Answer :
Sure, let's break this problem into its two parts: (a) finding the arithmetic progression and (b) finding the value of [tex]\( x \)[/tex] in the base equation.
### Part (a): Arithmetic Progression
#### Given:
1. The sum of the first 9 terms (S[tex]\(_9\)[/tex]) of an arithmetic progression (AP) is 72.
2. The sum of the next 4 terms is 71.
Let's assume:
- [tex]\( a \)[/tex] is the first term of the AP.
- [tex]\( d \)[/tex] is the common difference.
The sum of the first [tex]\( n \)[/tex] terms of an arithmetic progression is given by the formula:
[tex]\[ S_n = \frac{n}{2} (2a + (n-1)d) \][/tex]
1. Sum of the first 9 terms:
[tex]\[ S_9 = \frac{9}{2} (2a + 8d) = 72 \][/tex]
Simplifying, we get:
[tex]\[ 9(2a + 8d) = 144 \][/tex]
[tex]\[ 2a + 8d = 16 \][/tex]
[tex]\[ a + 4d = 8 \][/tex] [tex]\(\Rightarrow \text{(i)}\)[/tex]
2. Sum of the first 13 terms [tex]\( S_{13} \)[/tex]:
The sum of the first 13 terms minus the sum of the first 9 terms is the sum of the next 4 terms.
[tex]\[ S_{13} - S_9 = 71 \][/tex]
The sum of the first 13 terms is:
[tex]\[ S_{13} = \frac{13}{2} (2a + 12d) \][/tex]
[tex]\[ S_{13} = \frac{13}{2} (2a + 12d) \][/tex]
Therefore,
[tex]\[ \frac{13}{2} (2a + 12d) = 72 + 71 \][/tex]
[tex]\[ \frac{13}{2} (2a + 12d) = 143 \][/tex]
[tex]\[ 13(2a + 12d) = 286 \][/tex]
[tex]\[ 2a + 12d = 22 \][/tex] [tex]\(\Rightarrow \text{(ii)}\)[/tex]
Now we have a system of linear equations:
1. [tex]\( a + 4d = 8 \)[/tex] [tex]\(\Rightarrow \text{(i)}\)[/tex]
2. [tex]\( 2a + 12d = 22 \)[/tex] [tex]\(\Rightarrow \text{(ii)}\)[/tex]
Solving these two equations:
From [tex]\(\text{(i)}\)[/tex]:
[tex]\[ a + 4d = 8 \][/tex]
[tex]\[ a = 8 - 4d \][/tex]
Substitute [tex]\( a = 8 - 4d \)[/tex] into [tex]\(\text{(ii)}\)[/tex]:
[tex]\[ 2(8 - 4d) + 12d = 22 \][/tex]
[tex]\[ 16 - 8d + 12d = 22 \][/tex]
[tex]\[ 16 + 4d = 22 \][/tex]
[tex]\[ 4d = 6 \][/tex]
[tex]\[ d = \frac{6}{4} \][/tex]
[tex]\[ d = \frac{3}{2} \][/tex]
Substitute [tex]\( d = \frac{3}{2} \)[/tex] back into [tex]\( a = 8 - 4d \)[/tex]:
[tex]\[ a = 8 - 4 \left(\frac{3}{2}\right) \][/tex]
[tex]\[ a = 8 - 6 \][/tex]
[tex]\[ a = 2 \][/tex]
So, the first term [tex]\( a = 2 \)[/tex] and the common difference [tex]\( d = \frac{3}{2} \)[/tex].
The arithmetic progression is:
[tex]\[ 2, 2 + \frac{3}{2}, 2 + 2\left(\frac{3}{2}\right), 2 + 3\left(\frac{3}{2}\right), \ldots \][/tex]
[tex]\[ 2, 3.5, 5, 6.5, \ldots \][/tex]
### Part (b): Base Conversion
Given the equation:
[tex]\[ 21_4 + 111_2 = 31_x \][/tex]
First, let's convert each number to base 10:
- [tex]\( 21_4 \)[/tex] means [tex]\( 2 \times 4^1 + 1 \times 4^0 = 2 \times 4 + 1 \times 1 = 8 + 1 = 9 \)[/tex]
- [tex]\( 111_2 \)[/tex] means [tex]\( 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 1 \times 4 + 1 \times 2 + 1 \times 1 = 4 + 2 + 1 = 7 \)[/tex]
Adding these two numbers in base 10:
[tex]\[ 9 + 7 = 16 \][/tex]
Now, we need to find [tex]\( x \)[/tex] such that [tex]\( 31_x = 16_{10} \)[/tex].
[tex]\[ 31_x \][/tex] means [tex]\( 3 \times x + 1 \] So, \[ 3x + 1 = 16 \] \[ 3x = 15 \] \[ x = 5 \] Therefore, \( x = 5 \)[/tex].
