What number comes next in this series?
$$7, 16, 8, 27, 9,...$$
I thought it was $38$, but I'm wrong.
It is a multiple choice, and options are $27, 10, 40, 37$.
Don't worry - I'm not cheating on anything, but helping my daughter with homework.
And I can't help her since I can't figure it out myself!
On
One possibility is that the $1$st, $3$rd, $5$th, and so on, numbers are just increasing by one, so $$7,\_,8,\_,9,\_,10,\_,...$$
Now notice that $16=2 \cdot 8$, and $27=3 \cdot 9$. Therefore it would be reasonable to assume the next number to be $40=4 \cdot 10$.
On
The next term is again $27$; then we had a palindromic sequence $$ 7,16,8,27,9,27,8,16,7,\ldots, $$ repeating like this.
On
Another view, also giving $40$ as the next term: $2n+2;\frac{n}{2};3n+3;\frac{n}{3};4n+4;\frac{n}{4}\cdots$ \begin{align}\text{initial value}&=7\\ 7*2+2&=16\\ \frac{16}{2}&=8\\ 8*3+3&=27\\ \frac{27}{3}&=9\\ 9*4+4&=40\\ \frac{40}{4}&=10 \end{align} Or: $2(n+1);\frac{n}{2};3(n+1);\frac{n}{3};4(n+1);\frac{n}{4}\cdots$ \begin{align}\text{initial value}&=7\\ 2(7+1)&=16\\ \frac{16}{2}&=8\\ 3(8+1)&=27\\ \frac{27}{3}&=9\\ 4(9+1)&=40\\ \frac{40}{4}&=10 \end{align}
To me, it's $40$.
Indeed
$$16 = 2\cdot 8$$ $$27 = 3\cdot 9$$
thence
$$40 = 4\cdot 10$$
The even terms of the series may follow this path, so a possible series could be
$$7, 16, 8, 27, 9, 40, 10, 55, 11, 72, 12, \cdots$$