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.
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.