সমান্তর অনুক্রম

সমান্তর ধারা হচ্ছে এমন একটি ধারা যেখানে প্রতি ২ টি ক্রমিক সংখ্যার সাধারণ ব্যাবধান থাকবে।

ভালো! কিছু পরিভাষার জন্য, সমান্তর ধারার বৈশিষ্ঠই হচ্ছে ক্রমিক পদের সাধারণ ব্যাবধান

আপনি নিশ্চই এতক্ষণে বুঝতে পেরেছেন যে কোনো ধারা তখনই সমান্তর হয় যখন সেখানে ক্রমিক পদের মধ্যে সাধারণ ব্যাবধান থাকে।

অভিনন্দন! আশা করি আপনি সমান্তর ধারা বুঝতে পেরেছেন।

সমান্তর ধারা

সমান্তর ধারা হলো এমন একটি ধারা যেখানে সব সন্নিহিত পদের পার্থক্য সমান।

যদি প্রথম পদটি হয় $a$ আর পার্থক্য হয় $d$ তাহলে ধারাটি হবে $a+d,a+2d,a+3d \cdots$ এমন। পার্থক্য ধনাত্মক(positive) অথবা ঋণাত্মক(negative) দুইটাই হতে পারে। এমন কি $0$ ও হতে পারে।

উপরের পদ গুলর মধ্যে pattern রয়েছে। এখানে $n$-তম পদটি হলো $2^n$. তবে pattern থাকলেই কনো ধারা সমান্তর হবে না।

তোমরা অনেকেই জান যে উপরে ক্রমটিকে ফিবনাক্কি ধারা বলে। এখানেও একটি pattern রয়েছে। প্রতিটি পদ তার আগের দুই পদের সমষ্টি। তবে এখানে সন্নিহিত পদের পার্থক্য সমান নয়। তাই এটি সমান্তর ধারা নয়।

উদাহরণ ১ঃ $5, 9, 13, 17, 21, 25$ এই ধারার ৫০ নম্বর পদ কোনটি?

সমধানঃ এখানের $a_n$ নম্বর পদটিকে $4 \times n + 1$ আকারে লেখা যায়।

যেমনঃ ১ম পদটি হলোঃ $$a_1=4\times1+1=5$$

২য় পদটি হলোঃ $$a_2=4\times2+1=9$$

তাহলে ৫০ম পদটি হলো, $$a_{50}=4\times50+1=201$$

An arithmetic sequence is a sequence where the difference between consecutive terms is the same.

Good! For some terminology, the common difference of an arithmetic sequence is the difference between consecutive terms.

Good job! I hope you found this interesting :).

Arithmetic Sequences!

An arithmetic sequence is a sequence in which there is a common difference between the terms. If the first term is $a$ and the difference is $d$ then the sequence will be like $a+d, a+2d, a+ 3d, \cdots$. The difference can be both positive and negative. Let us try to identify some arithmetic sequences.

There is a clever way to find the amount of terms in an arithmetic sequence.

Let us go over what an arithmetic sequence is. It is a sequence with a common difference. So the terms of an arithmetic sequence will be $a, a+d, a+2d, \cdots$. It is called an arithmetic sequence since the difference between two terms is the same and the difference can be both positive and negative.

So let us look at the sequence of $a, a+d, a +2d, a+3d, \cdots, a+399d, a+ 400d$.

Notice how the first term has $(0)(d)$ added to to it, the second term has $(1)(d)$ which goes on and on. Each term has $(n-1)(d)$ added to it. So if we subtract the first term from the last term we get $400d$. So if this term is nth term we know that $(n-1)(d) = 400d$ so $n = \boxed{401}$.

Let us look at this with real numbers. If we have the sequence $3,8,13, \cdots 1498, 1503$, how many terms are there? We can observe that the difference is $5$. We can subtract $3$ from $1503$ and get $1500$. So by doing $\frac{1500}{5} = 300$ we can determine that the difference was added $300$ times. Thus, the amount of terms is $300+1 = \boxed{301}$.

Why don't we try some practice problems!

Now let us look at a more challenging question.

There is a simple way to find the $n$th term in an arithmetic sequence.

First we should make sure that we are clear in an arithmetic sequence.

An arithmetic sequence is called an arithmetic sequence since there is a common difference between each term. For example, if our first term is $a$ and our common difference is $d$ then our terms will be $a, a+d, a+2d, a+3d \cdots$. An important thing to remember is that $d$ can be both positive or negative!

Now let us look at some clever ways to explore arithmetic sequences.

