First of, I know this is quite easy but I can't really work it out.

I need find the rule for these sequences

$a.$ $2, 3, 4.5, 6.25...$

$b$ $54, 18, 6, 2...$

$c$ $0.01, 0.1, 1, 10...$

My Steps

What I did first was find the differences between the numbers.

For question $a$ the difference between numbers goes $1$, $1.5$, $1.75$

For question $b$ the differnce between numbers goes $-36$, $-12$, $-4$. In this sequence I have found out that the number is being divided by $3$.

For question $c$ the difference between numbers goes $+0.09$, $+0.9$, $+9$. In this the term $9$ is being multiplied by $10$.

I am not sure what to do after. I can do the $nth$ term when it comes to numbers that go up the same and also squared, cubed and triangle numbers.

I know the general formula for $nth$ term is:

nth term = difference x n + (first term - difference).

I can't really use it though as the difference varies. Thanks!

  • 1
    $\begingroup$ take ratios of each term to the previous $\endgroup$
    – Maesumi
    Apr 8, 2014 at 13:09
  • $\begingroup$ For $a$ note that in the sequence of differences, the distance to $2$ halves in each step. $\endgroup$ Apr 8, 2014 at 13:09
  • $\begingroup$ Infinite answers exist. Suppose a has values corresponding to a 3rd degree polynomial $ax^3+bx^2+cx+d$ at 1,2,3,4 Four equations four variables. Also, you can take any degree polynomial you wish and count at whatever you want... There are infinite answers. $\endgroup$
    – evil999man
    Apr 8, 2014 at 13:10
  • $\begingroup$ @Awesome You have to ignore this in order to meaningfully answer these questions. (Try to supply less "entropy" in your answer than given in the sequence information.) $\endgroup$ Apr 8, 2014 at 13:11
  • 1
    $\begingroup$ @PerfectNutter A geometric series takes the form $ar^{n-1}$, where $a$ is the first term and $r$ is the ratio between successive terms. (Be sure to test your guess by plugging in $n=1,2,3,4$ to make sure they match!) $\endgroup$ Apr 8, 2014 at 13:17

3 Answers 3


Instead of differences, try the quotients between consecutive terms.


a: Note that the difference of differences is halved at each step.

b: You already answered that - no need to look at the differences, you already stated a rule which generates the $(n+1)$-th term from the $n$-th term. You just need to turn this rule into a formula for the $n$-term.

c: Don't look at the differences here, that only obfuscates things. Just look at how the $(n+1)$-th term is generated from the $n$-th term. It's quite obvious.

  • $\begingroup$ Regarding $a$, you only have two data points for the difference of differences, so you're running low on entropy... $\endgroup$ Apr 8, 2014 at 13:14

$a.$ $2, 3, 4.5, 6.75...$

$a = 2$ $r = 1.5$ because $3÷2 = 1.5$, $4.5 ÷ 3 = 1.5$ and so on.

The $nth$ term rule is $54 × 1.5^n-1$.

$b.$ $54, 18, 6, 2...$

$a = 54$ $r = $$\frac{1}{3}$ because $18÷54 = $$\frac{1}{3}$, $6÷18 = $$\frac{1}{3}$ and so on.

The $nth$ term rule is $54 × $$\frac{1}{3}^n-1$.

$c.$ $0.01,0.1,1,10...$

$a = 0.01$ $r = 0.1$

The $nth$ term rule is $0.01 × $$\frac{1}{10}^n-1$

  • $\begingroup$ I hope this is right and sorry for the poor formatting. $\endgroup$
    – itshanks
    Apr 9, 2014 at 9:30

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