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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

Part A : Hint 1

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.

Hint 2

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.

Hint 3

So , $\frac{1}{m_k} > 0$.

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

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

Hint 4

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

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

Proceed with the above inequality and get $m_1 + 2 > k .$

Part B : Hint 1

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.

Hint 2

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.

Hint 3

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 $\frac{1}{60}.$

ISI Entrance Program at Cheenta

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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

Part A : Hint 1

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.

Hint 2

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.

Hint 3

So , $\frac{1}{m_k} > 0$.

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

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

Hint 4

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

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

Proceed with the above inequality and get $m_1 + 2 > k .$

Part B : Hint 1

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.

Hint 2

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.

Hint 3

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 $\frac{1}{60}.$

ISI Entrance Program at Cheenta

## Subscribe to Cheenta at Youtube

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