Answer:
the values of a and b are a = 3 and b = 123
Explanation:
To solve this problem, we can set up an arithmetic sequence and use the given information to find the values of a and b.
Let's consider the arithmetic sequence with the first term (a), the common difference (d), and the 31st term (b).
We know that there are 28 arithmetic means between a and b. Therefore, the total number of terms in the sequence is 31, including a, the 28 means, and b.
Using the formula for the nth term of an arithmetic sequence, we can express the terms as:
a, a + d, a + 2d, ..., a + 28d, 59, 63, a + 30d = b
The two means at the middle are 59 and 63, which correspond to the 15th and 16th terms in the sequence. Therefore, we have:
a + 14d = 59 (Equation 1)
a + 15d = 63 (Equation 2)
To find the values of a and b, we need to solve Equations 1 and 2 simultaneously.
Subtracting Equation 1 from Equation 2, we get:
(a + 15d) - (a + 14d) = 63 - 59
d = 4
Substituting the value of d = 4 into Equation 1, we can solve for a:
a + 14(4) = 59
a + 56 = 59
a = 59 - 56
a = 3
So, a = 3 and b is the 31st term, which is given by:
b = a + 30d
b = 3 + 30(4)
b = 3 + 120
b = 123
Therefore, the values of a and b are a = 3 and b = 123, respectively.