Arithmetic Progressions — Notes

5.1 Introduction

Mathematics helps us find patterns in numbers and use them to solve problems. We observe many patterns in nature and daily life. An arithmetic progression (AP) is a sequence of numbers where each term after the first is obtained by adding a fixed number to the previous term. This fixed number is called the common difference.

Let's look at some real-life examples that form a pattern:

• Example 1: Reena's Salary Reena gets a starting salary of ₹8000 with an annual increment of ₹500. Her salary over the years forms a sequence: 8000, 8500, 9000, ... Here, the first term is 8000. To get the next salary, ₹500 is added to the previous salary. So, the common difference is ₹500.

• Example 2: Ladder Rungs The lengths of the rungs of a ladder decrease by 2 cm from bottom to top. The bottom rung is 45 cm. The lengths of the rungs are: 45 cm, 43 cm, 41 cm, ... The first term is 45 cm. Each subsequent rung is 2 cm shorter, meaning we add -2 cm to get the next term. The common difference is -2 cm.

• Example 3: Shakila's Savings Shakila saves ₹100 on her daughter's first birthday and increases the amount by ₹50 each year. The amounts in the box are: 100, 150, 200, 250, ... This sequence starts with 100, and each year ₹50 is added. The common difference is ₹50.

These examples show sequences where there is a constant difference between consecutive terms. This is the defining characteristic of an arithmetic progression.

What is a Sequence?

A sequence is a list of numbers in a specific order. For example, 1, 2, 3, 4, 5 is a sequence. A sequence can be finite (like the ladder rungs example: 45, 43, ..., 31) or infinite (like Reena's salary if she continues working indefinitely: 8000, 8500, 9000, ...).

What is an Arithmetic Progression (AP)?

An arithmetic progression (AP) is a sequence where the difference between any two successive members is constant. This constant difference is called the common difference, denoted by '$d$'.

The sequence is usually denoted by $a_1,a_2,a_3,\dots ,a_n,\dots$. Here, $a_1$ is the first term. The common difference, $d$, is calculated as:

Important Points:

• The common difference '$d$' can be positive, negative, or zero.
• Positive '$d$': Terms increase (e.g., 2, 4, 6, ...)
• Negative '$d$': Terms decrease (e.g., 10, 7, 4, ...)
• Zero '$d$': All terms are the same (e.g., 5, 5, 5, ...)
• To form an AP, we need a first term and a common difference.

Let's analyze the examples given in the text using these terms:

Note: Examples (iii) and (iv) are not Arithmetic Progressions because the difference between consecutive terms is not constant. Example (vi) is the famous Fibonacci sequence, where each term is the sum of the two preceding ones.

Remember

An arithmetic progression is defined by a constant difference between consecutive terms. If the difference changes, it is not an AP.

Exam-smart

You might be given a list of numbers and asked to identify if it's an AP. You will need to check the difference between each pair of consecutive numbers. If all differences are the same, it's an AP. For example, you could be shown a sequence of temperatures over several days or a list of costs that increase by a fixed amount each month.

Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.

The terms in an AP are found by adding the common difference to the previous term.

General form of an AP

The general form of an Arithmetic Progression is represented as:

Here:

• $a$ is the first term.
• $d$ is the common difference.

Key Properties of an AP:

1. Constant Difference: The difference between any term ($a_{k+1}$) and its preceding term ($a_k$) is always the common difference ($d$). Mathematically, $a_{k+1}-a_k=d$.
2. Predictable Sequence: Knowing the first term ($a$) and the common difference ($d$) allows us to generate any term in the sequence.

Finite vs. Infinite APs:

• An AP with a limited number of terms is called a finite AP. It has a last term.
• An AP with an unlimited number of terms is called an infinite AP. It does not have a last term.

How to identify an AP:

To check if a list of numbers is an AP, calculate the difference between each consecutive pair of terms. If all these differences are the same, then the list forms an AP.

• The first term ($a$) is simply the first number in the list.
• The common difference ($d$) is found by subtracting any term from its immediately succeeding term (i.e., $a_{k+1}-a_k$).

Example Lists and their Analysis:

• (i) 1, 2, 3, 4, ...

• $2-1=1$

• $3-2=1$

• $4-3=1$

• This is an AP with $a=1$ and $d=1$.

