Not hard, but...
We have a sequence of 100 positive integers.
The average of 1st and 2nd term is 1.
The average of 2nd and 3rd term is 2.
The average of 3rd and 4th term is 3. The pattern continues.
The average of 99th and 100th term is 99. Find the 100th term.
The average of 1st and 2nd term is 1.
The average of 2nd and 3rd term is 2.
The average of 3rd and 4th term is 3. The pattern continues.
The average of 99th and 100th term is 99. Find the 100th term.
Re: Not hard, but...
Here are some hints.
Let's define the $i$-th element of the sequence as $a_i$.
Hint $1$:
Hint $2$:
Let's define the $i$-th element of the sequence as $a_i$.
Hint $1$:
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- Raiyan Jamil
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Re: Not hard, but...
I think the ans is 100 . I'm confused . Please someone tell me the correct ans .
Last edited by Raiyan Jamil on Wed Feb 12, 2014 10:28 am, edited 1 time in total.
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Re: Not hard, but...
There was a typo in Hint $1$.Here's a corrected version.Labib wrote: Hint $1$:
Hint $1$:
here it is =>
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"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
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- Raiyan Jamil
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Re: Not hard, but...
#Labib , how 99 ? I tried to make the sequence = 0 , 2 , 2 , 4 , 4 , 6 , 6 , 8 , 8 , 10 , 10 , .................... Like this I think the ans will be 100 . If not , it will be 98 . But 99 is a odd number and can't be contained in the sequence . So , how 99 ?
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Re: Not hard, but...
Positive integers start from $1$, not $0$.
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- Raiyan Jamil
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Re: Not hard, but...
Sorry I forgot . So the sequence will be 1 , 1 , 3 , 3 , 5 , 5 , 7 , 7 , 9 , 9 , 11 , 11................................ And the 100th term will be 99 .
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Re: Not hard, but...
Oooopppsss.....!sorry!I forgot that it must have to be an integer.My mistake........
Re: Not hard, but...
Thanks to all! The sequence is 1, 1, 3, 3,...and the 100th term is 99.