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# ISI 2015 Subjective Problem 8 | A Problem from Sequence

Try this beautiful Subjective Sequence Problem appeared in ISI Entrance - 2015.

## Problem

(b) For any integer $$k>0$$, give an example of a sequence of $$k$$ positive integers whose reciprocals are in arithmetic progression.

### Key Concepts

Sequence

Arithmetic Progression

## Suggested Book | Source | Answer

IIT Mathemathematics by Asit Dasgupta

ISI UG Entrance - 2015 , Subjective problem number - 8

Try to prove using the following hints.

## Try with 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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