Let us first see how we can find the nth term of a sequence. We can look at our original sequence of $a, a+d, a+2d, a+3d, \cdots$. Our first term is $a$, our second term is $a+d$, and our third term is $a+2d$. To find each of these, we can write in a different way. Our first term can be written as $a + (1-1)(d)$. Our second term can be written as $a+ (2-1)(d)$. Our third term can be written as $a+ (3-1)(d)$. We can observe that $a$ is the first term and $d$ is the difference. The number that is being multiplied by $d$ is the number of the term minus one, since the first term does not have the common difference added to it. Thus, we can observe that the nth term is $a+(n-1)(d)$.

Let us go over a question in this. What is the 523rd term in the sequence 2014,2017,2020... So let us first identify the first term which is $2014$. The term is the 523rd term. The difference is $3$. So we just plug in $2014 + (522)(3) = 3580$

Here is a question for you to try.

Finally, there is also a clever way to find the sum in an arithmetic sequence.

Note that you can make pairs in an arithmetic sequence.

$a, a+d, a+2d, a+3d$ can be our whole sequence. The first term and last term add up to the same thing as our second term and second to last term. $a+a+3d = a+d+a+2d = 2a+3d = 2a+3d$.

Now notice how it all adds up to $4a+6d$. We observe how each of the pairs add up to twice the average since the average is $a + 1.5d$. So the average of the two numbers in the pair is the average of the whole sequence. So basically, all we need to do is add the average the amount of numbers there are in the sequence.

For example if we have 1,3,5,7,9,11 as our sequence, our average is 6 and there are 6 terms. $6 \cdot 6 = 36$. We notice that $1+3+5+7+9+11 = 36$.

We can try a problem on this.

I hope you found my explanation on arithmetic sequences helpful and fun!

An arithmetic series (or sometimes called sequence) is the most basic of all kinds of series. Consider the following series. $$4+7+10+11+\ldots$$ If we ask ourselves, what's the next term, a common answer would be $14$ (let's pretend we don't know Lagrange's Interpolation or any interpolation formula right now). Which means, starting from $4$, we are always adding $3$ to the previous term to get the next term in the series. Similarly, in an arithmetic series, there is a certain constant difference (or simply difference) which can be added to any term to obtain the next. The basic structure of an arithmetic series is:

$a, a+d,a+2d, a+3d,\ldots$ Here, we started with the term $a$ and then we are adding $d$ to get the next term. So, if the first term is $a$, the next one will be $a+d$. The next one is $(a+d)+d=a+2d$. Similarly, the $n$th term will be $a+(n-1)d$.

An arithmetic sequence is a sequence where each successive term differs by a constant amount. The arithmetic sequence resembles the sequence: $a_n = a_{n-1} + d$ where $a_1$ is the first element and each term afterwards has the common difference term $d$ added to it, resulting in $a_1, a_1+d, a_1+2d...$ and to generalize, the nth term of this sequence is $a_n = a_1 + (n-1)d$.

An example would be the natural numbers where $a_1 =1, d= 1$ resulting in the progression: $1,2,3...$

Properties: $$S_n = \frac{n(a_1+a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}$$ The average of the elements in an arithmetic sequence is $$\bar{n} = \frac{(a_1+a_n)}{2}$$

An arithmetic sequence is a sequence where the difference between adjacent terms is constant.

The differences between the terms in the sequence above are: $2, 4, 8,\cdots$, which are not constant so this is not an arithmetic sequence!

The two main components of an arithmetic sequence are a starting term, $a$, and a common difference, $d$. From these two pieces of information alone, we can write out the entire arithmetic (though just the first few terms is fine): $a, a+d, a+2d, a+3d,$ etc. From this, we can write a formula to determine what the $n$th term of an arithmetic sequence is: $a_n=a_1+(n-1)d$.

Recall that an arithmetic sequence is in the form $x_k=d(k-1)+x_1$, where $d$ is the common difference, and $x_1$ is the first term.

Ways an arithmetic sequence ($x_1$, $x_2$, ...) can be defined:

1. By any term and its common difference

We first put in $d$ to the equation and then plug in ($k$, $x_k$) to find the formula.

Example:

An arithmetic sequence has $20$th term $14$ and common difference $2$. Find the formula of the sequence.

Plugging in $d$, we get $x_k=2(k-1)+x_1$. Plugging in ($k$, $x_k$)=($20$, $14$), we get $14=2(20-1)+x_1$, so $14=38+x_1$, and $x_1=-24$. Hence the formula of the sequence is $x_k=2(k-1)-24$.

Exercise:

2. By any $2$ terms (let's say $x_a$ and $x_b$)

We plug in ($a$, $x_a$) and ($b$, $x_b$) and then solve the equations for $x_1$ and $d$.

Example

An arithmetic sequence has $6$th term $20$ and $20$th term $90$. Find the formula of the sequence.

