Each of the series 13 + 15 + 17 + .... and 14 + 17 + 20 + ... is continued to 100 terms. Find how many terms are identical between the two series?

Options

A. 35
B. 34
C. 32
D. 33

Answer

To solve the problem of finding how many terms are identical between the two series, we first need to identify the terms of each series.

### Step 1: Identify the first series
The first series is:
\[ 13, 15, 17, 19, 21, \ldots \]
This is an arithmetic series where:
- The first term \( a_1 = 13 \)
- The common difference \( d_1 = 2 \)

The \( n \)-th term of an arithmetic series can be calculated using the formula:
\[ a_n = a_1 + (n-1) \cdot d \]

For the first series, the \( n \)-th term is:
\[ a_n = 13 + (n-1) \cdot 2 = 13 + 2n - 2 = 2n + 11 \]

### Step 2: Identify the second series
The second series is:
\[ 14, 17, 20, 23, 26, \ldots \]
This is also an arithmetic series where:
- The first term \( b_1 = 14 \)
- The common difference \( d_2 = 3 \)

The \( n \)-th term of this series is:
\[ b_n = 14 + (n-1) \cdot 3 = 14 + 3n - 3 = 3n + 11 \]

### Step 3: Set the two series equal to find common terms
To find the common terms, we need to set the \( n \)-th term of the first series equal to the \( m \)-th term of the second series:
\[ 2n + 11 = 3m + 11 \]

Subtracting 11 from both sides gives:
\[ 2n = 3m \]

Rearranging this gives:
\[ \frac{n}{m} = \frac{3}{2} \]

### Step 4: Express \( n \) in terms of \( m \)
From the equation \( 2n = 3m \), we can express \( n \) in terms of \( m \):
\[ n = \frac{3}{2}m \]

### Step 5: Determine the limits for \( n \) and \( m \)
Since both series continue to 100 terms:
- For the first series, \( n \) can take values from 1 to 100.
- For the second series, \( m \) can also take values from 1 to 100.

### Step 6: Find the maximum value of \( m \)
To ensure \( n \) remains an integer, \( m \) must be even (since \( n = \frac{3}{2}m \)). Let \( m = 2k \), where \( k \) is an integer.

Substituting \( m \) gives:
\[ n = \frac{3}{2}(2k) = 3k \]

### Step 7: Determine the limits for \( k \)
Now, we need to find the maximum value of \( k \) such that both \( n \) and \( m \) are within their limits:
- For \( n \leq 100 \): \( 3k \leq 100 \) implies \( k \leq \frac{100}{3} \approx 33.33 \) (so \( k \) can be at most 33)
- For \( m \leq 100 \): \( 2k \leq 100 \) implies \( k \leq 50 \)

The limiting factor is \( k \leq 33 \).

### Step 8: Count the number of common terms
The possible values of \( k \) are \( 1, 2, \ldots, 33 \), which gives us a total of 33 common terms.

### Final Answer
Thus, the number of identical terms between the two series is:
\[ \boxed{33} \]