ISI 2015 Subjective Problem 8 | A Problem from Sequence

IIT Mathemathematics by Asit Dasgupta

ISI UG Entrance - 2015 , Subjective problem number - 8

Try to prove using the following hints.

Assume , \(d\) is the common difference for the given AP.

Therefore,

\( d = \frac{1}{m_2} - \frac{1}{m_1} \leq \frac{1}{m_1+1} - \frac{1}{m_1} = d' (say)\)

[Equality holds when , \(m_2 = m_1 + 1 \) ]

Hence proceed.

Now , \( \frac{1}{m_k} = \frac{1}{m_1} + d(k-1) \leq \frac{1}{m_1} + d'(k-1)\)

Notice that till now we haven't used \(m_1 < m_2 < \ldots < m_k\) are positive integers.

So , \(\frac{1}{m_k} > 0 \).

Therefore , \(\frac{1}{m_1} + d'(k-1) > 0 \).

Now use \( d' = \frac{1}{m_1+1} - \frac{1}{m_1} \).

\[\frac{1}{m_1} + d'(k-1) > 0 \]

\[\Rightarrow \frac{1}{m_1} + ( \frac{1}{m_1+1} - \frac{1}{m_1})(k-1) > 0 \]

Proceed with the above inequality and get \(m_1 + 2 > k .\)

As \(m_1 < m_2 < \ldots < m_k\) ,

so \(\frac{1}{m_1}, \frac{1}{m_2}, \ldots, \frac{1}{m_k}\)

is a decreasing \(AP.\)

Think about the LCM of \(m_1 < m_2 < \ldots < m_k\) .

Then you proceed.

Suppose , \(L = LCM(m_1 < m_2 < \ldots < m_k).\)

Now multiply \(L\) with all the reciprocals

i.e. with \(\frac{1}{m_1}, \frac{1}{m_2}, \ldots, \frac{1}{m_k}.\)

Then observe the pattern and try get such a sequence.

E.g. if \(k=5\)

Take \(5,4,3,2,1\) and \(LCM(5,4,3,2,1)=60\).

So the AP : \(\frac{5}{60},\frac{4}{60} , \frac{3}{60} , \frac{2}{60}, \frac{1}{60}\)

with common difference \(\frac{1}{60}.\)

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