• (ii) 100, 70, 40, 10, ...

• $70-100=-30$

• $40-70=-30$

• $10-40=-30$

• This is an AP with $a=100$ and $d=-30$.

• (iii) –3, –2, –1, 0, ...

• $-2-(-3)=1$

• $-1-(-2)=1$

• $0-(-1)=1$

• This is an AP with $a=-3$ and $d=1$.

• (iv) 3, 3, 3, 3, ...

• $3-3=0$

• $3-3=0$

• This is an AP with $a=3$ and $d=0$.

• (v) –1.0, –1.5, –2.0, –2.5, ...

• $-1.5-(-1.0)=-0.5$

• $-2.0-(-1.5)=-0.5$

• $-2.5-(-2.0)=-0.5$

• This is an AP with $a=-1.0$ and $d=-0.5$.

• List: 1, 1, 2, 3, 5, ...

• $1-1=0$

• $2-1=1$

• Since the difference is not constant ($0\mathrm{\ne }1$), this is not an AP.

Example 1: Finding first term and common difference

Given: The AP: $\frac{3}{2},\frac{1}{2},-\frac{1}{2},-\frac{3}{2},\dots$ To find: The first term ($a$) and the common difference ($d$).

Solution:

• The first term ($a$) is the first number in the list.

• To find the common difference ($d$), subtract any term from its succeeding term. Let's use the first two terms:

Let's verify with the next pair:

Answer: $a=\frac{3}{2}$, $d=-1$.

Example 2: Identifying APs and finding next terms

Given: Lists of numbers. To find: Which lists form an AP, and if so, write the next two terms.

Solution:

(i) 4, 10, 16, 22, ...

• $10-4=6$
• $16-10=6$
• $22-16=6$
• The differences are constant ($d=6$). So, it's an AP.
• Next term: $22+6=28$
• Term after that: $28+6=34$ Answer: Forms an AP with $a=4,d=6$. Next two terms are 28 and 34.

(ii) 1, – 1, – 3, – 5, ...

• $-1-1=-2$
• $-3-(-1)=-3+1=-2$
• $-5-(-3)=-5+3=-2$
• The differences are constant ($d=-2$). So, it's an AP.
• Next term: $-5+(-2)=-7$
• Term after that: $-7+(-2)=-9$ Answer: Forms an AP with $a=1,d=-2$. Next two terms are -7 and -9.

(iii) – 2, 2, – 2, 2, – 2, ...

• $2-(-2)=2+2=4$
• $-2-2=-4$
• The differences are not constant ($4\mathrm{\ne }-4$). So, it's not an AP. Answer: Does not form an AP.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, ...

• $1-1=0$
• $1-1=0$
• $2-1=1$
• The differences are not constant ($0\mathrm{\ne }1$). So, it's not an AP. Answer: Does not form an AP.

Remember

A common mistake is to only check the first two terms for a constant difference. You must check all consecutive pairs to confirm it's an AP. Also, remember the common difference can be positive, negative, or zero.

Exam-smart

You might be given a list of numbers and asked if it's an AP. You'll need to calculate the differences between consecutive terms. If they are all the same, it's an AP, and you can then find the common difference. For example, you might see a problem like: "Consider the sequence 5, 8, 11, 14. Is this an AP? If yes, what is the common difference?" You would calculate $8-5=3$, $11-8=3$, and $14-11=3$. Since the difference is consistently 3, it is an AP with $d=3$.

Try this: For the list of numbers 7, 4, 1, -2, ..., determine if it is an AP. If it is, find the common difference. Answer: Yes, it is an AP. The common difference ($d$) is -3.

nth Term of an AP

An Arithmetic Progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.

We can find any term in an AP if we know the first term ($a$) and the common difference ($d$). Let's consider an AP: $a_1,a_2,a_3,\dots ,a_n,\dots$.

• The first term is $a_1=a$.
• The second term is $a_2=a_1+d=a+d$.
• The third term is $a_3=a_2+d=(a+d)+d=a+2d$.
• The fourth term is $a_4=a_3+d=(a+2d)+d=a+3d$.

