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 $\frac{1}{60}.$

ISI MStat 2020 PSB Problem 8 Solution

Problem

Assume that $X_{1}, \ldots, X_{n}$ is a random sample from $N(\mu, 1)$, with $\mu \in \mathbb{R}$. We want to test $H_{0}: \mu=0$ against $H_{1}: \mu=1$. For a fixed integer $m \in{1, \ldots, n}$, the following statistics are defined:

$\begin{aligned} T_{1} &= \frac{\left(X_{1}+\ldots+X_{m}\right)}{m} \\ T_{2} &= \frac{\left(X_{2}+\ldots+X_{m+1}\right)} {m} \\ \vdots &=\vdots \\ T_{n-m+1} &= \frac{\left(X_{n-m+1}+\ldots+X_{n}\right)}{m} . \end{aligned}$

Fix $\alpha \in(0,1)$.

Consider the test

Reject $H_{0}$ if $\max \{T_{i}: 1 \leq i \leq n-m+1\}>c_{m, \alpha}$

Find a choice of $c_{m, \alpha} \in \mathbb{R}$ in terms of the standard normal distribution function $\Phi$ that ensures that the size of the test is at most $\alpha$.

Give a free ISI MStat Diagonistic Test

Hint 1

Show that the problem is equivalent to finding that $P_{\mu = 0}(\max \{T_{i}: 1 \leq i \leq n-m+1\}\\>c_{m, \alpha}) \leq \alpha$

Hint 2

$P_{\mu = 0}(\max \{T_{i}: 1 \leq i \leq n-m+1\}\\>c_{m, \alpha})$

$= P_{\mu = 0}( T_1 > c_{m, \alpha} \cup T_2 > c_{m, \alpha} \cdots T_{n-m+1}\\ > c_{m, \alpha})$

Hint 3

Use Boole's Inequality o get

$P_{\mu = 0}( T_1 > c_{m, \alpha} \cup T_2 > c_{m, \alpha} \cdots T_{n-m+1}\\ > c_{m, \alpha}) \leq \sum_{i = 1}^{n-m+1} P(T_i > c_{m, \alpha}) = \alpha $

Hint 4

Show that under $H_0$, $T_i$ ~ $N(0,\frac{1}{m})$. Hence, find $c_{m, \alpha}$

See the full solution below.

Full Solution

Food For Thoughts

What if, $\lim _{n \rightarrow \infty} c_{m, \alpha}$?
What if strict $c_{m, \alpha}$ was required to find? Can you solve this using Jacobian. (You can give an approximate solution).

Give a free ISI MStat Diagonistic Test

IIT JAM MS 2020 Section A Problem 1 Solution

Problem

If $\{x_{n}\}_{n \geq 1}$ is a sequence of real numbers such that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0.001$, then

(A) $\{x_{n}\}_{n \geq 1}$ is a bounded sequence
(B)$\{x_{n}\}_{n \geq 1}$ is an unbounded sequence
(C) $\{x_{n}\}_{n \geq 1}$ is a convergent sequence
(D) $\{x_{n}\}_{n \geq 1}$ is a monotonically decreasing sequence

Hints

IIT JAM MS Coaching Courses

Hint 1

If $\{x_{n}\}_{n \geq 1}$ was bounded, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by sandwich theorem.

Hint 2

If $\{x_{n}\}_{n \geq 1}$ was convergent, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by algebra of limits.

Hint 3

If $\{x_{n}\}_{n \geq 1}$ was motonotically decreasing and bounded below, then it would have been convergent by Monotone Convergence Theorem.

Let's consider if it is not below below, i.e. $\lim _{n \rightarrow \infty} {x_{n}} = -\infty $

Take $ {x_{n}} = -n $
Take $ {x_{n}} = -n^2 $
Take $ {x_{n}} = - \sqrt{n} $

Find the limit in each of this case.

Hence, it will be unbounded. See the full solution and proof idea below.

Full Solution

Food For Thoughts

What if, $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}= 0$?
Find examples.

IIT JAM MS Coaching Courses

Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

Sequence and permutations - AIME II, 2015

Call a permutation $a_1,a_2,....,a_n$ of the integers 1,2,...,n quasi increasing if $a_k \leq a_{k+1} +2$ for each $1 \leq k \leq n-1$, find the number of quasi increasing permutations of the integers 1,2,....,7.

is 107
is 486
is 840
cannot be determined from the given information

Key Concepts

Sequence

Permutations

Integers

Check the Answer

Answer: is 486.

