Answer :
To solve this problem, we need to determine the values of \( a \) and \( b \) given that there are 6 arithmetic means between them. The second mean is 8, and the last mean (i.e., the sixth mean) is 20.
### Step-by-step solution:
1. Identify Variables:
- \( a_1 = 8 \) (the second arithmetic mean)
- \( a_6 = 20 \) (the sixth arithmetic mean)
- Number of terms, \( n = 8 \) (since there are 6 means plus the two ends, \( a \) and \( b \))
2. Finding the Common Difference (\( d \)) in the Arithmetic Series:
To find the common difference (\( d \)), we use the positions of the given means. In an arithmetic progression:
- The second mean is the 3rd term in the sequence (because \( a \) is the 1st term and 6 means follow \( a \))
- The sixth mean is the 7th term in the sequence
Using the formula for the \( k \)-th term of an arithmetic sequence, \( a_k = a + (k-1)d \):
[tex]\[ a_3 = a + 2d = 8 \][/tex]
[tex]\[ a_7 = a + 6d = 20 \][/tex]
Now we can create two equations:
[tex]\[ a + 2d = 8 \][/tex]
[tex]\[ a + 6d = 20 \][/tex]
3. Solving for \( d \):
Subtract the first equation from the second to solve for \( d \):
[tex]\[ (a + 6d) - (a + 2d) = 20 - 8 \][/tex]
Simplifying this:
[tex]\[ 4d = 12 \][/tex]
[tex]\[ d = 3 \][/tex]
4. Finding the First Term (\( a \)):
Substitute \( d \) back into the first equation:
[tex]\[ a + 2(3) = 8 \][/tex]
[tex]\[ a + 6 = 8 \][/tex]
[tex]\[ a = 8 - 6 \][/tex]
[tex]\[ a = 2 \][/tex]
5. Finding the Last Term (\( b \)):
The number of terms in the sequence is 8. The last term \( b \) can be calculated using the formula for the \( n \)-th term of an arithmetic sequence:
[tex]\[ a_n = a + (n-1)d \][/tex]
Here, \( n = 8 \):
[tex]\[ b = a + 7d \][/tex]
Substitute \( a \) and \( d \):
[tex]\[ b = 2 + 7(3) \][/tex]
[tex]\[ b = 2 + 21 \][/tex]
[tex]\[ b = 23 \][/tex]
### Conclusion:
- The value of \( a \) (the first term) is \( 2 \).
- The value of [tex]\( b \)[/tex] (the last term) is [tex]\( 23 \)[/tex].
### Step-by-step solution:
1. Identify Variables:
- \( a_1 = 8 \) (the second arithmetic mean)
- \( a_6 = 20 \) (the sixth arithmetic mean)
- Number of terms, \( n = 8 \) (since there are 6 means plus the two ends, \( a \) and \( b \))
2. Finding the Common Difference (\( d \)) in the Arithmetic Series:
To find the common difference (\( d \)), we use the positions of the given means. In an arithmetic progression:
- The second mean is the 3rd term in the sequence (because \( a \) is the 1st term and 6 means follow \( a \))
- The sixth mean is the 7th term in the sequence
Using the formula for the \( k \)-th term of an arithmetic sequence, \( a_k = a + (k-1)d \):
[tex]\[ a_3 = a + 2d = 8 \][/tex]
[tex]\[ a_7 = a + 6d = 20 \][/tex]
Now we can create two equations:
[tex]\[ a + 2d = 8 \][/tex]
[tex]\[ a + 6d = 20 \][/tex]
3. Solving for \( d \):
Subtract the first equation from the second to solve for \( d \):
[tex]\[ (a + 6d) - (a + 2d) = 20 - 8 \][/tex]
Simplifying this:
[tex]\[ 4d = 12 \][/tex]
[tex]\[ d = 3 \][/tex]
4. Finding the First Term (\( a \)):
Substitute \( d \) back into the first equation:
[tex]\[ a + 2(3) = 8 \][/tex]
[tex]\[ a + 6 = 8 \][/tex]
[tex]\[ a = 8 - 6 \][/tex]
[tex]\[ a = 2 \][/tex]
5. Finding the Last Term (\( b \)):
The number of terms in the sequence is 8. The last term \( b \) can be calculated using the formula for the \( n \)-th term of an arithmetic sequence:
[tex]\[ a_n = a + (n-1)d \][/tex]
Here, \( n = 8 \):
[tex]\[ b = a + 7d \][/tex]
Substitute \( a \) and \( d \):
[tex]\[ b = 2 + 7(3) \][/tex]
[tex]\[ b = 2 + 21 \][/tex]
[tex]\[ b = 23 \][/tex]
### Conclusion:
- The value of \( a \) (the first term) is \( 2 \).
- The value of [tex]\( b \)[/tex] (the last term) is [tex]\( 23 \)[/tex].
