# Finding the next term of a sequence [closed]

Find the next term in the sequence

$$\frac{10}{9},\frac13,\sqrt{\frac{20}{3}},\frac23$$

I have no idea how to proceed. The second, fourth, etc. terms might form an AP. But what about the other terms? Any hint on how to approach this problem will be helpful.

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There is no "right" answer mathematically, so this is not appropriate. –  Igor Rivin Dec 21 '13 at 16:27
What the heck is an "AP" –  Don Larynx Dec 21 '13 at 16:28
@DonLarynx: arithmetic progression. But unless that information is given, I don't see any evidence to support it. –  Ross Millikan Dec 21 '13 at 16:29
Generally for problems like this, you are due enough of a pattern to feel sure you have it right. Even then, you can't really be sure. Four terms is only enough if the pattern is very simple, which this is not. –  Ross Millikan Dec 21 '13 at 16:30
@RossMillikan: agreed, just giving suggestions as to how one might proceed... –  abiessu Dec 21 '13 at 16:45
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## closed as too broad by Igor Rivin, user7530, amWhy, mathlove, ronnoDec 21 '13 at 16:56

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

A brute force (but inelegant) answer is to suppose that the sequence terms are function values of some polynomial. Since you've been given $4$ points, then a degree $3$ polynomial will work. Suppose that $f(x)=ax^3+bx^2+cx+d,$ and that $f(1),f(2),f(3),f(4)$ are (respectively) the first four terms of your sequence. We can then solve the resulting system for $a,b,c,d$ to see: $$a=\frac{5-18\sqrt{15}}{54}\\b=\frac{7\sqrt{15}-1}3\\c=-\frac{23+252\sqrt{15}}{54}\\d=\frac{16+24\sqrt{15}}9$$
Every problem of this type can be "solved" in this way. In fact, we can make the next finite number of terms be anything we like by using a similar approach. Even for "obvious" sequences like $1,2,4,8,16,\underline{}$ we can make the next number be (for example) $31$ (rather than the $32$ we might expect) in this fashion, by making $g(n-1)$ be the $n$th term of the sequence, where $$g(x)=\frac1{24}x^4-\frac1{12}x^3+\frac{11}{24}x^2+\frac7{12}x+1.$$ Carl Linderholm discusses this in his essay "Mathematics Made Difficult" [1972].