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CAT 2019 Set-2 | Question: 99

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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

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