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Fill in the correct number in this sequence:

\begin{array} & 3 & 4 & 2^2 & 5 & 10 & 60 & ? \end{array}

The options are

\begin{array} 60/10 & 6 & 12/2 & 7 & 49/7 \end{array}

So, apparently, the answer is either $6$ or $7$, but why? Also, I do not understand why the 4 is written in terms of a square and is present 2 times in the sequence. I have no clue how to solve it.

Source: a publicly available practice test in a book for an IQ-test I got via a friend in the Netherlands.

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5
  • $\begingroup$ Well I think I might have noticed one thing: ROT13 Gur cnggrea fgnegf jvgu guerr, gura fxvccvat gjb fgrcf tbrf gb svir, fb gura fxvccvat gjb fgrcf ntnva, vg fubhyq tb gb frira. Abj jr whfg unir gb svaq n eryngvbafuvc orgjrra gur erfcrpgvir fxvcf fgrccrq naq gur ahzore cerprqvat rnpu bs gurz, V guvax... jung qb lbh erpxba? $\endgroup$
    – Mr Pie
    Dec 7, 2018 at 15:23
  • $\begingroup$ I have edited my comment (hope you understand)?I have to get to bed anyways. I hope this is right! Ooh, fingers crossed... g'night :D $\endgroup$
    – Mr Pie
    Dec 7, 2018 at 15:39
  • $\begingroup$ In the original puzzle, does it show 2^2 or $2^2$? $\endgroup$
    – Dr Xorile
    Jan 30, 2019 at 3:38
  • $\begingroup$ @Xorile The second one. $\endgroup$
    – Dennispuz
    Jan 30, 2019 at 5:04
  • $\begingroup$ Has a useful answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful. $\endgroup$
    – Rubio ♦
    Mar 7, 2019 at 4:45

2 Answers 2

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It's:

6

because:

the sequence is $3,4,X,5,X,X,6,X,X,X,7,\dots$. That is, start with $3,4,5,6,7\dots$ and insert $a_n-3$ $X$'s after each entry. The $X$'s can be anything - they only distract.

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We notice that \begin{array}{c|c}\text{number}&0&1&2&3&4&5&6\\\hline\#\text{occurences}&2&1&2&1&1&1&1\end{array} Going through each choice, the pattern of occurrences becomes \begin{array}{c|c}\text{choice}&60/10&6&12/2&7&49/7\\\hline\small\text{pattern}&\small4,2,2,1,1,1,2&\small2,1,2,1,1,1,2&\small2,2,4,1,1,1,1&\small2,1,2,1,1,1,1,1&\small2,1,2,1,2,1,1,1,0,1\end{array} and only the first two seem to have a pattern in the pattern of occurrences, since the first one has one $4$, two $2$s and three $1$s, and rises back up; the second one has odd numbers of consecutive $1$s broken by the $2$s. If we look back at the sequence, $3$ and $4$ are together, separation of $0$, and $4$ and $5$ are separated by $1$ - the $2^2$ as it is not written as a single-digit integer. Therefore, the obvious prediction is that two spaces along, it is a single-digit integer and is the next one along from $5$. So it is $6$, and the answer is the second choice.

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