Plugging in ($6$, $20$) and ($20$, $90$), we get $20=d(6-1)+x_1$ and $90=d(20-1)+x_1$. Thus $5d+x_1=20$ and $19d+x_1=90$. Subtracting the latter equation from the former, we get $14d=70$, so $d=5$. Plugging in $d=5$ to the first equation: $5d+x_1=20$, we get that $x_1=-5$. Hence the formula of the sequence is $\boxed{x_k=5(k-1)-5}$.

Great job!

Now we investigate another amazing property of arithmetic sequences, the sum of its terms.

There may be other ways, like polynomial interpolation after knowing that the degree of the polynomial must be $2$ (we will get to that later), and then using Lagrange Interpolation after calculating the partial sums of the first term, the first $2$ terms, and then the first $3$ terms.

However, there is a method of calculating it faster. This was first discovered by Gauss.

Since the common differences are the same, $x_1+x_n=x_2+x_{n-1}=x_3+x_{n-2}=...$. Thus we just need to find the sum of $x_1$ and $x_n$, divide that by $2$ to get the average of all of the terms up to $n$, then multiply that by $n$.

If we are not given the formula; instead we are given the first and last term, our job becomes even easier! We can just plug them in. However, if finding the first and last terms are too laborsome, we could just find the formula of the sequence by any of the $2$ methods above, then plug in the formula to find the sum of the first $n$ terms.

Example: A sequence has $3$rd term $4$ and $10$th term $18$. Find the sum of its first $2014$ terms.

By the $2$nd method for finding the formula of sequences, we could plug in ($3$, $4$) and ($10$, $18$) to get $4=d(3-1)+x_1$ and $18=d(10-1)+x_1$. Thus $4=2d+x_1$ and $18=9d+x_1$. Subtracting the latter from the former, we get $7d=14$, so $d=2$. Plugging in $d=2$ to the first equation, we get that $x_1=0$. Hence the formula is $x_k=2x$. Its first term is $0$ and its $2014$th term is $2(2014-1)+0=4026$, so the average of its first $2014$ terms is $\frac{0+4026}{2}=2013$. Thus the sum of the first $2014$ terms is $2013 \cdot 2014=\boxed{4054182}$.

Of course, the sum that we wish to find doesn't need to start from the first term. That same method works for sums starting from a term other than the first term.

Example:

A sequence has $3$rd term $4$ and common difference $3$. Find the sum of all the terms from its $10$th term to its $16$th term.

We use the $1$st method to get that $4=3(3-1)+x_1$, so $x_1=4-6=-2$. Hence the formula of the sequence is $3k-2$. Its $10$th term is $28$, and its $16$th term is $46$, so the average of all the terms from the $10$th to the $16$th is $\frac{28+46}{2}=37$. Since there are $16-10+1=7$ terms from the $10$ to the $16$th term, the sum of the terms involved is $37\cdot 7=\boxed{259}$.

CHALLENGE QUESTIONS

The challenge questions would require additional concepts from other fields of mathematics, but the main problem is related to the arithmetic sequence.

So the formula for an arithmetic sequence is $x_k=ak+b$,right?

The sum of its first $6$ terms is $(a+b)+(2a+b)+...+(6a+b)=21a+6b=48$, while the sum of its $8$th term to its $13$th term is $(8a+b)+(9a+b)+...+(13a+b)=63a+6b=132$

Subtracting, we get $42a=84$, so $a=2$.

1. (lvl. 3.5-5 Early-Middle AIME) Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

2. (lvl. 5-6.5 Pre-Olympiad) Prove that if an infinite arithmetic progression of positive integers contains a perfect square, then it contains an infinite number of perfect squares.

3. (lvl. 6.5-8.5 Easy-medium Olympiad, IMO 1991) Let $n>6$ be a positive integer and let $1=a_1<a_2<...<a_k$ be the sequence of all positive integers less than n which are relatively prime with n. Prove that if the sequence $a_1$, $a_2$, ..., $a_k$ is a non-trivial arithmetic progression, then n is either a prime number or a power of $2$. [A “non-trivial” arithmetic progression has length at least three].

4. (lvl. 8.5-10 Hard Olympiad, IMO 2009 #6)

Let $a_1, a_2, \ldots , a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s=a_1+a_2+\ldots+ =a_n.$ A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1, a_2, \ldots , a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M.$

Arithmetic sequences are sequences where, between each two consecutive terms, there is a common difference.

For example, $1,2,3,4,5,6,7,...$ is an arithmetic sequence because the distance between two terms is equal.

Arithmetic sequences can be written in the form $a, a+d, a+2d, a+3d, ...$. When they are in this form, d is the common difference and a is the first term.

