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This is a question appeared in a competitive exam. The question is:

Find the unknown term in $165,195,255,285,345,x$

1)375 $\ \ \ \ \ \ \ \ $ 2)420
3)435 $\ \ \ \ \ \ \ $ 4)390

My Research Effort

$$a_n = \begin{cases} a_{n-1}+30, & \text{if $n$ is even} \\ a_{n-1}+60, & \text{if $n$ is odd} \\ \end{cases}$$
where $n>1$
$a_1=165$. In this way the answer should be $a_6=375$. BUT this is not the correct answer. Any hints will be appreciated.
Thanks in advance.

Martin Sleziak
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user103816
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    That is a perfectly fine answer. Write an angry letter to whoever wrote this question ... – David Mitra Jun 03 '14 at 11:23
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    In fact, in all such question there is no 'one correct answer', there can be more than one answer. You can use [interpolation](http://en.wikipedia.org/wiki/Interpolation) to simply tell a relation between known and unknown value!. – hrkrshnn Jun 03 '14 at 11:35
  • @boywholived You are right there are 2 answers to this question. But I was not getting why 435 was assumed as the correct answer, so I asked it here. The book mentions that there is only 1 answer eligible and most of the questions have only 1 correct answer. The book seems to be wrong. – user103816 Jun 03 '14 at 13:41
  • @user31782. My point is all the four options are correct answer to the question. If you want a non-trivial relation between the terms you can use interpolation to find one. – hrkrshnn Jun 03 '14 at 14:49
  • @boywholived The question seems to be about _Number sequence_. Would it be ok to use interpolation? – user103816 Jun 03 '14 at 15:12
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    Generally, we are seeking the "simplest" explanation. That sounds like it can be a matter of taste, which is true. I like your version better than the accepted one. – Ross Millikan Jun 03 '14 at 15:54
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    @user31782, the issue is similar to that appear in data fitting. interpolation/fitting is acceptable when and only when the number of parameters required to fit the data is significantly less than the number of data points. In your case, Lagrange interpolation isn't that useful because it takes 5 parameters to fit your 5 known numbers. In certain sense, you method has 2 free parameters $30$ and $60$ while the one in Oleg567's answer only need 1 free parameter $15$. Both of these will be a much better choice/guess. – achille hui Jun 03 '14 at 16:06
  • @achillehui Do you know any other SE site where my question would be on-topic? – user103816 Jun 03 '14 at 16:24
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    @user31782, No idea. As other users have pointed out, there usually isn't a well defined answer for this sort of question. Since there are a lot of such questions recently, it is hard to find a site that welcome this. However, if you encounter a sequence when you study math or other science instead of just an exercise or puzzle from friend. You may still ask this sort of question provide you supplied enough evidence the sequence itself has real mathematical meanings. – achille hui Jun 03 '14 at 16:38
  • @achillehui The reason for closing this question is: "_This question is not about mathematics_". I read this question in an objective type book. They do not _explicitely_ mentions that the question is Mathematical. In the book it falls under _Quantitive aptitute_ category. So is there a SE site for _Quantitive aptitute_ questions? – user103816 Jun 03 '14 at 16:46
  • @user31782, once again, no idea. I don't visit other SE sites ;-p – achille hui Jun 03 '14 at 16:51
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    @user31782: Note that a simpler form of your answer is $a_n=a_{n-2}+90$. Of course, this also gives $375$ as the next term. – robjohn Aug 05 '14 at 13:18
  • @robjohn Thankyou for letting me know this. I guess it comes this way: $a_n=a_{n-1}+30$ and $a_{n-1}=a_{n-2}+60 \implies a_n=(a_{n-2}+60)+30=a_{n-2}+90$ for any $n$, odd or even( I know the other way $a_n=a_{n-1}+60\ ;a_{n-1}=a_{n-2}+30....)$ . Also Mr. Tunk Fey has given another similar looking pattern $a_{n}=a_{n-3}+(120+(n-4)30)$. – user103816 Aug 05 '14 at 15:17
  • @user31782: we can also solve it with $a_n=105+45n+15(1-(-1)^n)/2$ – robjohn Aug 05 '14 at 15:53
  • @user31782: this all goes to show that these kinds of problems have no unique solution. – robjohn Aug 05 '14 at 15:57
  • @robjohn Actually when I asked this and the [another similar one](http://math.stackexchange.com/q/819047/103816), I didn't know that these are ambiguous question. To conform the fact I asked another [related question](http://math.stackexchange.com/q/822544/103816) Now I understand this. – user103816 Aug 06 '14 at 14:46

