Let `f(n)=Sigma_(r=1)^(10n)(6+rd) and g(n)=Sigma_(r=1)^(n)(6+rd)`, where `n in N, d ne 0.` If `(f(n))/(g(n))` is independent of n, then d is equal to

Options

A. 12
B. `-6`
C. 6
D. `-12`

Answer

To solve the problem, we need to compute the functions \( f(n) \) and \( g(n) \) and then find the ratio \( \frac{f(n)}{g(n)} \) to determine the value of \( d \) such that this ratio is independent of \( n \).

### Step 1: Calculate \( f(n) \)

The function \( f(n) \) is defined as:

\[ f(n) = \sum_{r=1}^{10n} (6 + rd) \]

We can break this summation into two parts:

\[ f(n) = \sum_{r=1}^{10n} 6 + \sum_{r=1}^{10n} rd \]

Calculating the first part:

\[ \sum_{r=1}^{10n} 6 = 6 \cdot 10n = 60n \]

Now, for the second part, we have:

\[ \sum_{r=1}^{10n} rd = d \sum_{r=1}^{10n} r \]

Using the formula for the sum of the first \( m \) natural numbers, \( \sum_{r=1}^{m} r = \frac{m(m+1)}{2} \):

\[ \sum_{r=1}^{10n} r = \frac{10n(10n + 1)}{2} \]

Thus, we can write:

\[ \sum_{r=1}^{10n} rd = d \cdot \frac{10n(10n + 1)}{2} \]

Combining both parts, we have:

\[ f(n) = 60n + d \cdot \frac{10n(10n + 1)}{2} \]

### Step 2: Calculate \( g(n) \)

Now, we calculate \( g(n) \):

\[ g(n) = \sum_{r=1}^{n} (6 + rd) \]

Similarly, we can break this summation:

\[ g(n) = \sum_{r=1}^{n} 6 + \sum_{r=1}^{n} rd \]

Calculating the first part:

\[ \sum_{r=1}^{n} 6 = 6n \]

For the second part:

\[ \sum_{r=1}^{n} rd = d \sum_{r=1}^{n} r = d \cdot \frac{n(n + 1)}{2} \]

Thus, we can write:

\[ g(n) = 6n + d \cdot \frac{n(n + 1)}{2} \]

### Step 3: Find the ratio \( \frac{f(n)}{g(n)} \)

Now we find the ratio:

\[ \frac{f(n)}{g(n)} = \frac{60n + d \cdot \frac{10n(10n + 1)}{2}}{6n + d \cdot \frac{n(n + 1)}{2}} \]

### Step 4: Simplify the ratio

To simplify, we can factor out \( n \) from both the numerator and the denominator:

\[ \frac{f(n)}{g(n)} = \frac{n(60 + d \cdot \frac{10(10n + 1)}{2})}{n(6 + d \cdot \frac{(n + 1)}{2})} \]

Cancelling \( n \) (assuming \( n \neq 0 \)) gives:

\[ \frac{f(n)}{g(n)} = \frac{60 + d \cdot \frac{10(10n + 1)}{2}}{6 + d \cdot \frac{(n + 1)}{2}} \]

### Step 5: Set the ratio independent of \( n \)

For the ratio to be independent of \( n \), the coefficients of \( n \) in the numerator and denominator must be equal. Thus, we need to analyze the terms involving \( n \):

From the numerator, the coefficient of \( n \) is:

\[ d \cdot 50 \]

From the denominator, the coefficient of \( n \) is:

\[ d \cdot \frac{1}{2} \]

Setting these equal gives:

\[ 50d = \frac{d}{2} \]

### Step 6: Solve for \( d \)

Assuming \( d \neq 0 \), we can divide both sides by \( d \):

\[ 50 = \frac{1}{2} \]

This leads to:

\[ d = -12 \]

### Final Answer

Thus, the value of \( d \) is:

\[ \boxed{-12} \]