Number of identical terms in the sequence `2, 5, 8, 11, ...` up to 100 terms and 3, 5, 7, 9, 11, ... Up to 100 terms is

Options

A. 33
B. 17
C. 50
D. 147

Answer

To find the number of identical terms in the sequences \(2, 5, 8, 11, \ldots\) (up to 100 terms) and \(3, 5, 7, 9, 11, \ldots\) (up to 100 terms), we will follow these steps:

### Step 1: Identify the first sequence
The first sequence is an arithmetic sequence given by:
- First term \(a_1 = 2\)
- Common difference \(d_1 = 5 - 2 = 3\)

The \(n\)-th term of the first sequence can be expressed as:
\[ a_n = a_1 + (n - 1) \cdot d_1 = 2 + (n - 1) \cdot 3 \]

### Step 2: Find the 100th term of the first sequence
To find the 100th term:
\[ a_{100} = 2 + (100 - 1) \cdot 3 = 2 + 99 \cdot 3 = 2 + 297 = 299 \]
Thus, the first sequence is \(2, 5, 8, 11, \ldots, 299\).

### Step 3: Identify the second sequence
The second sequence is also an arithmetic sequence given by:
- First term \(b_1 = 3\)
- Common difference \(d_2 = 5 - 3 = 2\)

The \(n\)-th term of the second sequence can be expressed as:
\[ b_n = b_1 + (n - 1) \cdot d_2 = 3 + (n - 1) \cdot 2 \]

### Step 4: Find the 100th term of the second sequence
To find the 100th term:
\[ b_{100} = 3 + (100 - 1) \cdot 2 = 3 + 99 \cdot 2 = 3 + 198 = 201 \]
Thus, the second sequence is \(3, 5, 7, 9, \ldots, 201\).

### Step 5: Identify the identical terms
Now we need to find the common terms in both sequences. The common terms will be of the form:
- From the first sequence: \(2 + (n - 1) \cdot 3\)
- From the second sequence: \(3 + (m - 1) \cdot 2\)

Setting these equal to each other:
\[ 2 + (n - 1) \cdot 3 = 3 + (m - 1) \cdot 2 \]
Simplifying this gives:
\[ 3n - 3 = 2m - 1 \]
\[ 3n - 2m = 2 \]

### Step 6: Solve for integer values of \(n\) and \(m\)
Rearranging gives:
\[ 3n = 2m + 2 \quad \Rightarrow \quad n = \frac{2m + 2}{3} \]
For \(n\) to be an integer, \(2m + 2\) must be divisible by 3. This implies:
\[ 2m + 2 \equiv 0 \mod{3} \quad \Rightarrow \quad 2m \equiv 1 \mod{3} \]
Multiplying both sides by the modular inverse of 2 modulo 3 (which is 2):
\[ m \equiv 2 \mod{3} \]
Thus, \(m\) can take values of the form \(m = 3k + 2\) for integers \(k\).

### Step 7: Find the range of \(m\)
Since \(m\) must be such that \(b_m \leq 201\):
\[ 3k + 2 \leq 100 \quad \Rightarrow \quad 3k \leq 98 \quad \Rightarrow \quad k \leq 32.67 \]
Thus, \(k\) can take values from \(0\) to \(32\) (inclusive), giving us \(33\) possible values for \(m\).

### Conclusion
The number of identical terms in both sequences is \(33\).