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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.

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    $\begingroup$ That is a perfectly fine answer. Write an angry letter to whoever wrote this question ... $\endgroup$
    – David Mitra
    Commented Jun 3, 2014 at 11:23
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    $\begingroup$ In fact, in all such question there is no 'one correct answer', there can be more than one answer. You can use interpolation to simply tell a relation between known and unknown value!. $\endgroup$
    – hrkrshnn
    Commented Jun 3, 2014 at 11:35
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    $\begingroup$ 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. $\endgroup$
    – Ross Millikan
    Commented Jun 3, 2014 at 15:54
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    $\begingroup$ @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. $\endgroup$
    – achille hui
    Commented Jun 3, 2014 at 16:06
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    $\begingroup$ @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. $\endgroup$
    – achille hui
    Commented Jun 3, 2014 at 16:38

4 Answers 4

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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.

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    $\begingroup$ 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. $\endgroup$
    – user103816
    Commented Jun 3, 2014 at 15:52
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    $\begingroup$ 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).) $\endgroup$
    – user1729
    Commented Jun 10, 2014 at 12:46
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    $\begingroup$ @user1729PhD. Frankly, I don't understand "the game". $\endgroup$
    – hrkrshnn
    Commented Jun 10, 2014 at 17:28
  • $\begingroup$ The game is "guess the next number in this sequence which I am thinking about". $\endgroup$
    – user1729
    Commented Jun 11, 2014 at 10:13
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    $\begingroup$ @user1729 I have made progress with the game. See here $\endgroup$
    – hrkrshnn
    Commented Jul 15, 2014 at 9:37
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In this case:

consider sequence

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

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

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    $\begingroup$ Thankyou. The book mentions 435 as the answer. since 165 is also a valid answer, I guess the book is not trustworthy. $\endgroup$
    – user103816
    Commented Jun 3, 2014 at 13:22
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    $\begingroup$ Find the GCD of the numbers (15) and the sequence is revealed. $\endgroup$
    – Jesan Fafon
    Commented Jul 15, 2014 at 20:43
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Another approach but it's just mere an intuitive one.

enter image description here

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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

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    $\begingroup$ $375$ is not the correct answer. See the question. $\endgroup$
    – hola
    Commented Jul 20, 2014 at 19:21
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    $\begingroup$ 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? $\endgroup$
    – robjohn ♦
    Commented Aug 5, 2014 at 13:21

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