Consider a harmonic progression with distinct real terms. Of the infinite number of statements below regarding that progression, what is the maximum possible number of true statements?
1) The
1
st
term of this sequence is equal to 1.
2) The
2
nd
term of this sequence is equal to 2.
3) The
3
rd
term of this sequence is equal to 3.
4) The
4
th
term of this sequence is equal to 4.
⋮
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Exactly the solution I had in mind! An alternative solution is to apply the quadratic formula .
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Please post it using quadratic formula
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Hint : If for a certain n , the n th term is equal to n , make n as a subject in terms of the first term a and the common difference d . What do you get? A quadratic equation. Now what is the maximum number of roots of a quadratic equation?
Hi Pi han, while solving this question, I first misinterpreted it and thought that its asks for how many unit fractions are possible with the denominator a positive integer such that they are in A.P. This makes an interesting question and I found 1/20, 1/24, 1/30, 1/40,1/60,1/120 after a lot of number theory as I wanted to prove that there can be more than 5 terms so I was about to answer There is no maximum value but Then I realised that the number in the denominator is the corresponding term so then it took me only some seconds to solve it as it is extremely easy but if you post the above as a question then it can be a much better one.
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Why don't you post it as a note then? Since it's your question...
Yes, that would make a fun problem. Let me know when you posted it!
Same way,BTW nice solution using basics .+1
The infinite list of statements describes the linear function ℓ ( n ) = n . The harmonic progression describes a function h ( n ) = b + n a . The number of true statements is the number of intersection points of the graphs. A line and a hyperbola have at most two intersection points, so that the maximum number is less than or equal to 2.
OHHHHH! This is very very nice!! ;)
h ( n ) = a + ( n − 1 ) d k defines harmonic progression. With h(1)=1 => a=k, h(2)=2 => d=-k/2 we get:
h ( n ) = 3 − n 2
We realize that the progression has only the 1st and 2nd term of development since h(3) is not defined and negative members are invalid.
Hmmmm... I didn't say that the first statement must be true. I could have the 7th and the 12th statement to be true.
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And now the general case: Assuming we have h(n)=n we get n = a + ( n − 1 ) d k which is a quadratic equation for n. The known solution formula only permits maximum 2 values for n.
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Yeah, that's correct!
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@Pi Han Goh – Thank you, Sir! Please continue posting challenging problems!
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Suppose, to the contrary, that at least three of the statements are true. Then for some distinct indices a , b , c we have t a = a , t b = b , t c = c . Since this sequence is a harmonic sequence, its reciprocals make an arithmetic sequence. The common difference of the arithmetic sequence can be written as follows: d = a − b a 1 − b 1 = b − c b 1 − c 1 d = − a b 1 = − b c 1
Therefore a = c . However, this contradicts the assumption that the indices are different. Hence, at most two of the statements can be true.