The number of common terms in the two sequences: $15, 19, 23, 27,......., 415$ and $14, 19, 24, 29,........,464$ is

$19$

$20$

$21$

$18$

✅ Correct Option: 2

First sequence: $15, 19, 23, 27, \ldots, 415$

First term $(a_1) = 15$ , common difference $(d_1) = 19 - 15 = 4$

Second sequence: $14, 19, 24, 29, \ldots, 464$

First term $(a_2) = 14$ , common difference $(d_2) = 19 - 14 = 5$

We observe that $19$ appears in both sequences:

First sequence: $15, 19, 23, 27, \ldots$

Second sequence: $14, 19, 24, 29, \ldots$

The first common term is $19$ .

When two arithmetic progressions have common terms, these common terms also form an arithmetic progression.

The common difference of this new sequence equals the LCM of the original common differences.

First sequence increases by $4$ each time, second sequence increases by $5$ each time. For both to "meet" again, we need the smallest number that both $4$ and $5$ divide evenly.

$\text{LCM}(4, 5) = 20$

The common terms form the sequence: $19, 39, 59, 79, \ldots$

We can verify that $19 + 20 = 39$ and $39 + 20 = 59$ , confirming both sequences contain these terms.

The sequence of common terms is: $19, 39, 59, \ldots$

We need to find how many terms are $\leq 415$ (the last term of the first sequence).

Using the formula for the $n$ -th term: $a_n = a_1 + (n-1)d$

$19 + (n-1) \times 20 \leq 415$

$19 + 20n - 20 \leq 415$

$20n - 1 \leq 415$

$20n \leq 416$

$n \leq 20.8$

Since $n$ must be a whole number, $n = 20$ .

There are $20$ common terms in the two sequences.

The $20$ th common term is $19 + (20-1) \times 20 = 19 + 380 = 399$ , which is indeed $\leq 415$ .

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The number of common terms in the two sequences: 15,19,23,27,.......,41515, 19, 23, 27,......., 41515,19,23,27,.......,415 and 14,19,24,29,........,46414, 19, 24, 29,........,46414,19,24,29,........,464 is

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The number of common terms in the two sequences: $15, 19, 23, 27,......., 415$ and $14, 19, 24, 29,........,464$ is