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

Try with Hints

While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 $\times$ number of permutations of (n-1) elements

or, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements

......

or, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 $\times$ $3^{n-2}$

forming recurrence relation as the number of permutations =2 $\times$ $3^{n-2}$

for n=3 all six permutations taken and go up 18, 54, 162, 486

for n=7, here $2 \times 3^{5} =486.$

Header text

sds

ISI MStat 2018 PSA Problem 11 | Sequence & it's subsequence

This is a problem from ISI MStat 2018 PSA Problem 11 based on Sequence and subsequence.

Sequence & it's subsequence - ISI MStat Year 2018 PSA Problem 11

Let $ {a_{n}}_{n \geq 1}$ be a sequence such that $ a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq \cdots$
Suppose the subsequence $ {a_{2 n}}_{n \geq 1}$ is bounded. Then

(A) $ \{a_{2 n}\}_{n \geq 1}$ is always convergent but $ \{a_{2 n+1}\}_{n \geq 1} $ need not be convergent.
(B) both $ \{a_{2 n}\}_{n \geq 1}$ and $ \{a_{2 n+1}\}_{n \geq 1}$ are always convergent and have the same limit.
(C) $ \{a_{3 n}\}_{n \geq 1}$ is not necessarily convergent.
(D) both $ \{a_{2 n}\}_{n \geq 1}$ and $ \{a_{2 n+1}\}_{n \geq 1}$ are always convergent but may have different limits.

Key Concepts

Sequence

Subsequence

Check the Answer

Answer: is (B)

ISI MStat 2018 PSA Problem 11

Introduction to Real Analysis by Bertle Sherbert

Try with Hints

Given that $ a_{2n} $ is bounded . Again we have $ a_{1} \leq a_{2} \leq \cdots \leq a_{2n} \leq a_{2n+1} \leq a_{2n+2} \leq \cdots$ which shows that if $ a_{2n} $ is bounded then $ a_{2n+1} $ is also bounded .

Again both $ a_{2n} $ and $ a_{2n+1} $ both are monotonic sequence . Hence both converges .

Now we have to see whether they converges to same limit or not ?

As both $ a_{2n} $ and $ a_{2n+1} $ are bounded hence $ a_{n} $ is bounded and it's already given that it is monotonic . Hence $ a_{n} $ converges . So, it's subsequences must converges to same limit . Hence option (B) is correct .

Outstanding Statistics Program with Applications

ISI MStat 2018 PSA Problem 12 | Sequence of positive numbers

This is a problem from ISI MStat 2018 PSA Problem 12 based on Sequence of positive numbers

Sequence of positive numbers - ISI MStat Year 2018 PSA Question 12

Let $a_n $ ,$ n \ge 1$ be a sequence of positive numbers such that $a_{n+1} \leq a_{n}$ for all n, and $\lim {n \rightarrow \infty} a{n}=a .$ Let $p_{n}(x)$ be the polynomial $ p_{n}(x)=x^{2}+a_{n} x+1$ and suppose $p_{n}(x)$ has no real roots for every n . Let $\alpha$ and $\beta$ be the roots of the polynomial $p(x)=x^{2}+a x+1 .$ What can you say about $ (\alpha, \beta) $?

(A) $ \alpha=\beta, \alpha$ and $\beta$ are not real
(B) $ \alpha=\beta, \alpha$ and $\beta$ are real.
(C) $\alpha \neq \beta, \alpha$ and $\beta$ are real.
(D) $\alpha \neq \beta, \alpha$ and $\beta$ are not real

Key Concepts

Sequence

Quadratic equation

Discriminant

Check the Answer

Answer: is (D)

ISI MStat 2018 PSA Problem 12

Introduction to Real Analysis by Bertle Sherbert

Try with Hints

Write the discriminant. Use the properties of the sequence $ a_n $ .

Note that as $P_n$ has no real root so discriminant is $(a_n)^2-4<0$ so $|a_n|<2$ and $a_n$ 's are positive and decreasing so $0\leq a_n<2$ . So , what can we say about a ?

Therefore we can say that $ 0 \le a < 2 $ hence discriminant of P , $a^2-4 $ must be strictly negative so option D.

Outstanding Statistics Program with Applications

Telescopic Continuity | ISI MStat 2015 PSB Problem 1

This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.

Problem- Telescopic Continuity

Let $f: R \rightarrow R$ be a function which is continuous at 0 and $f(0)=1$
Also assume that $f$ satisfies the following relation for all $x$ :
$$
f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x
$$ Find $f(3)$.

