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$
$\Rightarrow \boxed{n=20}$
Correct Answer: D