To find the sum of the first 16 common terms between the two series \(3 + 7 + 11 + 15 + \ldots\) and \(1 + 6 + 11 + 16 + \ldots\), we can follow these steps:
### Step 1: Identify the two series
The first series is:
- \(3, 7, 11, 15, 19, 23, 27, 31, 35, 39, \ldots\)
This is an arithmetic series where:
- First term \(a_1 = 3\)
- Common difference \(d_1 = 4\)
The general term of this series can be expressed as:
\[
a_n = 3 + (n-1) \cdot 4 = 4n - 1
\]
The second series is:
- \(1, 6, 11, 16, 21, 26, 31, 36, 41, \ldots\)
This is also an arithmetic series where:
- First term \(b_1 = 1\)
- Common difference \(d_2 = 5\)
The general term of this series can be expressed as:
\[
b_n = 1 + (n-1) \cdot 5 = 5n - 4
\]
### Step 2: Find common terms
To find the common terms, we need to solve the equation:
\[
4n - 1 = 5m - 4
\]
Rearranging gives:
\[
4n - 5m = -3
\]
This is a linear Diophantine equation. We can find integer solutions for \(n\) and \(m\).
### Step 3: Find particular solutions
To find a particular solution, we can try different integer values for \(m\) and see if \(n\) is an integer:
- For \(m = 1\):
\[
4n - 5(1) = -3 \implies 4n = 2 \implies n = \frac{2}{4} \text{ (not an integer)}
\]
- For \(m = 2\):
\[
4n - 5(2) = -3 \implies 4n = 7 \implies n = \frac{7}{4} \text{ (not an integer)}
\]
- For \(m = 3\):
\[
4n - 5(3) = -3 \implies 4n = 12 \implies n = 3 \text{ (integer)}
\]
So one solution is \(n = 3\) and \(m = 3\).
### Step 4: General solution
The general solution for the equation \(4n - 5m = -3\) can be expressed as:
\[
n = 3 + 5k \quad \text{and} \quad m = 3 + 4k \quad \text{for integer } k
\]
### Step 5: Find the first 16 common terms
Now we can find the first 16 values of \(n\):
- For \(k = 0\): \(n = 3\)
- For \(k = 1\): \(n = 8\)
- For \(k = 2\): \(n = 13\)
- For \(k = 3\): \(n = 18\)
- For \(k = 4\): \(n = 23\)
- For \(k = 5\): \(n = 28\)
- For \(k = 6\): \(n = 33\)
- For \(k = 7\): \(n = 38\)
- For \(k = 8\): \(n = 43\)
- For \(k = 9\): \(n = 48\)
- For \(k = 10\): \(n = 53\)
- For \(k = 11\): \(n = 58\)
- For \(k = 12\): \(n = 63\)
- For \(k = 13\): \(n = 68\)
- For \(k = 14\): \(n = 73\)
- For \(k = 15\): \(n = 78\)
Now we can find the actual common terms by substituting these values of \(n\) into the formula for the first series:
- Common terms are:
\[
\begin{align*}
n = 3: & \quad 4(3) - 1 = 11 \\
n = 8: & \quad 4(8) - 1 = 31 \\
n = 13: & \quad 4(13) - 1 = 51 \\
n = 18: & \quad 4(18) - 1 = 71 \\
n = 23: & \quad 4(23) - 1 = 91 \\
n = 28: & \quad 4(28) - 1 = 111 \\
n = 33: & \quad 4(33) - 1 = 131 \\
n = 38: & \quad 4(38) - 1 = 151 \\
n = 43: & \quad 4(43) - 1 = 171 \\
n = 48: & \quad 4(48) - 1 = 191 \\
n = 53: & \quad 4(53) - 1 = 211 \\
n = 58: & \quad 4(58) - 1 = 231 \\
n = 63: & \quad 4(63) - 1 = 251 \\
n = 68: & \quad 4(68) - 1 = 271 \\
n = 73: & \quad 4(73) - 1 = 291 \\
n = 78: & \quad 4(78) - 1 = 311 \\
\end{align*}
\]
### Step 6: Calculate the sum of the first 16 common terms
Now we sum these common terms:
\[
S = 11 + 31 + 51 + 71 + 91 + 111 + 131 + 151 + 171 + 191 + 211 + 231 + 251 + 271 + 291 + 311
\]
Calculating this sum:
\[
S = 11 + 31 + 51 + 71 + 91 + 111 + 131 + 151 + 171 + 191 + 211 + 231 + 251 + 271 + 291 + 311 = 2576
\]
Thus, the sum of the first 16 common terms is \(2576\).
### Final Answer
The sum of the first 16 terms common between the series is \(2576\).
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