To find the first term [tex]\(a_1\)[/tex] and the common difference [tex]\(d\)[/tex] of the arithmetic sequence, we start by analyzing the information given.
1. The sum of the first 12 terms of the arithmetic sequence is 366.
2. The sum of the next 12 terms (terms 13 to 24) is 1086.
We know that the sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic sequence with the first term [tex]\( a \)[/tex] and common difference [tex]\( d \)[/tex] is given by the formula:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \][/tex]
For the first 12 terms:
[tex]\[ S_{12} = \frac{12}{2} \left( 2a + (12-1)d \right) = 366 \][/tex]
Simplify:
[tex]\[ 6 \left( 2a + 11d \right) = 366 \][/tex]
[tex]\[ 2a + 11d = 61 \quad \text{(Equation 1)} \][/tex]
For the next 12 terms, the terms from 13 to 24:
The sum from terms 1 to 24 is:
[tex]\[ S_{24} = \frac{24}{2} \left( 2a + (24-1)d \right) = \frac{24}{2} \left( 2a + 23d \right) = 12 \left( 2a + 23d \right) \][/tex]
We know the sum of the first 24 terms is the sum of the first 12 terms plus the sum of the next 12 terms:
[tex]\[ S_{24} = 366 + 1086 = 1452 \][/tex]
So:
[tex]\[ 12 \left( 2a + 23d \right) = 1452 \][/tex]
[tex]\[ 2a + 23d = 121 \quad \text{(Equation 2)} \][/tex]
We now have a system of linear equations:
[tex]\[ 2a + 11d = 61 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 2a + 23d = 121 \quad \text{(Equation 2)} \][/tex]
Subtract Equation 1 from Equation 2:
[tex]\[ (2a + 23d) - (2a + 11d) = 121 - 61 \][/tex]
[tex]\[ 12d = 60 \][/tex]
[tex]\[ d = 5 \][/tex]
Substitute [tex]\( d = 5 \)[/tex] back into Equation 1:
[tex]\[ 2a + 11(5) = 61 \][/tex]
[tex]\[ 2a + 55 = 61 \][/tex]
[tex]\[ 2a = 6 \][/tex]
[tex]\[ a = 3 \][/tex]
Thus, the first term [tex]\( a_1 \)[/tex] and the common difference [tex]\( d \)[/tex] are:
[tex]\[ a_1 = 3, \quad d = 5 \][/tex]
