Arithmetic Progressions

Class 10 . Mathematics . Mathematics . Arithmetic Progressions

Chapter 5: Arithmetic Progressions

Ch 5

Mathematics

Class 10

Lecture 1

Introduction & Arithmetic Progressions

Reference video · Lecture 1Introduction & Arithmetic Progressions

Patterns in Everyday Life

A ladder with rung lengths decreasing by a fixed amount.

Rabbit-pair counts follow 1, 1, 2, 3, 5, 8,. — Rabbit-pair counts follow $1, 1, 2, 3, 5, 8, \ldots$.

Arithmetic Progressions

Definition

Arithmetic progression (AP) — A list of numbers in which each term after the first is obtained by adding the same fixed number to the preceding term.

Terms and Common Difference

Terms: The numbers in the list are called terms; write them as $a_1, a_2, a_3, \ldots$.
First term: $a_1$ (often written $a$) is the first term; $a_n$ is the $n$th term.
Common difference: The fixed number added at each step is $d$; it can be positive, negative, or zero.
AP test: Consecutive differences must all be equal. For example, $1, 2, 3, 4, \ldots$ and $100, 70, 40, 10, \ldots$ are APs.
Zero difference: A constant list such as $3, 3, 3, \ldots$ is an AP with $d=0$.

Common Difference

$$d=a_{k+1}-a_k$$

Subtract a term from the term immediately after it.

Equal-differences test

A list is an AP if and only if $a_{k+1}-a_k$ has the same value for every pair of consecutive terms. Checking one difference is enough only when the list is already known to be an AP.

General Form of an AP

$$a,\ a+d,\ a+2d,\ a+3d,\ \ldots$$

Starting with first term $a$, add the common difference repeatedly.

Finite AP vs Infinite AP

Point	Finite AP	Infinite AP
Terms	Has a fixed number of terms	Continues without end
Last term	Has a last term	Has no last term
Examples	$147, 148, \ldots, 157$	$1, 2, 3, 4, \ldots$

Recognising APs in Context

Temperatures: $-3.1, -3.0, -2.9, \ldots, -2.5$ has $d=0.1$.
Loan balance: Paying Rs. 50 from Rs. 1000 each month gives $950, 900, 850, \ldots, 50$, with $d=-50$.
Prizes: $200, 250, 300, \ldots, 750$ has $d=50$.
Savings totals: Saving Rs. 50 each month gives totals $50, 100, 150, \ldots, 500$, with $d=50$.

Test Whether a List Is an AP

Find the difference between the second term and the first.
$$a_2-a_1$$
Find the differences for each following consecutive pair.
$$a_3-a_2,\ a_4-a_3,\ \ldots$$
If all differences are equal, the list is an AP; that shared value is $d$.
$$a_{k+1}-a_k=d$$

Worked example

Does $4, 10, 16, 22, \ldots$ form an AP? If so, find the next two terms.

Check the consecutive differences in the list $4, 10, 16, 22, \ldots$.
$$10-4=6$$
Compare the next differences; they are equal, so the list is an AP with common difference $6$.
$$16-10=6,\qquad 22-16=6$$
Add $6$ twice to find the next two terms.
$$22+6=28,\qquad 28+6=34$$

Answer: Yes. The common difference is $6$, and the next two terms are $28$ and $34$.

⚠ Common mistake

Subtract in the order “following term minus preceding term.” For $6, 3, 0, -3, \ldots$, $d=3-6=-3$, not $6-3$.

Practice

For $1, -1, -3, -5, \ldots$, determine whether it is an AP and find the next two terms.

Hint: Compare consecutive differences.

Key step: $-1-1=-2$, $-3-(-1)=-2$, and $-5-(-3)=-2$.

Solution: It is an AP with $d=-2$. The next terms are $-7$ and $-9$.

Practice

Does $-2, 2, -2, 2, -2, \ldots$ form an AP?

Hint: Check the first two consecutive differences.

Key step: $2-(-2)=4$, whereas $-2-2=-4$.

Solution: No. The consecutive differences are not equal.

Practice

For the AP $\frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \ldots$, state the first term and common difference.

Hint: The first term is the first listed number; subtract it from the next term to get $d$.

Key step: $\frac{1}{2}-\frac{3}{2}=-1$.

Solution: The first term is $\frac{3}{2}$ and the common difference is $-1$.

