What is the appropriate number that should be inserted in the following series?

Question

What is the appropriate number that should be inserted in the following series of numbers arranged in a logical order?

$6,5,10,8,\text{____},9,12,8,10\\ $

$ \text{(A)} 10 \\ \text{(B)} 11 \\ \text{(C)} 12 \\ \text{(D)} 13 \\ \text{(E)} 14 $

In order to solve this problem I first tried to apply the formula for Arithmetic Sequences, Finding the nth term.

$a_n = a_1+d(n-1)\\ d = \text{common distance} , a_n= \text{nth term in the sequence},n = \text{the number of the term.}.$

However after careful inspection I realized that there was no common distance among the integers.

One thing I have noticed is the pattern subtract,add, subtract.

The answer to this question is (C) $12$. What pattern or formula is used to find solve this question?

Answer 1

It appears you first -1, then +5, then -2, then +4, etc.

So an iterative formula would be $a_1 = 6$ and for $n >1$: $$a_n = \begin{cases}a_{n-1}-\frac n2, & n \text{ even}, \\ a_{n-1} + 6-\frac{n-1}{2} & n \text{ odd}. \end{cases}$$

I don't have a general formula to hand, but perhaps another contributor could, or explain why not.

Answer 2

This kind of questions is hideous... You can come up with an infinite number of "logical" orders that would make any of the options true. For example, with the following expressions you can get any of the answers:

(A) $$ a_n=\frac{23 n^8}{3360}-\frac{23 n^7}{84}+\frac{3317 n^6}{720}-\frac{339 n^5}{8}+\frac{333463 n^4}{1440}-\frac{6107 n^3}{8}+\frac{3691333 n^2}{2520}-\frac{61697 n}{42}+580 $$

(B) $$ a_n =\frac{173 n^8}{20160}-\frac{173 n^7}{504}+\frac{8309 n^6}{1440}-\frac{7627 n^5}{144}+\frac{830791 n^4}{2880}-\frac{136291 n^3}{144}+\frac{9100991 n^2}{5040}-\frac{151219 n}{84}+706 $$

(C) $$ a_n = \frac{13 n^8}{1260}-\frac{26 n^7}{63}+\frac{104 n^6}{15}-\frac{572 n^5}{9}+\frac{10361 n^4}{30}-\frac{10166 n^3}{9}+\frac{2704829 n^2}{1260}-\frac{44761 n}{21}+832 $$

(D) $$ a_n = \frac{27 n^8}{2240}-\frac{27 n^7}{56}+\frac{11659 n^6}{1440}-\frac{3559 n^5}{48}+\frac{1158521 n^4}{2880}-\frac{63007 n^3}{48}+\frac{12537641 n^2}{5040}-\frac{206869 n}{84}+958 $$

(E)

$$ a_n=\frac{139 n^8}{10080}-\frac{139 n^7}{252}+\frac{6667 n^6}{720}-\frac{6101 n^5}{72}+\frac{661193 n^4}{1440}-\frac{107693 n^3}{72}+\frac{7127983 n^2}{2520}-\frac{117347 n}{42}+1084 $$