Observing the pattern, the $n^{th}$ term ($a_n$) can be expressed as:

This formula is crucial because it allows us to directly calculate any term in an AP without having to list all the preceding terms. The term $a_n$ is also referred to as the general term of the AP. If an AP has a finite number of terms, say $m$ terms, then the last term is $a_m$, which is often denoted by $l$.

Example 1

Find the 10th term of the AP: 2, 7, 12, ...

Given: The AP is 2, 7, 12, ... First term, $a=2$ Common difference, $d=7-2=5$ We need to find the 10th term, so $n=10$.

To find: The 10th term ($a_{10}$).

Solution: Using the formula for the $n^{th}$ term of an AP: $a_n=a+(n-1)d$

Answer: The 10th term of the given AP is 47.

Example 2

Which term of the AP: 21, 18, 15, ... is -81? Also, determine if 0 is a term in this AP.

Given: The AP is 21, 18, 15, ... First term, $a=21$ Common difference, $d=18-21=-3$ We need to find which term is -81. Let this be the $n^{th}$ term, so $a_n=-81$.

To find: The value of $n$ for which $a_n=-81$, and if $a_n=0$ for any $n$.

Solution: First, to find $n$ for $a_n=-81$: Using the formula $a_n=a+(n-1)d$:

So, the 35th term of the AP is -81.

Next, to check if 0 is a term: Let $a_n=0$. Using the formula $a_n=a+(n-1)d$:

Since $n=8$ is a positive integer, 0 is indeed a term in this AP. It is the 8th term.

Answer: The 35th term of the AP is -81. The 8th term is 0.

Example 3

Determine the AP whose 3rd term is 5 and the 7th term is 9.

Given: 3rd term of an AP, $a_3=5$ 7th term of an AP, $a_7=9$

To find: The AP.

Solution: Using the formula $a_n=a+(n-1)d$: For the 3rd term: $a_3=a+(3-1)d⟹5=a+2d$ (Equation 1) For the 7th term: $a_7=a+(7-1)d⟹9=a+6d$ (Equation 2)

Now we solve the system of linear equations (1) and (2): Subtract Equation 1 from Equation 2:

Substitute the value of $d$ into Equation 1:

The first term is $a=3$ and the common difference is $d=1$. The AP is $a,a+d,a+2d,\dots$ So, the AP is 3, $3+1$, $3+2(1)$, ... which is 3, 4, 5, ...

Answer: The AP is 3, 4, 5, 6, ...

Example 4

Check whether 301 is a term of the list of numbers: 5, 11, 17, 23, ...

Given: The list of numbers is 5, 11, 17, 23, ... We need to check if 301 is a term in this sequence.

To find: If 301 belongs to the AP.

Solution: First, let's confirm if this list is an AP and find its common difference.

The common difference $d=6$. The first term $a=5$. Now, let's assume 301 is the $n^{th}$ term of this AP, so $a_n=301$. Using the formula $a_n=a+(n-1)d$:

Since $n$ must be a positive integer (representing the term number), and $\frac{151}{3}$ is not an integer, 301 is not a term of this AP.

Answer: 301 is not a term of the given AP.

Example 5

How many two-digit numbers are divisible by 3?

Given: We need to find the count of two-digit numbers divisible by 3.

To find: The number of terms ($n$).

Solution: The smallest two-digit number is 10. The largest two-digit number is 99. The two-digit numbers divisible by 3 form an AP: The first two-digit number divisible by 3 is 12 ($3\times 4$). The last two-digit number divisible by 3 is 99 ($3\times 33$). So, the AP is 12, 15, 18, ..., 99. Here, the first term $a=12$. The common difference $d=15-12=3$. The last term $a_n=99$.

Using the formula $a_n=a+(n-1)d$:

There are 30 two-digit numbers divisible by 3.

Answer: There are 30 two-digit numbers divisible by 3.

Example 6

Find the 11th term from the last term of the AP: 10, 7, 4, ..., -62.

Given: The AP is 10, 7, 4, ..., -62. First term, $a=10$. Common difference, $d=7-10=-3$. Last term, $l=-62$.

To find: The 11th term from the last.

Solution: Method 1: Find the total number of terms first. Using $l=a+(n-1)d$:

So, there are 25 terms in the AP.

