Here is a straightforward example that knowing the first $k$ terms (no matter how big $k$ is) of a sequence $(x_n)$ tells you very little about the term $x_{k+1}$: I can generate a sequence whose first $k$ terms are $1,2,\dots,k$ but whose $(k+1)^{{\rm st}}$ term is an arbitrary $m$ of my choosing. For the given $k$, let $x_n = n + \frac {m-n}{k!}(n-1)(n-2)\dots(n-k)$.
Now, for any $n$ up to $k$, a factor in the product is $0$, forcing the entire product to be $0$, leaving $x_n=n$. Meanwhile, at $n=k+1$, the product $(k+1-1)\dots(k+1-k)=k!$, and straightforward arithmetic shows $x_{k+1}=m$.
Much more in fact is true. Linear algebra can be used to show, given completely arbitrary first terms $a_1,\dots,a_{k+1}\in\mathbb R$, there is a sequence $(x_n)$ for which $x_1=a_1,\dots,x_{k+1}=a_{k+1}$, and the generic term $x_n$ of $(x_n)$ is given as a polynomial in $n$ with degree at most $k$. (Indeed, the terms don't even need to be the ``first'' terms, we just need to know which terms they are.)
Without knowing something more about a sequence of real numbers than just ``these are some of the terms'', nothing of substance can be concluded about the sequence.
Edit (30 May 2019): Many standardized tests have questions along the lines of ``If a sequence starts with the numbers $1$, $2$, $4$, $8$, what is the next term in the sequence?" then proceeds to have a multiple choice answer, providing four numbers/answers. One main point behind this example is that all four answers are correct.