The sequence form by common terms of `17,21,25,"….."` & `16,21,26,"…."` has

Options

A. common difference 20
B. sum of first 100 terms is 100 100
C. sum of first 1000 terms is 101100
D. tenth term is 201.

Answer

To find the common terms of the sequences \(17, 21, 25, \ldots\) and \(16, 21, 26, \ldots\), we will follow these steps:

### Step 1: Identify the sequences
The first sequence is an arithmetic progression (AP) with:
- First term \(a_1 = 17\)
- Common difference \(d_1 = 4\)

The second sequence is also an AP with:
- First term \(a_2 = 16\)
- Common difference \(d_2 = 5\)

### Step 2: Write the general terms of both sequences
The \(n\)-th term of the first sequence can be expressed as:
\[ T_1(n) = a_1 + (n-1)d_1 = 17 + (n-1)4 = 4n + 13 \]

The \(m\)-th term of the second sequence can be expressed as:
\[ T_2(m) = a_2 + (m-1)d_2 = 16 + (m-1)5 = 5m + 11 \]

### Step 3: Set the general terms equal to find common terms
To find the common terms, we set \(T_1(n) = T_2(m)\):
\[ 4n + 13 = 5m + 11 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 4n - 5m = -2 \]
This can be rewritten as:
\[ 4n = 5m - 2 \]

### Step 5: Find integer solutions
To find integer solutions for \(n\) and \(m\), we can express \(m\) in terms of \(n\):
\[ m = \frac{4n + 2}{5} \]
For \(m\) to be an integer, \(4n + 2\) must be divisible by 5.

### Step 6: Solve for \(n\)
We can analyze the congruence:
\[ 4n + 2 \equiv 0 \mod 5 \]
This simplifies to:
\[ 4n \equiv -2 \equiv 3 \mod 5 \]
Multiplying both sides by the modular inverse of 4 modulo 5, which is 4, we get:
\[ n \equiv 4 \cdot 3 \equiv 12 \equiv 2 \mod 5 \]
Thus, \(n\) can be expressed as:
\[ n = 5k + 2 \quad \text{for integers } k \]

### Step 7: Find the common terms
Substituting \(n\) back into the formula for \(T_1(n)\):
\[ T_1(5k + 2) = 4(5k + 2) + 13 = 20k + 8 + 13 = 20k + 21 \]
Thus, the common terms are of the form:
\[ 21, 41, 61, 81, \ldots \]
This is another arithmetic progression with:
- First term \(21\)
- Common difference \(20\)

### Step 8: Calculate the sum of the first 100 terms
The sum \(S_n\) of the first \(n\) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For our common sequence:
- \(n = 100\)
- \(a = 21\)
- \(d = 20\)

Substituting these values:
\[ S_{100} = \frac{100}{2} \times (2 \cdot 21 + (100 - 1) \cdot 20) = 50 \times (42 + 1980) = 50 \times 2022 = 101100 \]

### Conclusion
The sequence formed by the common terms of the two sequences has a sum of the first 100 terms equal to \(101100\).