The 11th term from the last is the same as the $(n-11+1{)}^{th}$ term from the beginning. Term from beginning = $25-11+1={15}^{th}$ term. Now, find the 15th term ($a_{15}$):

Method 2: Reverse the AP. If we reverse the AP, the last term becomes the first term and the common difference changes its sign. The reversed AP is -62, -59, -56, ..., 10. New first term, $a^{\mathrm{\prime }}=-62$. New common difference, $d^{\mathrm{\prime }}=(-59)-(-62)=3$. We need to find the 11th term of this reversed AP. Using $a_n^{\mathrm{\prime }}=a^{\mathrm{\prime }}+(n-1)d^{\mathrm{\prime }}$:

Both methods yield the same result.

Answer: The 11th term from the last is -32.

Example 7

A sum of ₹1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years using this fact.

Given: Principal amount, P = ₹1000 Rate of interest, R = 8% per annum

To find: Interest at the end of each year, whether it forms an AP, and the interest at the end of 30 years.

Solution: The formula for Simple Interest (SI) is $SI=\frac{P\times R\times T}{100}$, where T is the time in years.

Interest at the end of 1st year (T=1):

Interest at the end of 2nd year (T=2):

Interest at the end of 3rd year (T=3):

The sequence of interests at the end of each year is 80, 160, 240, ... Let's check if this is an AP:

Yes, this is an AP with the first term $a=80$ and the common difference $d=80$.

Now, to find the interest at the end of 30 years (which is the 30th term of this AP): We need to find $a_{30}$. Using $a_n=a+(n-1)d$:

The interest at the end of 30 years is ₹2400.

Answer: The interests form an AP: 80, 160, 240, ... with $a=80$ and $d=80$. The interest at the end of 30 years is ₹2400.

Example 8

In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

Given: Number of rose plants in rows: 23, 21, 19, ..., 5. The last row has 5 plants.

To find: The number of rows ($n$).

Solution: The sequence of rose plants in the rows is 23, 21, 19, ..., 5. This is an AP. First term, $a=23$. Common difference, $d=21-23=-2$. The last term (number of plants in the last row), $a_n=5$.

Using the formula $a_n=a+(n-1)d$:

There are 10 rows in the flower bed.

Answer: There are 10 rows in the flower bed.

Remember

The $n^{th}$ term formula $a_n=a+(n-1)d$ requires $n$ to be a positive integer. If you get a non-integer or a negative value for $n$ when checking if a number belongs to an AP, it means that number is not a term of that AP.

Exam-smart

Questions often ask to find a specific term, determine if a given number is part of an AP, or find the position of a term. Sometimes, you might be given two terms of an AP and asked to find the first term or the common difference, which leads to solving a system of linear equations. Case studies may involve real-life scenarios like salary increments, savings, or plant arrangements forming an AP.

Try this: Find the 25th term of the AP: 4, 10, 16, ... Answer: 148

Sum of First n Terms of an AP

The sum of the first $n$ terms of an Arithmetic Progression (AP) is the total when you add up the first $n$ terms of that sequence. Instead of manually adding each term, which can be very time-consuming for a large number of terms, we use a formula to find this sum efficiently.

Consider an AP with the first term $a$, a common difference $d$, and $n$ terms. The terms are: $a,a+d,a+2d,\dots ,a+(n-1)d$.

Let $S_n$ denote the sum of the first $n$ terms. We can write this sum as: $S_n=a+(a+d)+(a+2d)+\cdots +[a+(n-1)d]$ (Equation 1)

To derive a formula, we can write the sum in reverse order: $S_n=[a+(n-1)d]+[a+(n-2)d]+\cdots +(a+d)+a$ (Equation 2)

Now, add Equation 1 and Equation 2 term by term:

Notice that each pair of terms sums to the same value, $[2a+(n-1)d]$, and there are $n$ such pairs (since there are $n$ terms in the AP). So, $2S_n=n[2a+(n-1)d]$

This gives us the primary formula for the sum of the first $n$ terms of an AP:

This formula is useful when you know the first term ($a$), the common difference ($d$), and the number of terms ($n$).

