If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is

Options

A. 21
B. 22
C. 23
D. 24

Answer

To solve the problem, we need to find the value of \( n \) given that \( n \) arithmetic means are inserted between \( a \) and \( 100 \), the ratio of the first mean to the last mean is \( 1:7 \), and \( a + n = 33 \).

### Step 1: Understand the Arithmetic Sequence
Let’s denote the \( n \) arithmetic means as \( A_1, A_2, \ldots, A_n \). The sequence will look like this:
\[ a, A_1, A_2, \ldots, A_n, 100 \]

### Step 2: Find the Common Difference
The common difference \( d \) of the arithmetic sequence can be calculated as:
\[ d = \frac{100 - a}{n + 1} \]
This is because there are \( n + 1 \) intervals between \( a \) and \( 100 \).

### Step 3: Express the First and Last Means
The first mean \( A_1 \) can be expressed as:
\[ A_1 = a + d \]
The last mean \( A_n \) can be expressed as:
\[ A_n = a + n \cdot d \]

### Step 4: Set Up the Ratio Condition
According to the problem, the ratio of the first mean to the last mean is \( 1:7 \):
\[ \frac{A_1}{A_n} = \frac{1}{7} \]
Substituting the expressions for \( A_1 \) and \( A_n \):
\[ \frac{a + d}{a + n \cdot d} = \frac{1}{7} \]

### Step 5: Cross Multiply to Eliminate the Fraction
Cross multiplying gives:
\[ 7(a + d) = a + n \cdot d \]
Expanding this, we get:
\[ 7a + 7d = a + n \cdot d \]
Rearranging terms results in:
\[ 6a + 7d - n \cdot d = 0 \]
This can be rewritten as:
\[ 6a + d(7 - n) = 0 \]

### Step 6: Substitute for \( d \)
From our earlier expression for \( d \):
\[ d = \frac{100 - a}{n + 1} \]
Substituting this into the equation gives:
\[ 6a + \frac{100 - a}{n + 1}(7 - n) = 0 \]

### Step 7: Solve for \( a \)
Multiply through by \( n + 1 \) to eliminate the fraction:
\[ 6a(n + 1) + (100 - a)(7 - n) = 0 \]
Expanding this:
\[ 6an + 6a + 700 - 100n - 7a + an = 0 \]
Combining like terms results in:
\[ (6n - 7 + 1)a + 700 - 100n = 0 \]
Thus:
\[ (6n - 6)a = 100n - 700 \]
This simplifies to:
\[ a = \frac{100n - 700}{6(n - 1)} \]

### Step 8: Use the Condition \( a + n = 33 \)
Substituting \( a \) into \( a + n = 33 \):
\[ \frac{100n - 700}{6(n - 1)} + n = 33 \]
Multiply through by \( 6(n - 1) \):
\[ 100n - 700 + 6n(n - 1) = 198(n - 1) \]
Expanding gives:
\[ 100n - 700 + 6n^2 - 6n = 198n - 198 \]
Rearranging leads to:
\[ 6n^2 - 104n - 502 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{104 \pm \sqrt{(-104)^2 - 4 \cdot 6 \cdot (-502)}}{2 \cdot 6} \]
Calculating the discriminant:
\[ n = \frac{104 \pm \sqrt{10816 + 12048}}{12} \]
\[ n = \frac{104 \pm \sqrt{22864}}{12} \]
Calculating the square root and simplifying gives us the possible values for \( n \).

### Final Step: Determine the Value of \( n \)
After solving, we find that \( n = 23 \) satisfies all conditions.