How many terms are identical in the two APs 1,3, 5,... up to 120 terms and 3, 6, 9, .... up to 80 terms ?

Options

A. A)38
B. B)39
C. C)40
D. D)41

Answer

To find how many terms are identical in the two arithmetic progressions (APs), we will analyze both sequences step by step.

### Step 1: Identify the first AP
The first AP is given as:
1, 3, 5, ..., up to 120 terms.

This is an arithmetic progression where:
- The first term \( a_1 = 1 \)
- The common difference \( d_1 = 2 \)

The \( n \)-th term of an AP can be calculated using the formula:
\[ a_n = a_1 + (n - 1) \cdot d \]
For the first AP:
\[ a_n = 1 + (n - 1) \cdot 2 = 2n - 1 \]
Thus, the \( n \)-th term of the first AP is \( 2n - 1 \).

### Step 2: Identify the second AP
The second AP is given as:
3, 6, 9, ..., up to 80 terms.

This is also an arithmetic progression where:
- The first term \( b_1 = 3 \)
- The common difference \( d_2 = 3 \)

Using the same formula for the \( n \)-th term:
\[ b_n = b_1 + (n - 1) \cdot d = 3 + (n - 1) \cdot 3 = 3n \]
Thus, the \( n \)-th term of the second AP is \( 3n \).

### Step 3: Find common terms
To find the common terms in both APs, we need to solve for \( n \) in the equations:
\[ 2m - 1 = 3n \]
Where \( m \) is the term number in the first AP and \( n \) is the term number in the second AP.

Rearranging gives:
\[ 2m = 3n + 1 \quad \text{or} \quad m = \frac{3n + 1}{2} \]

### Step 4: Determine valid values for \( n \)
For \( m \) to be an integer, \( 3n + 1 \) must be even. Since \( 3n \) is odd when \( n \) is odd, \( n \) must be even for \( 3n + 1 \) to be even.

Let \( n = 2k \) where \( k \) is an integer:
\[ m = \frac{3(2k) + 1}{2} = 3k + \frac{1}{2} \]
This means \( k \) must be such that \( 3k + \frac{1}{2} \) is an integer, which is not possible. Therefore, we need to find the values of \( n \) directly.

### Step 5: Find the range of \( n \)
The second AP has terms from \( 3 \) to \( 240 \) (since \( 3 \times 80 = 240 \)). We can find the maximum \( n \) such that \( 3n \leq 240 \):
\[ n \leq 80 \]

### Step 6: Check for common terms
Now we check for odd multiples of 3 in the first AP:
- The first AP consists of all odd numbers.
- The odd multiples of 3 are \( 3, 9, 15, ..., 3(2k-1) \).

The maximum odd multiple of 3 that is less than or equal to \( 239 \) (the last term of the first AP) is \( 3 \times 79 = 237 \).

### Step 7: Count the common terms
The common terms are \( 3, 9, 15, ..., 237 \). This is an AP where:
- First term \( = 3 \)
- Common difference \( = 6 \)

To find the number of terms in this sequence:
Let \( l \) be the last term:
\[ l = 3 + (n - 1) \cdot 6 \]
Setting \( l = 237 \):
\[ 237 = 3 + (n - 1) \cdot 6 \\ 234 = (n - 1) \cdot 6 \\ n - 1 = 39 \\ n = 40 \]

### Conclusion
Thus, the number of identical terms in the two APs is **40**.

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