Question 1195799: What is the next number in this series ?
Series : 1, 54, 375, 1372, 3645, ?
Looking forward to your prompt and positive reply 🙂
Found 3 solutions by ikleyn, greenestamps, math_tutor2020:
Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website! .
https://www.ibpsguide.com/practice-quantitative-aptitude-questions-for-ibps-2017-exams-missing-wrong-number-series/
N 7.
Nonsense, as all other similar problems of this type.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
As the other tutor said, any problem like this is nonsense if you are expected to get the "right" answer.
However, in ANY problem like this, you can get AN answer that is mathematically justifiable by assuming the given sequence is formed by a polynomial equation. In that case, you can find "A" next number using the method of finite differences.
The given sequence has 5 terms, so there is a unique polynomial of degree 4 that produces the sequence. And in that polynomial of degree 4, the 4th differences are constant. (If you know some basic calculus, that is saying that the 4th derivative of a polynomial of degree 4 is a constant.)
Use the method of finite differences to find the constant 4th difference
1 54 375 1372 3645
53 321 997 2273
268 676 1276
408 600
192
The constant 4th difference is 192. Use that constant difference to build back up the array to find the next number in the sequence.
1 54 375 1372 3645 7986
53 321 997 2273 4341
268 676 1276 2068
408 600 792
192 192
The next number in the sequence, assuming a polynomial of degree 4, is 7986.
That is one of the answer choices; so apparently this sequence is a polynomial sequence.
Some of the other sequences at the URL referenced by the other tutor MIGHT be polynomial sequences, but others might not.
In the end, you can NEVER know if any answer you get to a problem like this is the "right" answer.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Here's a similar problem.
Consider the sequence 1, 2, 4, ...
One assumption is that the next term could be 7 because:
The jump from 1 to 2 is +1
The jump from 2 to 4 is +2
If we kept this pattern going, then the next jump would be +3 and we'd get 4+3 = 7
After that, the next term is 7+4 = 11, and so on.
So we might have 1, 2, 4, 7, 11, ...
Side note: The generating polynomial for this would be y = 0.5x^2 - 0.5x + 1, if the previous paragraph's assumptions were true.
Or, the next term could be 8 since the original three terms appear to show the powers of 2
2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
and so on
Put another way, the jump from term to term might be "multiply by 2".
Or maybe there's some recursive method we could use to build those first three terms.
Maybe we could say: To generate the next term, add the previous two terms, then add 1
term3 = (term1)+(term2)+1
term3 = (1)+(2)+1
term3 = 4
then,
term4 = (term2)+(term3)+1
term4 = (2)+(4)+1
term4 = 7
and,
term5 = (term3)+(term4)+1
term5 = (4)+(7)+1
term5 = 12
Under this process we have the sequence 1, 2, 4, 7, 12, ...
It somewhat resembles the Fibonacci sequence.
I'm sure you could get creative in a number of ways to generate the next term.
To summarize, we could have these sequences
1, 2, 4, 7, 11, ...
1, 2, 4, 8, 16, ...
1, 2, 4, 7, 12, ...
Likely there are infinitely others.
The punchline here is that problems asking about the next term are often too vague.
When things are vague, multiple answers are possible. It's like saying "I'm thinking of a round thing", and another person asking "is that a rock? a ball? a marble? a wheel?".
There's simply not enough information.
It's why the other tutors consider this problem to be "nonsense". I agree somewhat.
If there was context like "this sequence is arithmetic" or "the sequence is polynomial", then it would be much more clear and would provide one single definitive answer.
With that said, there might be the implication the sequence could be polynomial. In that it follows a fourth degree polynomial form that @greenestamps mentioned. Of course this is purely speculative based on my experience with problems of this nature.
I would ask your teacher for further clarification.
|
|
|
