Understanding the Problem: Finding Common Terms in Two Arithmetic Progressions

The question asks us to identify the number of terms that appear in both of the given Arithmetic Progressions (APs). We have two APs:

• AP 1: Starts with 19, common difference is 2 (\(21 - 19 = 2\)), and it has 110 terms.
• AP 2: Starts with 19, common difference is 3 (\(22 - 19 = 3\)), and it has 75 terms.

To find the identical terms, we need to determine the sequence formed by these common terms.

Analyzing the First Arithmetic Progression (AP 1)

The first term (\(a_1\)) of AP 1 is 19.

The common difference (\(d_1\)) of AP 1 is \(21 - 19 = 2\).

The number of terms (\(n_1\)) in AP 1 is 110.

We can find the last term (\(a_{110}\)) of AP 1 using the formula for the n-th term of an AP: \(a_n = a_1 + (n-1)d\).

Calculating the last term of AP 1:

\(a_{110} = 19 + (110 - 1) \times 2\)

\(a_{110} = 19 + (109) \times 2\)

\(a_{110} = 19 + 218\)

\(a_{110} = 237\)

So, AP 1 ranges from 19 to 237.

Analyzing the Second Arithmetic Progression (AP 2)

The first term (\(b_1\)) of AP 2 is 19.

The common difference (\(d_2\)) of AP 2 is \(22 - 19 = 3\).

The number of terms (\(n_2\)) in AP 2 is 75.

We can find the last term (\(b_{75}\)) of AP 2 using the same formula: \(b_n = b_1 + (n-1)d\).

Calculating the last term of AP 2:

\(b_{75} = 19 + (75 - 1) \times 3\)

\(b_{75} = 19 + (74) \times 3\)

\(b_{75} = 19 + 222\)

\(b_{75} = 241\)

So, AP 2 ranges from 19 to 241.

Identifying the Common Terms' Progression

The first term common to both APs is clearly 19, as both sequences start with it.

Any term common to both APs must satisfy the conditions of both sequences. The common difference of the sequence formed by these common terms will be the Least Common Multiple (LCM) of the individual common differences (\(d_1\) and \(d_2\)).

Common difference of AP 1 (\(d_1\)) = 2

Common difference of AP 2 (\(d_2\)) = 3

LCM(\(d_1\), \(d_2\)) = LCM(2, 3) = 6.

Therefore, the sequence of common terms is also an Arithmetic Progression with:

• First term (\(c_1\)) = 19
• Common difference (\(d_c\)) = 6

The terms in this common AP are \(19, 19+6, 19+2\times6, ...\), which are \(19, 25, 31, ...\)

Calculating the Number of Common Terms

The common terms must exist within the range of both APs. The maximum value a common term can take is limited by the smaller of the two last terms.

Last term of AP 1 = 237

Last term of AP 2 = 241

The maximum possible value for a common term is \(\min(237, 241) = 237\).

Let \(k\) be the number of common terms. The \(k\)-th term of the common AP (\(c_k\)) can be represented as:

\(c_k = c_1 + (k-1)d_c\)

\(c_k = 19 + (k-1) \times 6\)

We need to find the largest integer value of \(k\) such that \(c_k \le 237\).

\(19 + (k-1) \times 6 \le 237\)

Subtract 19 from both sides:

\((k-1) \times 6 \le 237 - 19\)

\((k-1) \times 6 \le 218\)

Divide by 6:

\(k-1 \le \frac{218}{6}\)

\(k-1 \le 36.333...\)

Since \(k-1\) must be an integer (representing the number of steps from the first term), the largest possible integer value for \(k-1\) is 36.

\(k-1 = 36\)

Solving for \(k\):

\(k = 36 + 1\)

\(k = 37\)

Therefore, there are 37 identical terms in the two given APs.