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

Step 1: Analyze the given Arithmetic Progressions (A.P.s)

• First A.P.: $$3, 7, 11, \dots, 407$$
  • First term ($$a_1$$) = 3
  • Common difference ($$d_1$$) = $$7 - 3 = 4$$
  • Last term = 407
• Second A.P.: $$2, 9, 16, \dots, 709$$
  • First term ($$a_2$$) = 2
  • Common difference ($$d_2$$) = $$9 - 2 = 7$$
  • Last term = 709

Step 2: Find the first common term

Let's write out the first few terms of each sequence to spot the first match:

• First A.P.: 3, 7, 11, 15, 19, 23, 27, 31...
• Second A.P.: 2, 9, 16, 23, 30, 37...

The first common term ($$a$$) is 23.

Step 3: Find the common difference of the new A.P.

The common terms will form a new A.P. The common difference ($$d$$) of this new sequence is the Least Common Multiple (LCM) of the common differences of the two original sequences.

$$d = \text{LCM}(d_1, d_2) = \text{LCM}(4, 7) = 28$$

So, the new A.P. of common terms is: 23, 51, 79, ...

Step 4: Determine the maximum possible limit

The common terms cannot exceed the last term of the shorter sequence.

$$\text{Maximum Limit} = \min(407, 709) = 407$$

Step 5: Calculate the number of common terms

Let the common A.P. have $$n$$ terms. The $$n$$-th term must be less than or equal to the maximum limit (407). Using the standard formula $$T_n = a + (n-1)d$$:

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

$$28(n-1) \le 407 - 23$$

$$28(n-1) \le 384$$$$n-1 \le \frac{384}{28}$$

$$n-1 \le \frac{96}{7}$$$$n-1 \le 13.71$$

$$n \le 14.71$$

Since the number of terms ($$n$$) must be an integer, we take the largest whole number that satisfies the inequality.

Therefore, $$n = 14$$. The number of terms common to the two A.P.s is 14.