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2 \times 3
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2^{34}
234
a_{i-1}
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\frac{2}{3}
32
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Although the intended answer here is C, you should know that there is no correct answer to this. I could easily say the answer is B.
You can ask me why. Well, my explanation is: the nth term of this sequence is f(n) where f(n)=−24n5−15n4+85n3−213n2+262n−720.Plug in n=1,2,3,4,5 and get f(n)=25,24,22,19,15 [your sequence]. Plug in n=6 and you'll get f(n)=5. So, B is a perfectly valid answer.
I know that this is not the answer you want and you're probably getting angry at me right now! What I'm trying to say is there's no correct answer.
@Iitian Singh
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Tushar's many accounts have been banned. Please in the future email me at discussions@brilliant.org to report cheating. I deleted the comment because, I thought it distracted from the point Mursalin was trying to make.
The nth term of this series is An=An−1−n where 25 is the 0th term; hence the 5th term which is what your looking for is A5=A4−5. Therefore A5=15−5 which is 10 hence the answer is C.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Although the intended answer here is C, you should know that there is no correct answer to this. I could easily say the answer is B.
You can ask me why. Well, my explanation is: the nth term of this sequence is f(n) where f(n)=−24n5−15n4+85n3−213n2+262n−720.Plug in n=1,2,3,4,5 and get f(n)=25,24,22,19,15 [your sequence]. Plug in n=6 and you'll get f(n)=5. So, B is a perfectly valid answer.
I know that this is not the answer you want and you're probably getting angry at me right now! What I'm trying to say is there's no correct answer.
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How did u get this f(n)
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why was that comment deleted i wrote in it that tushar is a cheater who cheats from many id's and has atleast 20 ids on brillaint
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I used Lagrange interpolation. But this is not the only way.
For example: you could add (n−1)(n−2)(n−3)(n−4)(n−5) to the polynomial to get a different value for the sixth term.
This way you can practically get anything (even complex numbers) for the sixth term!
The nth term of this series is An=An−1−n where 25 is the 0th term; hence the 5th term which is what your looking for is A5=A4−5. Therefore A5=15−5 which is 10 hence the answer is C.
10, because 25-1=24, 24-2=22,22-3=19,19-4=15, so 15-5=10
10
10
we can do this in this way to: here consecutive nos are being subtracted.i from 25 then 2 from 24 and then 3 from 22 and so on..........
so the ultimate answer is 15-5(as it is the 5th term in sequence)=10 (c)
C,10
Well, It is necessary to ask you what part of Mathematics is it after we will conclude an answer.
10
c
It's C. 10