If `(3 + 5 + 7 +...+n)/(5 + 8 + 11 +....+ "10 terms") = 7`, the value of n is

Options

A. 35
B. 36
C. 37
D. 40

Answer

To solve the problem, we need to find the value of \( n \) given the equation:

\[ \frac{(3 + 5 + 7 + \ldots + n)}{(5 + 8 + 11 + \ldots + \text{10 terms})} = 7 \]

### Step 1: Identify the first series
The first series is \( 3 + 5 + 7 + \ldots + n \). This is an arithmetic progression (AP) where:
- First term \( a = 3 \)
- Common difference \( d = 2 \)

### Step 2: Find the number of terms in the first series
To find the number of terms \( k \) in the series, we can express \( n \) in terms of \( k \):
\[ n = a + (k-1)d = 3 + (k-1) \cdot 2 \]
\[ n = 2k + 1 \]

### Step 3: Calculate the sum of the first series
The sum \( S_k \) of the first \( k \) terms of an AP is given by:
\[ S_k = \frac{k}{2} \cdot (2a + (k-1)d) \]
Substituting the values:
\[ S_k = \frac{k}{2} \cdot (2 \cdot 3 + (k-1) \cdot 2) = \frac{k}{2} \cdot (6 + 2k - 2) = \frac{k}{2} \cdot (2k + 4) = k(k + 2) \]

### Step 4: Identify the second series
The second series is \( 5 + 8 + 11 + \ldots \) for 10 terms. This is also an AP where:
- First term \( a = 5 \)
- Common difference \( d = 3 \)

### Step 5: Calculate the sum of the second series
The sum \( S_{10} \) of the first 10 terms is:
\[ S_{10} = \frac{n}{2} \cdot (2a + (n-1)d) = \frac{10}{2} \cdot (2 \cdot 5 + (10-1) \cdot 3) \]
\[ = 5 \cdot (10 + 27) = 5 \cdot 37 = 185 \]

### Step 6: Set up the equation
Now we substitute \( S_k \) and \( S_{10} \) into the original equation:
\[ \frac{k(k + 2)}{185} = 7 \]

### Step 7: Solve for \( k \)
Multiplying both sides by 185 gives:
\[ k(k + 2) = 1295 \]
Rearranging:
\[ k^2 + 2k - 1295 = 0 \]

### Step 8: Factor the quadratic equation
To factor \( k^2 + 2k - 1295 \), we look for two numbers that multiply to \(-1295\) and add to \(2\). The numbers \(35\) and \(-37\) work:
\[ (k + 37)(k - 35) = 0 \]

### Step 9: Find the values of \( k \)
Setting each factor to zero gives:
\[ k + 37 = 0 \quad \Rightarrow \quad k = -37 \quad (\text{not valid since } k \text{ must be positive}) \]
\[ k - 35 = 0 \quad \Rightarrow \quad k = 35 \]

### Step 10: Find the value of \( n \)
Now substituting \( k = 35 \) back to find \( n \):
\[ n = 2k + 1 = 2 \cdot 35 + 1 = 70 + 1 = 71 \]

Thus, the value of \( n \) is \( \boxed{71} \).