Quick recap

An AP is a list with a constant difference between consecutive terms.
The common difference is $d=a_{k+1}-a_k$; it may be positive, negative, or zero.
An AP is determined by its first term $a$ and common difference $d$.
The general form is $a, a+d, a+2d, a+3d, \ldots$.
Finite APs have a last term; infinite APs do not.
A list with changing consecutive differences, such as $1, 1, 2, 3, 5, \ldots$, is not an AP.

Lecture 2

nth Term Of An AP

Reference video · Lecture 2nth Term Of An AP

General term of an AP

nth-term formula

The $n$th term of an AP is the first term plus $n-1$ common differences.

$$a_n=a+(n-1)d$$

nth term and related forms

$$a_n=a+(n-1)d,\qquad d=a_{n+1}-a_n,\qquad l=a+(m-1)d$$

Use the general term to find a term, determine its position, or find the number of terms.

Using the nth-term formula

Identify the first term $a$, common difference $d$, and the required position $n$ (or given term value).
Substitute the known values into $a_n=a+(n-1)d$.
Simplify; if finding the position, solve for $n$ and check that it is a positive integer.

Worked example

Find the 10th term of $2,7,12,\ldots$.

Find the 10th term of the AP $2,7,12,\ldots$. The first term is $2$ and the common difference is $7-2=5$.
Apply the nth-term formula.
$$a_{10}=2+(10-1)\times5$$
Simplify.
$$a_{10}=2+45=47$$

Answer: The 10th term is $47$.

⚠ Common mistake

Do not use $nd$ in place of $(n-1)d$: the first term already contributes $a$, so only $n-1$ common differences are added.

Finding a term's position and checking membership

Practice

In the AP $21,18,15,\ldots$, which term is $-81$? Is $0$ a term?

Hint: Use $a=21$ and $d=-3$; set the general term equal to each target value.

Key step: Solve $x=21+(n-1)(-3)$ and require $n$ to be a positive integer.

Solution: For $x=-81$, $-81=21-3(n-1)$ gives $n=35$, so $-81$ is the 35th term. For $x=0$, $0=21-3(n-1)$ gives $n=8$, so $0$ is the 8th term.

Practice

The 3rd term of an AP is $5$ and the 7th term is $9$. Find the AP.

Hint: Write one equation for each known term using $a_n=a+(n-1)d$.

Key step: $a+2d=5$ and $a+6d=9$.

Solution: Subtracting gives $4d=4$, so $d=1$. Then $a+2=5$, so $a=3$. The AP is $3,4,5,6,\ldots$.

Applications of the nth term

Counting from the last term

$$\text{term }k\text{ from the end}=a_{m-k+1}=l-(k-1)d$$

To find the $k$th term from the end, either count back from the total number of terms or reverse the AP.

Example: counting from the end

Given AP: $10,7,4,\ldots,-62$ has $a=10$, $d=-3$, and $l=-62$.
Number of terms: $-62=10+(n-1)(-3)$ gives $n=25$.
11th from the end: Its position from the beginning is $25-11+1=15$; the term is $a_{15}=10+14(-3)=-32$.
Reverse method: The reversed AP starts at $-62$ with common difference $3$; its 11th term is $-62+10(3)=-32$.

Practice

Check whether $301$ is a term of $5,11,17,23,\ldots$.

Hint: Find $n$ from $301=5+(n-1)6$.

Key step: The resulting position must be a positive integer.

Solution: $d=6$. Solving $301=5+6(n-1)$ gives $n=302/6=151/3$, not an integer; therefore, $301$ is not a term.

Quick recap

For an AP with first term $a$ and common difference $d$, $a_n=a+(n-1)d$.
Find the common difference by subtracting consecutive terms.
To test whether a value is in an AP, solve for $n$ and check that it is a positive integer.
Use known terms to form equations and determine unknown values of $a$ and $d$.
If there are $m$ terms, the last term is $l=a+(m-1)d$.
The $k$th term from the end is $a_{m-k+1}=l-(k-1)d$.

Lecture 3

Sum Of First N Terms Of An A

Reference video · Lecture 3Sum Of First N Terms Of An A

Definition

Arithmetic progression (AP) — An AP is a sequence in which each term after the first is obtained by adding the same common difference $d$ to the preceding term.

Pairing Terms to Find a Sum

The pairing idea

Pairing: In an AP, adding terms from opposite ends gives equal pair sums because the first-plus-last, second-plus-second-last, and so on are equal.
Example: For $1+2+\cdots+100$, each opposite-end pair sums to $101$, so the total is $\frac{100\times101}{2}=5050$.

Sum of the First n Terms

Definition

Sum of the first n terms — The sum of the first $n$ terms of an AP is denoted by $S_n$.