In the arithmetic sequence $2,5,8,11,14, ..$ we have $a=2, d=3$.

Notice that if an arithmetic sequence has n terms, then the last term is $a+(n-1)d$. In general, this expression equals the nth term of the sequence.

(Video) Hard Arithmetic and Geometric Progression Questions

by Eddie Woo

Here's a video by Eddie Woo where he goes over some harder questions involving arithmetic progressions and geometric progressions.

If you need a quick reminder of sequences, click here.

The first question he looks at asks:

Show that if the first, second, and fourth terms of an arithmetic progression form a geometric sequence. If that's the case, show that either (1) the sequence is a constant sequence or (2) that the terms are the positive integer multiples of the first term.

We start by writing the general form of an arithmetic progression. There's the first term $a$ and then each term after differs by a difference of $d$. $$a,a+d,a+2d,a+3d,\dots$$ The question is interested in the first, second, and fourth terms. $$\boxed{\color{red}{a}},\boxed{\color{red}{a+d}},a+2d,\boxed{\color{red}{a+3d}},\dots$$ It is saying that these three form a geometric sequence. To prove that something is a geometric progression you need to show that: $$\frac{T_{n+1}}{T_{n}}=\frac{T_n}{T_{n-1}}$$ So we want to prove that: $$\frac{a+d}{a}=\frac{a+3d}{a+d}$$

Before we continue proving the above let's understand what the two cases at the end mean.

Constant Sequence

A constant sequence is a when the sequence as a whole is constant. That means all the terms are the same. $$a,a,a,a,a,\dots$$ This would mean, for the arithmetic progression, the difference would be zero, $d=0$.

Positive Integer Multiples

If the terms are the positive integer multiples of the first then that means they're the positive integer multiples of $a$. $$a,2a,3a,4a,\dots$$ We can rewrite this as how we define an arithmetic sequence. $$a,a+a,a+2a,a+3a,\dots$$ In this case, the common difference is the first term: $d=a$.

So what we want to do is prove that either:

1. $d=0$ or
2. $d=a$

So now we go back to our equation and attempt to solve for $d$. $$\begin{align} \frac{a+d}{a}&=\frac{a+3d}{a+d}\\ (a+d)(a+d)&=a(a+3d)\\ a^2+2ad+d^2&=a^2+3ad\\ d^2-ad&=0\\ d(d-a)&=0 \end{align}$$

There is a simple way to find the nth term in an arithmetic sequence.

First we should make sure that we are clear on what an arithmetic sequence is.

An arithmetic sequence is called an arithmetic sequence since there is a common difference between each term. For example, if our first term is $a$ and our common difference is $y$, then our terms will be $a, a+d, a+2d, a+3d, \dots$. An important thing to remember is that $d$ can be both positive or negative!

Now let us look at some clever ways to explore arithmetic sequences.

Let us first see how we can find the nth term of a sequence. We can look at our original sequence of $$a, a+d, a+2d, a+3d, \cdots$$ Our first term is $a$, our second term is $a+d$, and our third term is $a+2d$. To find each of these, we can write in a different way. Our first term can be written as $a + (1-1)(d)$. Our second term can be written as $a+ (2-1)(d)$. Our third term can be written as $a+ (3-1)(d)$.

We can observe that $a$ is the first term and $d$ is the difference. The number that is being multiplied by $d$ is the number of the term minus one, since the first term does not have the common difference added to it. Thus, we can observe that the nth term is $a+(n-1)(d)$.

I hope you enjoyed this!

There is a clever way to find the amount of terms in an arithmetic sequence.

Let us go over what an arithmetic sequence is. It is a sequence with a common difference. So the terms of an arithmetic sequence will be $a, a+d, a+2d, \cdots$. It is called an arithmetic sequence since the difference between two terms is the same and the difference can be both positive and negative.

So let us look at the sequence of $a, a+d, a +2d, a+3d, \cdots, a+399d, a+ 400d$.

Notice how the first term has $(0)(d)$ added to to it, the second term has $(1)(d)$ which goes on and on. Each term has $(n-1)(d)$ added to it. So if we subtract the first term from the last term we get $400d$. So if this term is nth term we know that $(n-1)(d) = 400d$ so $n = \boxed{401}$.

Let us look at this with real numbers. If we have the sequence $3,8,13, \cdots 1498, 1503$, how many terms are there? We can observe that the difference is $5$. We can subtract $3$ from $1503$ and get $1500$. So by doing $\frac{1500}{5} = 300$ we can determine that the difference was added $300$ times. Thus, the amount of terms is $300+1 = \boxed{301}$.

I hope you learned about arithmetic sequences from this!