Certainly! Let's solve the problem in a detailed, step-by-step manner.
We are given that three arithmetic means are inserted between \(a\) and \(b\), and the first and third means are 16 and 34 respectively. We'll find the values of \(a\) and \(b\).
#### Step 1: Understand the sequence
The sequence has \(a\), 16, a mean, 34, and \(b\). Thus, we can denote the sequence as:
[tex]\[ a, 16, \text{mean}, 34, b \][/tex]
#### Step 2: Define the properties of an arithmetic sequence
In an arithmetic sequence, each term is obtained by adding a common difference (\(d\)) to the previous term. If we let \(d\) be this common difference, we can express the terms in terms of \(a\):
[tex]\[ a_2 = a + d = 16 \][/tex]
[tex]\[ a_4 = a + 3d = 34 \][/tex]
#### Step 3: Find the common difference (\(d\))
From the above equations:
[tex]\[ a + d = 16 \][/tex]
[tex]\[ a + 3d = 34 \][/tex]
Subtract the first equation from the second:
[tex]\[ (a + 3d) - (a + d) = 34 - 16 \][/tex]
[tex]\[ 2d = 18 \][/tex]
[tex]\[ d = 9 \][/tex]
#### Step 4: Find the value of \(a\)
Substitute the value of \(d\) back into \(a + d = 16\):
[tex]\[ a + 9 = 16 \][/tex]
[tex]\[ a = 7 \][/tex]
#### Step 5: Find the value of \(b\)
Now, calculate \(b\) using the fact that \(b = a + 4d\):
[tex]\[ b = 7 + 4 \times 9 \][/tex]
[tex]\[ b = 7 + 36 \][/tex]
[tex]\[ b = 43 \][/tex]
#### Step 6: Verify the sequence
The sequence should be:
[tex]\[ a, a+d, a+2d, a+3d, a+4d \][/tex]
So it becomes:
[tex]\[ 7, 16, 25, 34, 43 \][/tex]
Thus, the values are:
- \(a = 7\)
- The second mean (the one in the middle) is \(25\)
- \(b = 43\)
Therefore, the values of the terms are:
- The first mean: \(16\)
- The second mean: \(25\)
- The third mean: \(34\)
- \(a = 7\)
- [tex]\(b = 43\)[/tex]
