The number of terms common to the A.P.’s 3, 7, 11, … 407 and 2, 9, 16, … 709 is.

Q: Why is the common difference of the common terms the LCM of the original differences?

Think of the common difference as the 'step size'. If the first A.P. jumps by $4$ and the second by $7$, a term that is common to both must be reachable by taking some number of $4$-steps and some number of $7$-steps from the first common term. To land on the same number again, you must have taken a number of steps that is a multiple of both $4$ and $7$. The smallest such step is the Least Common Multiple, which is $\text{LCM}(4, 7) = 28$.

Q: Why do we use the minimum of the two last terms as the upper bound?

An arithmetic progression is a finite set in this problem. A term can only be 'common' if it exists in both sets. If the first sequence stops at $407$ and the second at $709$, any term greater than $407$ cannot possibly be in the first sequence, and therefore cannot be a common term. Thus, the common sequence must terminate at or before the smaller of the two limits.

Q: What if the first common term isn't obvious by just listing them?

Great question! If the sequences were longer, you would use the general term equations: $T_{n1} = 3 + (n_1-1)4$ and $T_{n2} = 2 + (n_2-1)7$. Setting them equal, $3 + 4n_1 - 4 = 2 + 7n_2 - 7$, simplifies to $4n_1 - 1 = 7n_2 - 5$, or $4n_1 = 7n_2 - 4$. You can solve this Diophantine equation for the smallest positive integers $n_1$ and $n_2$ to find the first common term.

Question

The number of terms common to the A.P.’s 3, 7, 11, … 407 and 2, 9, 16, … 709 is _______.

Sigma Percentile · Accepted Answer

### The Dance of Two Sequences

Imagine you are standing on a long, straight path. On this path, there are two sets of markers.

The first set of markers is placed at $3, 7, 11, \dots$ and stops at $407$. The second set of markers is placed at $2, 9, 16, \dots$ and stops at $709$.

Your goal is to find how many locations have a marker from *both* sets. This is the essence of finding common terms in arithmetic progressions (A.P.s).

### Phase 1: Decoding the DNA of the Sequences

Every A.P. has a unique DNA defined by its first term ($a$) and its common difference ($d$). For our first sequence, we start at $a_1 = 3$ and jump by $d_1 = 4$.

The second sequence is a bit more aggressive; it starts at $a_2 = 2$ and takes larger leaps of $d_2 = 7$. To solve this, we must first identify the very first point where these two rhythms synchronize.

By listing the first few terms, we see the pattern emerge:

- **A.P. 1:** $3, 7, 11, 15, 19, \mathbf{23}, 27, \dots$
- **A.P. 2:** $2, 9, 16, \mathbf{23}, 30, \dots$

There it is! The number $23$ is the first common term, which we denote as $a_c = 23$.

### Phase 2: The Birth of the 'Super-Sequence'

Here is where the beauty of mathematics shines. The common terms are not random; they form their own arithmetic progression.

Since the first sequence jumps by $4$ and the second by $7$, the common terms must jump by a value that is a multiple of both. This is the Least Common Multiple (LCM).

Calculating $d_c = 	ext{LCM}(4, 7) = 28$, we realize that every $28$ units, the two sequences will collide again. We have created a new A.P. with $a_c = 23$ and $d_c = 28$.

### Phase 3: The Boundary Constraint

We cannot march forever. The first sequence ends at $407$, and the second ends at $709$.

A common term must exist in both, so it cannot exceed the smaller of these two limits. Thus, our 'super-sequence' is bounded by $\min(407, 709) = 407$.

We are looking for the number of terms $n$ such that the $n$-th term $T_n \leq 407$.

### Phase 4: The Final Calculation

We use the standard formula for the $n$-th term of an A.P.: $T_n = a_c + (n-1)d_c$. Substituting our known values, we get the inequality:

$$23 + (n-1)28 \leq 407$$

First, we subtract $23$ from both sides to isolate the term involving $n$:

$$(n-1)28 \leq 384$$

Next, we divide by $28$:

$$n-1 \leq \frac{384}{28} \approx 13.71$$

Adding $1$ to both sides, we find $n \leq 14.71$. Since $n$ must be a whole number, the maximum integer value for $n$ is $14$.

### Conclusion

There you have it! By understanding the rhythm of the sequences and respecting the boundaries of the problem, we have determined that there are exactly **14** common terms.

Mathematics is not just about numbers; it is about finding the hidden order in the chaos. Keep practicing, and soon, these patterns will become second nature to you!

Animated Solution for Mathematics - Sequence and Series: The number of terms common to the A.P.’s 3, 7, 11, … 407 and 2, 9, 16, … 709 is _______.

Enter Numerical Value:

Visualized Solution

Analyze the First A.P.

Analyze the Second A.P.

Finding the First Common Term

Logic of Common Difference

Determining the Upper Bound

Setting up the Inequality

Solving: Step 1

Solving: Step 2

Finding the Maximum 'n'

Final Conclusion

The Sigma Insight: Arithmetic Progression (A.P.)

The Dance of Two Sequences

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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