Let's solve this step-by-step:
### (a) Write the common difference [tex]\(d\)[/tex] in terms of number of means [tex]\(n\)[/tex].
Given:
- First term [tex]\(a = 1\)[/tex]
- Last term [tex]\(31\)[/tex]
- There are [tex]\(n\)[/tex] arithmetic means between 1 and 31.
The sequence will have [tex]\(n + 2\)[/tex] terms (1, [tex]\(A_1\)[/tex], [tex]\(A_2\)[/tex], ..., [tex]\(A_n\)[/tex], 31).
The formula for the [tex]\(k\)[/tex]-th arithmetic mean [tex]\(A_k\)[/tex] in an arithmetic sequence is:
[tex]\[ A_k = a + kd \][/tex]
Here, the last term in the sequence which is [tex]\(31\)[/tex]:
[tex]\[ 31 = 1 + (n+1)d \][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[ 31 = 1 + (n + 1)d \][/tex]
[tex]\[ 31 - 1 = (n + 1)d \][/tex]
[tex]\[ 30 = (n + 1)d \][/tex]
[tex]\[ d = \frac{30}{n + 1} \][/tex]
So, we have:
[tex]\[ d = \frac{30}{n + 1} \][/tex]
### (b) Find the value of [tex]\(n\)[/tex].
Given the ratio of the seventh mean to the [tex]\((n + 1)\)[/tex]-th mean is 5:9.
Seventh mean [tex]\(A_7\)[/tex]:
[tex]\[ A_7 = 1 + 7d \][/tex]
[tex]\((n + 1)\)[/tex]-th mean (which is the last term, 31):
[tex]\[ A_{n+1} = 31 \][/tex]
Given ratio:
[tex]\[ \frac{A_7}{A_{n+1}} = \frac{5}{9} \][/tex]
Substitute [tex]\(A_7\)[/tex] and [tex]\(A_{n+1}\)[/tex]:
[tex]\[ \frac{1 + 7d}{31} = \frac{5}{9} \][/tex]
Using [tex]\(d = \frac{30}{n + 1}\)[/tex]:
[tex]\[ \frac{1 + 7 \left(\frac{30}{n + 1}\right)}{31} = \frac{5}{9} \][/tex]
[tex]\[ \frac{1 + \frac{210}{n + 1}}{31} = \frac{5}{9} \][/tex]
[tex]\[ 9 \left(1 + \frac{210}{n+1}\right) = 5 \times 31 \][/tex]
[tex]\[ 9 + \frac{1890}{n + 1} = 155 \][/tex]
[tex]\[ \frac{1890}{n + 1} = 155 - 9 \][/tex]
[tex]\[ \frac{1890}{n + 1} = 146 \][/tex]
[tex]\[ 1890 = 146(n + 1) \][/tex]
[tex]\[ 1890 = 146n + 146 \][/tex]
[tex]\[ 1890 - 146 = 146n \][/tex]
[tex]\[ 1744 = 146n \][/tex]
[tex]\[ n = \frac{1744}{146} \][/tex]
[tex]\[ n = 12 \][/tex]
So, the value of [tex]\(n\)[/tex] is 12.
### (c) Compute the ratio of the third mean to the eighth mean.
Using [tex]\(d = \frac{30}{n + 1}\)[/tex], and with [tex]\(n = 12\)[/tex]:
[tex]\[ d = \frac{30}{12 + 1} = \frac{30}{13} \approx 2.31 \][/tex]
Third mean [tex]\(A_3\)[/tex]:
[tex]\[ A_3 = 1 + 3d = 1 + 3 \left(\frac{30}{13}\right) = 1 + \frac{90}{13} \][/tex]
[tex]\[ A_3 = 1 + 6.92 = 7.92 \][/tex]
Eighth mean [tex]\(A_8\)[/tex]:
[tex]\[ A_8 = 1 + 8d = 1 + 8 \left(\frac{30}{13}\right) = 1 + \frac{240}{13} \][/tex]
[tex]\[ A_8 = 1 + 18.46 = 19.46 \][/tex]
The ratio:
[tex]\[ \frac{A_3}{A_8} = \frac{7.92}{19.46} = \frac{7}{17} \][/tex]
So, the ratio of the third mean to the eighth mean is:
[tex]\[ \frac{7}{17} \][/tex]