Alternative Formula for the Sum of an AP

The formula $S_n=\frac{n}{2}[2a+(n-1)d]$ can also be rewritten by separating one $a$ from $2a$:

We know that the $n$th term of an AP is given by $a_n=a+(n-1)d$. If the AP has exactly $n$ terms, then the last term, $l$, is equal to the $n$th term, $a_n$. So, we can substitute $l$ for $a+(n-1)d$:

This formula is particularly useful when you know the first term ($a$), the last term ($l$), and the number of terms ($n$), but not necessarily the common difference ($d$).

Example 1: Sum of the first 21 terms of an AP

Shakila's daughter's money box has amounts: 100, 150, 200, ... on her birthdays. We need to find the total amount collected by her 21st birthday.

Given: The sequence is an AP. First term, $a=100$ Common difference, $d=150-100=50$ Number of terms, $n=21$

To find: The sum of the first 21 terms, $S_{21}$.

Solution: Using the formula $S_n=\frac{n}{2}[2a+(n-1)d]$:

Answer: The total amount collected by her 21st birthday is ` 12600.

Example 2: Sum of the first 22 terms of an AP

Find the sum of the first 22 terms of the AP: 8, 3, –2, ...

Given: The AP is 8, 3, –2, ... First term, $a=8$ Common difference, $d=3-8=-5$ Number of terms, $n=22$

To find: The sum of the first 22 terms, $S_{22}$.

Solution: Using the formula $S_n=\frac{n}{2}[2a+(n-1)d]$:

Answer: The sum of the first 22 terms is –979.

Example 3: Finding the 20th term when sum and first term are given

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Given: Sum of first 14 terms, $S_{14}=1050$ First term, $a=10$ Number of terms for sum, $n=14$

To find: The 20th term, $a_{20}$.

Solution: First, we need to find the common difference ($d$) using the sum formula $S_n=\frac{n}{2}[2a+(n-1)d]$:

Divide both sides by 7:

Now that we have the common difference $d=10$ and the first term $a=10$, we can find the 20th term using the formula $a_n=a+(n-1)d$:

Answer: The 20th term is 200.

Example 4: Number of terms when sum and AP are given

How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?

Given: The AP is 24, 21, 18, ... First term, $a=24$ Common difference, $d=21-24=-3$ Sum of $n$ terms, $S_n=78$

To find: The number of terms, $n$.

Solution: Using the formula $S_n=\frac{n}{2}[2a+(n-1)d]$:

Multiply both sides by 2:

Rearrange into a quadratic equation:

Divide by 3 to simplify:

Factor the quadratic equation: We need two numbers that multiply to 52 and add to -17. These are -4 and -13.

This gives two possible values for $n$: $n=4$ or $n=13$

Both values are valid. This means that taking either 4 terms or 13 terms will result in a sum of 78. This can happen if the sum of the terms between the 4th and 13th term is zero (i.e., positive terms are cancelled out by negative terms).

Answer: The number of terms can be either 4 or 13.

Example 5: Sum of the first n positive integers

Find the sum of the first $n$ positive integers.

Given: The sequence is 1, 2, 3, ..., $n$. This is an AP. First term, $a=1$ Common difference, $d=1$ Number of terms, $n$ The last term, $l=n$

To find: The sum of the first $n$ terms, $S_n$.

Solution: We can use either formula. Using $S_n=\frac{n}{2}[a+l]$:

Alternatively, using $S_n=\frac{n}{2}[2a+(n-1)d]$:

Answer: The sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$.

Example 6: Sum of first 24 terms of a sequence given by its nth term

Find the sum of the first 24 terms of the list of numbers whose $n$th term is given by $a_n=3+2n$.

Given: The $n$th term of the sequence is $a_n=3+2n$.

To find: The sum of the first 24 terms, $S_{24}$.

Solution: First, determine if the sequence is an AP and find its first term and common difference. For $n=1$, $a_1=3+2(1)=5$ For $n=2$, $a_2=3+2(2)=7$ For $n=3$, $a_3=3+2(3)=9$

The sequence starts with 5, 7, 9, ... The difference between consecutive terms is $a_2-a_1=7-5=2$, and $a_3-a_2=9-7=2$. Since the difference is constant, this is an AP. First term, $a=5$ Common difference, $d=2$ Number of terms, $n=24$

Now, use the sum formula $S_n=\frac{n}{2}[2a+(n-1)d]$:

Answer: The sum of the first 24 terms is 672.