Equal sums from opposite ends

Writing the AP forwards and backwards gives corresponding sums of $a+l$, where $a$ is the first term and $l$ the last term. Thus, $2S_n=n(a+l)$.

$$2S_n=n(a+l)$$

Sum of the first n terms

$$S_n=\frac{n}{2}\left[2a+(n-1)d\right]=\frac{n}{2}(a+l)$$

Use the first form when $a$ and $d$ are known; use the second when the first and last terms are known.

Finding the sum of an AP

Identify the first term, common difference and number of terms.
$$a,\quad d,\quad n$$
Substitute these values into the sum formula.
$$S_n=\frac{n}{2}\left[2a+(n-1)d\right]$$
Simplify and state the sum with its appropriate units, if any.
$$S_n=\text{calculated sum}$$

Worked example

Shakila deposits Rs. 100 on her daughter's first birthday, increasing the deposit by Rs. 50 each birthday. Find the total deposited by the 21st birthday.

Identify the AP values.
$$a=100,\quad d=50,\quad n=21$$
Substitute into the sum formula.
$$S_{21}=\frac{21}{2}\left[2(100)+(21-1)(50)\right]$$
Simplify.
$$S_{21}=\frac{21}{2}(200+1000)=12600$$

Answer: The total deposited is Rs. 12,600.

Using the Sum Formula

Term from consecutive sums

The $n$th term is the difference between the sum of the first $n$ terms and the sum of the first $n-1$ terms.

$$a_n=S_n-S_{n-1}$$

Unknown quantities and special cases

Finding an unknown: The sum formula contains $S_n$, $a$, $d$ and $n$; knowing any three lets you find the fourth.
Positive integers: $1+2+\cdots+n=\frac{n(n+1)}{2}$; the first $1000$ positive integers sum to $500500$.
Given nth-term rule: If $a_n=3+2n$, the terms begin $5,7,9,\ldots$; this is an AP with $a=5$, $d=2$, and $S_{24}=672$.
Uniform increase: If production rises by a fixed amount each year, yearly production forms an AP. Given $a_3=600$ and $a_7=700$, $a=550$, $d=25$, $a_{10}=775$, and $S_7=4375$.

Practice

Find the sum of the first $22$ terms of $8,3,-2,\ldots$.

Hint: Find $d$ from consecutive terms, then use the sum formula.

Key step: $a=8$, $d=-5$, and $n=22$.

Solution: $S_{22}=\frac{22}{2}[2(8)+21(-5)]=11(16-105)=-979$.

Practice

An AP has first term $10$ and $S_{14}=1050$. Find its 20th term.

Hint: Use $S_{14}$ to find $d$, then use the nth-term formula.

Key step: $1050=\frac{14}{2}[2(10)+13d]$, so $d=10$.

Solution: $a_{20}=10+19(10)=200$.

Practice

How many terms of $24,21,18,\ldots$ must be taken so their sum is $78$?

Hint: Substitute $a=24$, $d=-3$ and $S_n=78$; solve the resulting quadratic.

Key step: $78=\frac{n}{2}[48-3(n-1)]$ gives $n^2-17n+52=0$.

Solution: $(n-4)(n-13)=0$, so $n=4$ or $n=13$. Both work: the terms from the fifth through thirteenth sum to zero because positive and negative terms cancel.

Arithmetic Mean

Arithmetic mean in an AP

If $a,b,c$ are in AP, the middle term $b$ is the arithmetic mean of the other two.

$$b=\frac{a+c}{2}$$

Quick recap

Pairing opposite ends of an AP gives equal sums.
Use $S_n=\frac{n}{2}[2a+(n-1)d]$ when the common difference is known.
Use $S_n=\frac{n}{2}(a+l)$ when the last term is known.
Find a term from consecutive sums using $a_n=S_n-S_{n-1}$.
The first $n$ positive integers sum to $\frac{n(n+1)}{2}$.
If $a,b,c$ are in AP, then $b=\frac{a+c}{2}$.

Formula sheet

Common Difference

$$d=a_{k+1}-a_k$$

General Form of an AP

$$a,\ a+d,\ a+2d,\ a+3d,\ \ldots$$

nth term and related forms

$$a_n=a+(n-1)d,\qquad d=a_{n+1}-a_n,\qquad l=a+(m-1)d$$

Counting from the last term

$$\text{term }k\text{ from the end}=a_{m-k+1}=l-(k-1)d$$

Sum of the first n terms

$$S_n=\frac{n}{2}\left[2a+(n-1)d\right]=\frac{n}{2}(a+l)$$

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