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

1. $18$
2. $19$
3. $21$
4. $20$

Given that,

• $15,{\color{Red}{19}},23,27,31,35,{\color{Red}{39}},\dots, 415 \quad \longrightarrow (1)$
• $14,{\color{Red}{19}},24,29,34,{\color{Red}{39}}, \dots ,464 \quad \longrightarrow (2)$

Let $’n’$ be a number of common terms in the two sequences.

Common difference in sequence $1 = 4,$ and in sequence $2 = 5.$

$\therefore$ The LCM of $(4,5)=20\quad$ (Common difference of new sequnce)

Now, the sequence will be,

$19,39,59,79,99,\dots , l \left(\leqslant 415\right) \quad \longrightarrow (3)$

We know that, last term $l=a+(n-1)d ,$ where $a =$ first term, $d =$ common difference, $n =$ number of terms.

Now, $l \leqslant 415$

$\Rightarrow 19+(n-1)20 \leqslant 415\quad [\because a =19,d = 20]$

$\Rightarrow 19+20n-20 \leqslant 415$

$\Rightarrow 20n-1 \leqslant 415$

$\Rightarrow 20n \leqslant 416$

$\Rightarrow n \leqslant \frac{416}{20}$

$\Rightarrow n \leqslant 20.8$

$\Rightarrow \boxed{n=20}$

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