### Part (a): Arithmetic Progression
#### Given:
1. The sum of the first 9 terms (S[tex]\(_9\)[/tex]) of an arithmetic progression (AP) is 72.
2. The sum of the next 4 terms is 71.
Let's assume:
- [tex]\( a \)[/tex] is the first term of the AP.
- [tex]\( d \)[/tex] is the common difference.
The sum of the first [tex]\( n \)[/tex] terms of an arithmetic progression is given by the formula:
[tex]\[ S_n = \frac{n}{2} (2a + (n-1)d) \][/tex]
1. Sum of the first 9 terms:
[tex]\[ S_9 = \frac{9}{2} (2a + 8d) = 72 \][/tex]
Simplifying, we get:
[tex]\[ 9(2a + 8d) = 144 \][/tex]
[tex]\[ 2a + 8d = 16 \][/tex]
[tex]\[ a + 4d = 8 \][/tex] [tex]\(\Rightarrow \text{(i)}\)[/tex]
2. Sum of the first 13 terms [tex]\( S_{13} \)[/tex]:
The sum of the first 13 terms minus the sum of the first 9 terms is the sum of the next 4 terms.
[tex]\[ S_{13} - S_9 = 71 \][/tex]
The sum of the first 13 terms is:
[tex]\[ S_{13} = \frac{13}{2} (2a + 12d) \][/tex]
[tex]\[ S_{13} = \frac{13}{2} (2a + 12d) \][/tex]
Therefore,
[tex]\[ \frac{13}{2} (2a + 12d) = 72 + 71 \][/tex]
[tex]\[ \frac{13}{2} (2a + 12d) = 143 \][/tex]
[tex]\[ 13(2a + 12d) = 286 \][/tex]
[tex]\[ 2a + 12d = 22 \][/tex] [tex]\(\Rightarrow \text{(ii)}\)[/tex]
Now we have a system of linear equations:
1. [tex]\( a + 4d = 8 \)[/tex] [tex]\(\Rightarrow \text{(i)}\)[/tex]
2. [tex]\( 2a + 12d = 22 \)[/tex] [tex]\(\Rightarrow \text{(ii)}\)[/tex]
Solving these two equations:
From [tex]\(\text{(i)}\)[/tex]:
[tex]\[ a + 4d = 8 \][/tex]
[tex]\[ a = 8 - 4d \][/tex]
Substitute [tex]\( a = 8 - 4d \)[/tex] into [tex]\(\text{(ii)}\)[/tex]:
[tex]\[ 2(8 - 4d) + 12d = 22 \][/tex]
[tex]\[ 16 - 8d + 12d = 22 \][/tex]
[tex]\[ 16 + 4d = 22 \][/tex]
[tex]\[ 4d = 6 \][/tex]
[tex]\[ d = \frac{6}{4} \][/tex]
[tex]\[ d = \frac{3}{2} \][/tex]
Substitute [tex]\( d = \frac{3}{2} \)[/tex] back into [tex]\( a = 8 - 4d \)[/tex]:
[tex]\[ a = 8 - 4 \left(\frac{3}{2}\right) \][/tex]
[tex]\[ a = 8 - 6 \][/tex]
[tex]\[ a = 2 \][/tex]
So, the first term [tex]\( a = 2 \)[/tex] and the common difference [tex]\( d = \frac{3}{2} \)[/tex].
The arithmetic progression is:
[tex]\[ 2, 2 + \frac{3}{2}, 2 + 2\left(\frac{3}{2}\right), 2 + 3\left(\frac{3}{2}\right), \ldots \][/tex]
[tex]\[ 2, 3.5, 5, 6.5, \ldots \][/tex]
### Part (b): Base Conversion
Given the equation:
[tex]\[ 21_4 + 111_2 = 31_x \][/tex]
First, let's convert each number to base 10:
- [tex]\( 21_4 \)[/tex] means [tex]\( 2 \times 4^1 + 1 \times 4^0 = 2 \times 4 + 1 \times 1 = 8 + 1 = 9 \)[/tex]
- [tex]\( 111_2 \)[/tex] means [tex]\( 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 1 \times 4 + 1 \times 2 + 1 \times 1 = 4 + 2 + 1 = 7 \)[/tex]
Adding these two numbers in base 10:
[tex]\[ 9 + 7 = 16 \][/tex]
Now, we need to find [tex]\( x \)[/tex] such that [tex]\( 31_x = 16_{10} \)[/tex].
[tex]\[ 31_x \][/tex] means [tex]\( 3 \times x + 1 \] So, \[ 3x + 1 = 16 \] \[ 3x = 15 \] \[ x = 5 \] Therefore, \( x = 5 \)[/tex].