4 Answers4

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I'm saying that $435$ is the answer to the question. Why? Consider the polynomial $$ p(x)=-\frac{3x^5}{2}+\frac{55x^4}{2}-\frac{375x^3}{2}+\frac{1175x^2}{2}-786x+525$$

Here is a table of values of $p(x)$ on consecutive $x$.

$\begin{array}{|l|c|c|c|c|c|c|}\hline x & 1 & 2& 3& 4& 5&6\\ \hline p(x)&165&195&255&285&345&\color{red}{435}\\\hline \end{array} $


My friend is saying that $390$ is the answer. Why? Consider the polynomial $$g(x)=-\frac{15x^5}{8}+\frac{265x^4}{8}-\frac{1755x^3}{8}+\frac{5375x^2}{8}-\frac{3555x}{4}+570$$

Here is a table of values of $g(x)$ for $x=1,2,3,4,5,6$

$\begin{array}{|l|c|c|c|c|c|c|}\hline x & 1 & 2& 3& 4& 5&6\\ \hline g(x)&165&195&255&285&345&\color{red}{390}\\\hline \end{array} $


How did I calculate the polynomials?

We are calculating a polynomial $p(x)$ that attains values $165,195,255,285,345,k$ (here $k$ is any number number) when $x=1,2,3,4,5,6$. I used a principle known as Lagrange Interpolation and the tool Wolframalpha interpolation calculator. Similarly one can construct even more complex relations using various interpolation techniques.

Conclusion: There is no unique "next term of the sequence", since for arbitrary number $\lambda$, you can always form a relation in which $\lambda$ should be the next term, although some relations may look more natural than others.

hrkrshnn
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    Your answer is very very very useful to me. Using this method the whole exam can be proven faulty. In India they ask this type of questions for Bank jobs, under the heading "_Quantitative aptitude_". Thanks a lot. – user103816 Jun 03 '14 at 15:52
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    I downvoted because it does not answer the question. Not really. With such questions the asker is wanting the answerer to "play a game". Both parties understand the game exists, and some may or may not understand that you can rise above this game and give any answer. To gain true understanding, you must learn to both play the game and to rise above it. (I would have upvoted your post if you had mentioned this "game", but as it stands you do not acknowledge the game (and this led to the OPs comment, as well as their comment on the other answer).) – user1729 Jun 10 '14 at 12:46
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    @user1729PhD. Frankly, I don't understand "the game". – hrkrshnn Jun 10 '14 at 17:28
  • The game is "guess the next number in this sequence *which I am thinking about*". – user1729 Jun 11 '14 at 10:13
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    @user1729 I have made progress with the game. See [here](http://math.stackexchange.com/a/867790/63095) – hrkrshnn Jul 15 '14 at 09:37
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In this case:

consider sequence

$$ a_n = 15 \cdot p_n, $$ where $p_n$ is $n$-th prime number.

$a_n$: $\color{gray}{30, 45, 75, 105,} 165, 195, 255, 285, 345, \color{red}{435}, ...$

Oleg567
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15

Another approach but it's just mere an intuitive one.

enter image description here

Tunk-Fey
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4

375 is the answer. Though I don't treat such questions as mathematical question. But they do appear in competitive exams and are base on kind of pattern recognition. Look at the sequence: 165,(+30)=195 ,(+60)=225,(+30)=285,(+60)=345,(+30)=375

WhoKnowsWho
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    $375$ is not the correct answer. See the question. – hola Jul 20 '14 at 19:21
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    This is the same reasoning as presented in the question. I've already commented to the question that this is also $a_n=a_{n-2}+90$. Did you have anything more to add? – robjohn Aug 05 '14 at 13:21