Example 7: Production of TV sets

A manufacturer produced 600 TV sets in the third year and 700 sets in the seventh year. Assuming production increases uniformly, find: (i) production in the 1st year (ii) production in the 10th year (iii) total production in the first 7 years.

Given: Production figures form an AP. Production in the 3rd year, $a_3=600$ Production in the 7th year, $a_7=700$

To find: (i) Production in the 1st year ($a_1$) (ii) Production in the 10th year ($a_{10}$) (iii) Total production in first 7 years ($S_7$)

Solution: We use the formula $a_n=a+(n-1)d$. For the 3rd year: $a_3=a+(3-1)d\Rightarrow 600=a+2d$ (Equation 1) For the 7th year: $a_7=a+(7-1)d\Rightarrow 700=a+6d$ (Equation 2)

(i) To find the production in the 1st year ($a$): Subtract Equation 1 from Equation 2:

The common difference is 25.

Substitute $d=25$ into Equation 1:

So, the production in the 1st year is 550.

(ii) To find the production in the 10th year ($a_{10}$): Using $a_n=a+(n-1)d$:

So, the production in the 10th year is 775.

(iii) To find the total production in the first 7 years ($S_7$): Using $S_n=\frac{n}{2}[2a+(n-1)d]$:

So, the total production in the first 7 years is 4375.

Answer: (i) The production in the 1st year is 550. (ii) The production in the 10th year is 775. (iii) The total production in the first 7 years is 4375.

Relationship between $a_n$ and $S_n$

The $n$th term of an AP can also be found by taking the difference between the sum of the first $n$ terms and the sum of the first $(n-1)$ terms:

This is because $S_n$ includes all terms up to $a_n$, and $S_{n-1}$ includes all terms up to $a_{n-1}$. Their difference leaves only $a_n$.

Example 8: Finding terms using $S_n-S_{n-1}$

If the sum of the first $n$ terms of an AP is given by $S_n=4n-n^2$, find the first term, the sum of the first two terms, the second term, the 3rd term, the 10th term, and the $n$th term.

Given: The sum of the first $n$ terms is $S_n=4n-n^2$.

To find: $a_1$, $S_2$, $a_2$, $a_3$, $a_{10}$, $a_n$.

Solution:

1. First term ($a_1$): The sum of the first term ($S_1$) is just the first term itself.

So, $a_1=3$.

2. Sum of the first two terms ($S_2$): Substitute $n=2$ into the formula for $S_n$.

So, the sum of the first two terms is 4.

3. Second term ($a_2$): Use the relationship $a_n=S_n-S_{n-1}$. For $n=2$:

So, the second term is 1.

4. Third term ($a_3$): First, find $S_3$:

Now, find $a_3$:

So, the third term is -1.

5. Tenth term ($a_{10}$): First, find $S_{10}$:

Now, find $S_9$ to calculate $a_{10}$:

Then, $a_{10}=S_{10}-S_9$

So, the tenth term is -15.

6. Nth term ($a_n$): Use the general formula $a_n=S_n-S_{n-1}$ for $n>1$.

Now, $a_n=S_n-S_{n-1}$:

Let's check this formula for the terms we found: For $n=1$: $a_1=-2(1)+5=3$ (Matches) For $n=2$: $a_2=-2(2)+5=-4+5=1$ (Matches) For $n=3$: $a_3=-2(3)+5=-6+5=-1$ (Matches) For $n=10$: $a_{10}=-2(10)+5=-20+5=-15$ (Matches)

Answer: The first term is 3. The sum of the first two terms is 4. The second term is 1. The third term is -1. The tenth term is -15. The $n$th term is $a_n=5-2n$.

Remember

The formula $S_n=\frac{n}{2}[2a+(n-1)d]$ is fundamental. Always identify $a$, $d$, and $n$ correctly from the problem statement. For problems involving the last term, $S_n=\frac{n}{2}[a+l]$ is often more direct.

Exam-smart

Case/source-based questions might provide a sequence of earnings over years, savings patterns, or costs that form an AP. You'll need to identify $a$, $d$, and $n$ from the text or a table and apply the sum formula to find total earnings, savings, or costs. Some questions may ask for the number of terms required to reach a certain total, leading to a quadratic equation in $n$.