Problems on Integers: Q1-Q3 - Solutions Needed

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The discussion focuses on three math problems involving sequences and fractions. For the first question, participants clarify the sequence definition and determine that all terms exceed 6, thus none fall between 1/100 and 39/100. The second question about irreducible fractions between 10 and 20 with a denominator of 3 leads to confusion, with one participant suggesting the answer might be infinite. The third question involves finding pairs of natural numbers with a GCD of 5 and an LCM of 105, but no specific solutions are provided. Overall, the thread highlights the challenges faced in solving these integer-related problems.
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Q 1:-
Given a sequence kn=[(1+(-1)^n)+1]/5n+6..
find the no of terms of the sequence kn which will satisfy the condition kn lies between 1/100 and 39/100.

Q 2:-
Find the sum of all the irreducable fractions between 10 and 20 with a denominator of 3

Q 3:-
Find all pairs of natural no s whose greatest common divisor is 5 and L.C.M is 105

i could really not do much about these so please help! Any assistance appreciated ..
 
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Well, what did you do? And why was this not posted under "homework help"?
 
See i don't know anything about forums..like where do we even get the section "homework help"..! anyways that's not the issue here.
For the first question i tried to make the denominator 100 for which i got n in fraction now that if i put in numerator then in all probability it becomes a question of complex numbers which i don't know.For the second i am getting the answer uncountable or infinity and i don't know third
 
kn=[(1+(-1)^n)+1]/5n+6
Find the no of terms of the sequence kn which will satisfy the condition kn lies between 1/100 and 39/100.

There are none, as all the terms in the sequence are greater than 6.
 
I suppose you mean kn=[(1+(-1)n)+1]/(5n+6) .

The parentheses are important.

What you wrote literally means: k_n=\frac{(1+(-1)^n)+1}{5n}+6\,, which is how uart likely interpreted it.

There are many numbers between 1/100 and 39/100 which don't have a denominator of 100.

How many terms of the sequence are between 0.01 and 0.39 ?

The terms of the sequence with n odd look much different from the terms with n even.
 
There are many numbers between 1/100 and 39/100 which don't have a denominator of 100.

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yeah but i was trying to first find that n for which i shall get the limiting values;i mean 0.01 and 0.39. And i did not understand what you said in the second part
 
If n is odd, (-1)n = -1 .

If n is even, (-1)n = 1 .
 
yeah right i also got till there but what when n is in decimal??
 
n is a positive integer.

It's a sequence we're talking about.
 
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