How is this a number sequence $58, 26, 16, 14, 10$ I recently had a IQ Test taken and we all got stuck on the same question. The question was:
What comes next in the following sequence?
$$58, 26, 16, 14,\_\_$$
The answer given in the answer sheet was $10$. My question is why? What pattern exists in those numbers?
 A: I get $6$ for the fifth term:
$T(n)=\frac{378-272n+75n^2-7n^3}{3}$  gives the general term.
$58=T(1)$
$26=T(2)$
$16=T(3)$
$14=T(4)$
$6=T(5)$
A: $5+8 = 13$ and twice $13$ is $26$, etc.
A: I will probably be accused of pedantry for this answer, but I cannot help but point out that a question like this is not a math question.  There are many sequences that begin 58, 26, 16, 14, 10.  There are many patterns in the sequence 58, 26, 16, 14, 10, by which I mean that there are many computer programs that compute a sequence that begins this way, and they do not all finish the sequence in the same way.
I understand that what the OP is looking for is the simplest pattern in the sequence, which gives the simplest continuation of the sequence.  My objection is that "simple" here has no mathematical meaning (Kolmogorov complexity will not work because it is only defined up to a constant.)  
That being said, here is my preferred answer: 58, 26, 16, 14, 10, 666, 666, 666, 666, 666,...
The pattern is that Satan put the first five numbers there to mislead clever people and the rest of them are 666.
A: This has a very simple arithmetical interpretation as casting out nines from $\rm\:2^k n,\:$ performed by interleaved casting and doubling, i.e. repeatedly: sum the digits of $\rm\,n\,$ then double.  Using this to compute $\:2^4\cdot 58\pmod 9$ we obtain the following computation
$$\begin{eqnarray}\rm cast &\rm out&\rm 9'\!s\quad\ double\qquad\\ \hline \end{eqnarray} $$
$$\begin{eqnarray}
 5 + 8 &=& 13,\quad 2&\cdot& 13 &\,=\,& \color{#0A0}{26} \\
\color{#0A0}2 + \color{#0A0}6  & = &\ \  8,\quad 2&\cdot&\ \ 8 &\,=\,& \color{#C00}{16} \\
\color{#C00}1 + \color{#C00}6 &=&\ \ 7,\quad 2&\cdot&\ \ 7 &\,=\,&\color{blue}{ 14} \\
\color{blue}1 + \color{blue}4 &=&\ \ 5,\quad 2&\cdot&\ \ 5 &\,=\,& 10
\end{eqnarray}$$
Hence $\: 2^4\cdot 58\equiv 10 \pmod 9.\ $ Notice that $\,58,\,\color{#0A0}{26},\,\color{#C00}{16},\,\color{blue}{14},10,\ $ the sequence of intermediate results above, is precisely the original sequence.
A: The Answer is pretty simple once you notice it.
58, 26, 16, 14 ....


*

*(5+8)*2=26

*(2+6)*2=16

*(1+6)*2=14

*(1+4)*2=10

*(1+0)*2=2


So just take the first number which is 58 and change it to 5+8, then multiply by 2 to get the next number from which you just follow the same steps, adding the two numbers together then multiply by 2.
A: I got 13.5
2^5 = 32
2^3 = 8
2^1 = 2
2^(-1) = 1/2
58 - 2^5 = 26
26 - 2^3 = 16
16 - 2^1 = 14
14 - 2^(-1) = 13.5
