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.

@Shaun I've checked it. It says ?=128 which is wrong. I've found the nth term. — user103816, Jun 03 '14 at 10:39
@Shaun this question was asked in SBI clerk exam in India on 27/05/2012. — user103816, Jun 03 '14 at 10:43
Maybe your question doesn't show research effort, is unclear or not useful...+1 for your answer — draks ..., Jun 03 '14 at 10:43
Yes, but one could be perfectly forgiven for thinking it's $128$ and that $127$ is a typo. (I think just about any number could be valid if your reason for choosing it is sophisticated enough.) — Shaun, Jun 03 '14 at 10:46
@draks 1. What is unclear? 2. Why is it not useful? 3. What kind of research effort is expected? — user103816, Jun 03 '14 at 10:48
I just wrote what the popup on the downvote button says. Maybe you could have come up with the differences yourself. Further for me it was clear, but, ok, not useful... — draks ..., Jun 03 '14 at 10:51
@Shaun No it is not 128. There is no typo. How any other answer can be valid? This question was asked in an exam. There was only one answer eligible. Can you show how any other answer can be made valid? — user103816, Jun 03 '14 at 10:52
@draks I would come up with the difference if I had knew it. — user103816, Jun 03 '14 at 10:53
@Shaun Do you mean that we can make a sophisticated nth term expression to prove other answers valid? If this is possible then the whole exam will become incorrect because their are plenty of these type of question. People are elected for the job of clerk in SBI bank in India. Could you really somehow show the other option valid? — user103816, Jun 03 '14 at 11:06
Well, there's 4 very sophisticated sequences in that link above for which the answer is $128$. It's not my fault that exam question is faulty. If it's "guess the number and explain your reasoning", that's fine, but if it's "guess the number", it's not, and for the reason above. — Shaun, Jun 03 '14 at 11:38
@Shaun I am not getting you. Your first comment links to the sequence 87, 89, 95, 107, 128, 152. My question's sequence is 87,89,95,107,?,157 — user103816, Jun 03 '14 at 13:11
@Shaun You are somehow right. This type of questions suck. One can show all the 4 options correct. Someone has explained me this. In this question 3rd option 139 can be shown correct as: http://www.wolframalpha.com/input/?i=interpolating+polynomial+calculator&f1=%7B87%2C89%2C95%2C107%2C139%2C157%7D&f=InterpolatingPolynomialCalculator.data_%7B87%2C89%2C95%2C107%2C139%2C157%7D&a=FVarOpt.1-_InterpolatingPolynomialCalculator.data--.InterpolatingPolynomialCalculator.data2---.-- — user103816, Jun 03 '14 at 16:48
Without the necessary context, such "guess-the-number-exercises" are ill-posed. You can never be sure whether you have hit the intention of the "inventer". The interpolating-argument however does not convince me. Moreover, for a reasonable guess, more numbers should be given. The more numbers are given, the higher the probability that the "reasonable guess" is correct. — Peter, Dec 06 '18 at 14:25

robjohn · Accepted Answer · 2014-06-04T17:24:08.523

11

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.

edited Jun 04 '14 at 17:24

answered Jun 04 '14 at 12:21

robjohn

345,667

I do not understand your answer. Would you elaborate it? – user103816 Jun 04 '14 at 16:13
@user31782: I have added some extra explanation. If you still have questions, let me know what they are. – robjohn Jun 04 '14 at 17:24
As far I know $\dbinom nk, = {^n}C_r$, where $n \geq r$. How do I calculate $f_1(1)$, that is $f_1(1)$ contains terms like $\dbinom 12,$, that is $-1!$. – user103816 Jun 04 '14 at 17:39
1

@user31782: No. For $k,n\in\mathbb{Z}$, $\displaystyle\binom{n}{k}=0$ if $k\lt0$ or $0\le n\lt k$. To be precise, $$\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$ – robjohn Jun 04 '14 at 17:43

score 7 · Answer 2 · answered Jun 03 '14 at 10:32

7

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 $20$?

answered Jun 03 '14 at 10:32

Hakim

10,213

score 2 · Answer 3 · answered Jun 03 '14 at 10:32

2

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.

answered Jun 03 '14 at 10:32

Marc

6,861

score 2 · Answer 4 · answered Jun 03 '14 at 10:40

2

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 $b_1=2$.

answered Jun 03 '14 at 10:40

user103816

3,831

score 2 · Answer 5 · answered Jun 03 '14 at 10:44

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) $127$.

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

5 Answers5

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