Prerequisites

Telescopic Sum
Continuity of a Function

Solution

$$
f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3
$$

$$
f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}
$$

$$
f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}
$$

$ \cdots $

$$
f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}
$$

Add them all up. That's the telescopic elegance.

$$
f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]
$$

Observe that $ a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1$ since, $f(x)$ is continuous at $x=0$.

Hence take limit $ n \to \infty $ on $[*]$, and we get $ f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15 $.

Food for Thought

Find the general function from the given condition.
$$f(x)-f(\frac{x}{2})=g(x)$$ and g(x) is continuous, then prove that $g(0) =0$.
What if $g(x)$ is not continuous?

Combination of Sequence | B.Stat Objective | TOMATO 79

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Combination of Sequence.

Logic and combination of sequence (B.Stat Objective)

The two sequence of numbers {1,4,16,64,.....} and {3,12,48,192,.....} are mixed as follows {1,3,4,12,16,48,64,192,....}. One of the numbers in the mixed series is 1048576. Then the number immediately preceeding it is

262144
786432
814572
786516

Key Concepts

Logic

Sequence

Integers

Check the Answer

Answer: 786432.

B.Stat Objective Question 79

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

The first series is of form $4^{r}$ for $r \geq 0$ $r \in$ set of natural numbers the second series is of form $3 \times 4^{r}$ for $r \geq 0$ $r \in$ set of natural numbers and the third series is of $4^{r}$,$3 \times 4^{r}$ in alternate element form for $r \geq 0$ $r \in$ set of natural numbers

given that 1048576=$4^{r}$=$4^{10}$

then preceeding term $3 \times 4^{9}$=(3)(262144)=786432.

Geometric Sequence Problem | AIME I, 2009 | Question 1

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on geometric sequence.

Geometric Sequence Problem - AIME 2009

Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

is 500
is 250
is 840
cannot be determined from the given information

Key Concepts

Sequence

Series

Real Analysis

Check the Answer

Answer: is 840.

AIME, 2009

Introduction to Real Analysis, 4th Edition by Robert G. Bartle, Donald R. Sherbert

Try with Hints

3-digit sequence a, ar, $ar^{2}$. The largest geometric number must have a<=9.

ar $ar^{2}$ less than 9 r fraction less than 1 For a=9 is $\frac{2}{3}$ then number 964.

a>=1 ar and $ar^{2}$ greater than 1 r is 2 and number is 124. Then difference 964-124=840.

Composite number Problem | B.Stat Objective | TOMATO 75

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Sequence and composite number.

Composite number Problem (B.Stat Objective)

Consider the sequence $a_1$=101, $a_2$=10101,$a_3$=1010101 and so on. Then $a_k$ is a composite number ( that is not a prime number)

if and only if $k \geq 2$ and $11|(10^{k+1}+1)$
if and only if $k \geq 2$ and k-2 is divisible by 3
if and only if $k \geq 2$ and $11|(10^{k+1}-1)$
if and only if $k \geq 2$

Key Concepts

Logic

Sequence

Composite number

Check the Answer

Answer: if and only if $k \geq 2$ and k-2 is divisible by 3

B.Stat Objective Question 75

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

for $a_k$ $k \geq 2$ may be prime also then not considering this here

for $a_{8}$ $10^{9}-1$ and $10^{9}+1$ not divisible by 11

8-2 is divisible by 3 and $a_{8}$ is composite number then $a_{k}$ is composite if and only if $k \geq 2$ and k-2 is divisible by 3.

Problem

Key Concepts

Suggested Book | Source | Answer

Try with Hints

Other Useful Links

Related Program

Subscribe to Cheenta at Youtube

ISI MStat 2020 PSB Problem 8 Solution

Problem

Hint 1

Hint 2

Hint 3

Hint 4

Full Solution

Food For Thoughts

IIT JAM MS 2020 Section A Problem 1 Solution

Problem

Hints

Hint 1

Hint 2

Hint 3

Full Solution

Food For Thoughts

Sequence and permutations - AIME II, 2015

Key Concepts

Check the Answer

Try with Hints

Header text

Header text

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Sequence & it's subsequence - ISI MStat Year 2018 PSA Problem 11

Key Concepts

Check the Answer

Try with Hints

Similar Problems and Solutions

Related Program

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube

Sequence of positive numbers - ISI MStat Year 2018 PSA Question 12

Key Concepts

Check the Answer

Try with Hints

Similar Problems and Solutions

Related Program

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube

Problem- Telescopic Continuity

Prerequisites

Solution

Food for Thought

Logic and combination of sequence (B.Stat Objective)

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Geometric Sequence Problem - AIME 2009

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Composite number Problem (B.Stat Objective)

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube