How to solve the sequence: $87, 89, 95, 107, ?, 157$

Question

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

Q. Find the unknown term in $87,89,95,107,?,157$

$1$.) $127$ $\ \ \ \ \ \ \ \ $ $2.$) $122$
$3$.) $139$ $\ \ \ \ \ \ \ \ $ $4$.) $140$

I tried very much to solve it but can't find the correct answer. Any hints will be appreciated.
Thanks in advance.

Edit:

My effort that I did before asking this question:

I calculated the difference b/w consecutive terms. The diff is $2,6,12,20$. I was not able to recognize any pattern. After this I subtracted each term, which is less than $100$, from $100$ and, subtracted $100$ from each term which is greater than $100$. The resultant sequence is is $13,17,11,5,7,x,y$. Again I could not recognize any pattern.

Source

It was asked in SBI Clerical Exam on $\operatorname{May} 27, 2012$. It is the $110$th question of that exam.

Answer 1

The expected answer is probably something like this:

Each column is the difference of the previous column

   A   B   C   D
  --------------
  87
       2
  89       4
       6       2
  95       6
      12       2
 107       8
      20       2
?127?     10
      30 
 157

---

However, we can fit any number in the unknown spot with the proper function. Each of the following polynomials has $f_k(0)=87$, $f_k(1)=89$, $f_k(2)=95$, $f_k(3)=107$, $f_k(5)=157$, yet $$\begin{array}{l} f_1(n)=87\binom{n}{0}+2\binom{n}{1}+4\binom{n}{2}+2\binom{n}{3}&\text{gives }f_1(4)=127\\ f_2(n)=87\binom{n}{0}+2\binom{n}{1}+4\binom{n}{2}+2\binom{n}{3}-5\binom{n}{4}+25\binom{n}{5}&\text{gives }f_2(4)=122\\ f_3(n)=87\binom{n}{0}+2\binom{n}{1}+4\binom{n}{2}+2\binom{n}{3}+12\binom{n}{4}-60\binom{n}{5}&\text{gives }f_3(4)=139\\ f_4(n)=87\binom{n}{0}+2\binom{n}{1}+4\binom{n}{2}+2\binom{n}{3}+13\binom{n}{4}-65\binom{n}{5}&\text{gives }f_4(4)=140\\ \end{array} $$ All the possible answers have an explanation in terms of these polynomials.

Answer 2

Hint:
The difference between each term goes like this: $2,6,12.$ Can you notice any pattern? Based on it, can you prove that the next term in it is $20$?

Answer 3

The difference of the difference between the numbers is increased by two each time. Therefore the answer is probably 127, although as always, if you're creative enough you could probably invent some story to motivate any of the other answers.

Answer 4

I've recognized the patteren. The general term is
$a_n=a_{n-1} +b_n$ for $n>1$, where $b_n=b_{n-1}+2n$ for $n>1$. $a_1=87$ and $b_1=2$.

Answer 5

Take the difference between the terms (1st difference): that's $2,6,12,..$

Now take the difference between those (2nd difference): that's $4,6...$

Guess the likely second difference continuation: $4,6,8,10,...$

Hence the likely first difference continuation would be: $2,6,12,20,30...$

Finally, the sequence would be: $87,89,95,107,127,157$. You're on the right track because the final term matches up and the second last term is one of the given options.

So the expected answer is 1) $127$.