].
Given the arithmetic sequence:
[tex]\[ (x + 5),\ (37 - x),\ (x + 13) \][/tex]
For the sequence to be arithmetic, the difference between consecutive terms must be the same.
The common difference between the first and second terms:
[tex]\[ (37 - x) - (x + 5) = 37 - x - x - 5 = 32 - 2x \][/tex]
The common difference between the second and third terms:
[tex]\[ (x + 13) - (37 - x) = x + 13 - 37 + x = 2x - 24 \][/tex]
Set these two differences equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ 32 - 2x = 2x - 24 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 32 + 24 = 2x + 2x \][/tex]
[tex]\[ 56 = 4x \][/tex]
[tex]\[ x = 14 \][/tex]
So, the value of [tex]\( x \)[/tex] is [tex]\( 14 \)[/tex].
#### (2) Determine the general term of the sequence in the form [tex]\( T_n = \ldots \)[/tex].
The first term [tex]\( T_1 \)[/tex] is:
[tex]\[ T_1 = (x + 5) = 14 + 5 = 19 \][/tex]
The common difference [tex]\( d \)[/tex] is:
[tex]\[ d = (37 - x) - (x + 5) = 32 - 2x = 32 - 2 \cdot 14 = 32 - 28 = 4 \][/tex]
The general term of an arithmetic sequence is given by:
[tex]\[ T_n = a + (n - 1)d \][/tex]
Here, [tex]\( a = 19 \)[/tex] and [tex]\( d = 4 \)[/tex], so:
[tex]\[ T_n = 19 + (n - 1) \cdot 4 = 19 + 4n - 4 = 4n + 15 \][/tex]
Thus, the general term is:
[tex]\[ T_n = 4n + 15 \][/tex]
### Part (b)
The sum of the first three terms of a geometric sequence is 91 and its common ratio is 3. Let [tex]\( a \)[/tex] be the first term.
The terms are:
[tex]\[ a, 3a, 9a \][/tex]
The sum of the first three terms is:
[tex]\[ a + 3a + 9a = 13a \][/tex]
Given that the sum is 91:
[tex]\[ 13a = 91 \][/tex]
Solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{91}{13} = 7 \][/tex]
So, the first term [tex]\( a \)[/tex] is [tex]\( 7 \)[/tex].
### Part (c)
Given that [tex]\( S_2 = 90 \)[/tex] and [tex]\( S_{\infty} = \frac{375}{4} \)[/tex], we need to find the first term [tex]\( a \)[/tex] and the common ratio [tex]\( r \)[/tex].
The sum of the first two terms in a geometric series is:
[tex]\[ S_2 = a + ar = a(1 + r) \][/tex]
Given:
[tex]\[ a(1 + r) = 90 \quad \text{(1)} \][/tex]
The sum to infinity of a convergent geometric series is:
[tex]\[ S_{\infty} = \frac{a}{1 - r} \][/tex]
Given:
[tex]\[ \frac{a}{1 - r} = \frac{375}{4} \quad \text{(2)} \][/tex]
We now have two equations:
[tex]\[ a(1 + r) = 90 \][/tex]
[tex]\[ \frac{a}{1 - r} = \frac{375}{4} \][/tex]
Let's solve these equations simultaneously.
From equation (1):
[tex]\[ a = \frac{90}{1 + r} \][/tex]
Substitute [tex]\( a \)[/tex] into equation (2):
[tex]\[ \frac{\frac{90}{1 + r}}{1 - r} = \frac{375}{4} \][/tex]
[tex]\[ \frac{90}{(1 + r)(1 - r)} = \frac{375}{4} \][/tex]
[tex]\[ \frac{90}{1 - r^2} = \frac{375}{4} \][/tex]
[tex]\[ 360 = 375(1 - r^2) \][/tex]
[tex]\[ 360 = 375 - 375r^2 \][/tex]
[tex]\[ 375r^2 = 15 \][/tex]
[tex]\[ r^2 = \frac{15}{375} = \frac{1}{25} \][/tex]
[tex]\[ r = \frac{1}{5} \quad \text{(since \( r \) must be positive for the series to be convergent)} \][/tex]
Substituting [tex]\( r = \frac{1}{5} \)[/tex] back into equation (1) to find [tex]\( a \)[/tex]:
[tex]\[ a(1 + \frac{1}{5}) = 90 \][/tex]
[tex]\[ a \cdot \frac{6}{5} = 90 \][/tex]
[tex]\[ a = 90 \cdot \frac{5}{6} = 75 \][/tex]
Thus, the first term [tex]\( a \)[/tex] is [tex]\( 75 \)[/tex] and the common ratio [tex]\( r \)[/tex] is [tex]\( \frac{1}{5} \)[/tex].
