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Sagot :
To determine the number of arithmetic means [tex]\( n \)[/tex] inserted between 5 and 35, given that the ratio of the second mean to the last mean is 1:4, we'll proceed with a systematic, step-by-step approach.
Let’s define:
- [tex]\( a_1 = 5 \)[/tex] (the first term).
- [tex]\( a_{n+2} = 35 \)[/tex] (the last term, where there are [tex]\( n \)[/tex] arithmetic means in between).
The terms of the arithmetic sequence are denoted as [tex]\( a_1, a_2, a_3, \ldots, a_{n+1}, a_{n+2} \)[/tex].
### Step 1: Establish the general formula for arithmetic progression
The nth term of an arithmetic progression can be expressed as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
where [tex]\( d \)[/tex] is the common difference.
### Step 2: Determine the expressions for the means
Since [tex]\( a_1 = 5 \)[/tex] and [tex]\( a_{n+2} = 35 \)[/tex], we can write:
[tex]\[ a_{n+2} = a_1 + (n+1)d \][/tex]
[tex]\[ 35 = 5 + (n+1)d \][/tex]
From this equation, we can solve for [tex]\( d \)[/tex]:
[tex]\[ (n+1)d = 30 \][/tex]
[tex]\[ d = \frac{30}{n+1} \][/tex]
### Step 3: Set up the ratios for the second and last mean
The second mean is [tex]\( a_2 \)[/tex]:
[tex]\[ a_2 = a_1 + d = 5 + \frac{30}{n+1} \][/tex]
The last mean is [tex]\( a_{n+1} \)[/tex]:
[tex]\[ a_{n+1} = a_1 + nd = 5 + n \cdot \frac{30}{n+1} \][/tex]
Given that the ratio of the second mean to the last mean is 1:4, we set up the following ratio:
[tex]\[ \frac{a_2}{a_{n+1}} = \frac{1}{4} \][/tex]
### Step 4: Substitute the expressions and solve for [tex]\( n \)[/tex]
[tex]\[ \frac{5 + \frac{30}{n+1}}{5 + n \cdot \frac{30}{n+1}} = \frac{1}{4} \][/tex]
Simplifying the equation:
[tex]\[ \frac{5(n+1) + 30}{5(n+1) + 30n} = \frac{1}{4} \][/tex]
[tex]\[ \frac{5n + 5 + 30}{5n + 5 + 30n} = \frac{1}{4} \][/tex]
[tex]\[ \frac{5n + 35}{35n + 5} = \frac{1}{4} \][/tex]
Cross-multiplying to clear the fraction:
[tex]\[ 4(5n + 35) = 35n + 5 \][/tex]
[tex]\[ 20n + 140 = 35n + 5 \][/tex]
[tex]\[ 140 - 5 = 35n - 20n \][/tex]
[tex]\[ 135 = 15n \][/tex]
[tex]\[ n = 9 \][/tex]
### Conclusion
The number of arithmetic means [tex]\( n \)[/tex] inserted between 5 and 35 such that the ratio of the second mean to the last mean is 1:4 is:
[tex]\[ n = 9 \][/tex]
Let’s define:
- [tex]\( a_1 = 5 \)[/tex] (the first term).
- [tex]\( a_{n+2} = 35 \)[/tex] (the last term, where there are [tex]\( n \)[/tex] arithmetic means in between).
The terms of the arithmetic sequence are denoted as [tex]\( a_1, a_2, a_3, \ldots, a_{n+1}, a_{n+2} \)[/tex].
### Step 1: Establish the general formula for arithmetic progression
The nth term of an arithmetic progression can be expressed as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
where [tex]\( d \)[/tex] is the common difference.
### Step 2: Determine the expressions for the means
Since [tex]\( a_1 = 5 \)[/tex] and [tex]\( a_{n+2} = 35 \)[/tex], we can write:
[tex]\[ a_{n+2} = a_1 + (n+1)d \][/tex]
[tex]\[ 35 = 5 + (n+1)d \][/tex]
From this equation, we can solve for [tex]\( d \)[/tex]:
[tex]\[ (n+1)d = 30 \][/tex]
[tex]\[ d = \frac{30}{n+1} \][/tex]
### Step 3: Set up the ratios for the second and last mean
The second mean is [tex]\( a_2 \)[/tex]:
[tex]\[ a_2 = a_1 + d = 5 + \frac{30}{n+1} \][/tex]
The last mean is [tex]\( a_{n+1} \)[/tex]:
[tex]\[ a_{n+1} = a_1 + nd = 5 + n \cdot \frac{30}{n+1} \][/tex]
Given that the ratio of the second mean to the last mean is 1:4, we set up the following ratio:
[tex]\[ \frac{a_2}{a_{n+1}} = \frac{1}{4} \][/tex]
### Step 4: Substitute the expressions and solve for [tex]\( n \)[/tex]
[tex]\[ \frac{5 + \frac{30}{n+1}}{5 + n \cdot \frac{30}{n+1}} = \frac{1}{4} \][/tex]
Simplifying the equation:
[tex]\[ \frac{5(n+1) + 30}{5(n+1) + 30n} = \frac{1}{4} \][/tex]
[tex]\[ \frac{5n + 5 + 30}{5n + 5 + 30n} = \frac{1}{4} \][/tex]
[tex]\[ \frac{5n + 35}{35n + 5} = \frac{1}{4} \][/tex]
Cross-multiplying to clear the fraction:
[tex]\[ 4(5n + 35) = 35n + 5 \][/tex]
[tex]\[ 20n + 140 = 35n + 5 \][/tex]
[tex]\[ 140 - 5 = 35n - 20n \][/tex]
[tex]\[ 135 = 15n \][/tex]
[tex]\[ n = 9 \][/tex]
### Conclusion
The number of arithmetic means [tex]\( n \)[/tex] inserted between 5 and 35 such that the ratio of the second mean to the last mean is 1:4 is:
[tex]\[ n = 9 \][/tex]
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