Arithmetic Progressions: nth Term and Sum

Hello! Welcome to the first lesson of our module on "Progressions, Functions, and Logarithms."

Having built a strong foundation in algebraic equations and expressions, we now shift our focus to sequences of numbers that follow specific patterns. Understanding these patterns is crucial for solving a whole category of CAT questions efficiently.

Today’s lesson addresses the learning outcome: Solve problems on Arithmetic Progressions (AP), including nth term and sum. We will explore what an Arithmetic Progression is, learn the formulas to find any term in the sequence, and calculate the sum of its terms. Crucially, we will focus on both the standard formulas and intuitive, time-saving methods that are vital for a competitive exam like the CAT.

1. What is an Arithmetic Progression (AP)?

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.

For example, the sequence 5, 8, 11, 14, ... is an AP because each term is obtained by adding 3 to the previous one.

The first term, denoted by $a$ , is 5.
The common difference, denoted by $d$ , is 3.

This iterative structure, where the next term is defined based on the previous one ( $t_{n} = t_{n - 1} + d$ ), might feel familiar from your work with loops and sequences in computer science.

To get a clear visual and formal introduction, let's watch a short video segment.

Arithmetic Sequences and Arithmetic Series - Basic Introduction

Watch this introductory part from the video 'Arithmetic Sequences and Arithmetic Series - Basic Introduction' by The Organic Chemistry Tutor. It clearly defines an AP and distinguishes it from another type of progression we'll see later.

Watch the video from the beginning up to the 1:52 mark. Focus on understanding the concept of a 'common difference' and how it defines an arithmetic sequence.

Here are the key formulas we will be working with in this lesson.

Arithmetic Progression Formulas Summary — A concise summary of the key formulas for an Arithmetic Progression, including the nth term and the sum of n terms.

2. Finding the nth Term of an AP

How do we find a specific term in an AP without writing out the entire sequence? Let's build the pattern:

1st term ( $t_{1}$ ) = $a$
2nd term ( $t_{2}$ ) = $a + d$
3rd term ( $t_{3}$ ) = $a + 2 d$
4th term ( $t_{4}$ ) = $a + 3 d$

Notice that the coefficient of $d$ is always one less than the term number. This leads to the general formula for the nth term:

$t_{n} = a + (n - 1) d$

A Faster Way to Think for CAT

For exam purposes, you can think of this more intuitively. To get to the nth term from the first term, you need to take $(n - 1)$ "jumps," where each jump has a size of $d$ .

Example: Find the 31st term of the AP: 4, 10, 16, 22, ...

First term ( $a$ ): 4
Common difference ( $d$ ): $10 - 4 = 6$
Term number ( $n$ ): 31

Mental Calculation:
To get to the 31st term, we start at 4 and need to make $31 - 1 = 30$ jumps of size 6.

Total value of jumps: $30 \times 6 = 180$
31st term: $4 + 180 = 184$

Using the Formula:
$t_{31} = a + (31 - 1) d = 4 + (30) \times 6 = 4 + 180 = 184$

Both methods yield the same result, but the "jumps" approach can be quicker for mental math.

Test your understanding!

The 7th term of an AP is 34 and the 13th term is 64. What is the first term ( $a$ ) and the common difference ( $d$ )?

Hint: Think about the number of 'jumps' between the 7th and 13th terms.

Show answer

Find the common difference (d):
To go from the 7th term to the 13th term, we need to make $13 - 7 = 6$ jumps.
The total change in value is $64 - 34 = 30$ .
So, $6 d = 30$ , which means $d = 5$ .
Find the first term (a):
We know the 7th term is 34. Using the formula $t_{7} = a + (7 - 1) d$ :
$34 = a + 6 d$
Substitute $d = 5$ :
$34 = a + 6 (5)$
$34 = a + 30$
$a = 4$

The first term is 4 and the common difference is 5. The AP starts 4, 9, 14, ...

3. Calculating the Sum of an AP

What if you need to find the sum of the first 50 terms of an AP? Adding them one by one is not feasible. We have two powerful formulas for this.

The Intuitive 'Average' Method

This is a fantastic shortcut for competitive exams. The sum of any AP is simply:

$Sum = Number of terms \times Average of the sequence$

A special property of an AP is that its average is always the average of the first and last terms.
$Average = \frac{First Term + Last Term}{2} = \frac{a + t _{n}}{2}$

This gives us the first formula for the sum ( $S_{n}$ ):
$S_{n} = n \times (\frac{a + t _{n}}{2})$

This video explains this "average funda" very clearly.

Arithmetic Progression 1 | CAT Exam Preparation | Algebra | Quantitative Aptitude

The following video from Rodha demonstrates a powerful, logic-based approach to finding the sum of an AP using its average. This is often much faster than plugging values into a complex formula.

Watch the segment from 6:31 to 12:09. Pay close attention to how the average of an AP is always the middle term, or simply (first term + last term) / 2.

The Standard Formula

What if you don't know the last term? You can substitute the formula for the nth term, $t_{n} = a + (n - 1) d$ , into the average-based sum formula:

$S_{n} = \frac{n}{2} (a + [a + (n - 1) d])$

This simplifies to the standard formula for the sum of the first $n$ terms:

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

This formula is essential when you have $a$ , $d$ , and $n$ , but not the last term.

Let's ground these concepts in a quick study of a reference text.

Progressions

The 'Progressions' handbook by CETking provides a concise summary of these formulas and some practical, exam-oriented advice.

Please read the following sections: Start at the beginning with 'ARITHMETIC PROGRESSIONS' and read up to the end of the formula for the rth term Tr = a + (r – 1)d. Read the subsection 'To Find the Sum of the given Number of Terms in an Arithmetic Progression' to see the standard sum formulas. Finally, read the section that starts with 'However, you need to understand that questions on A.P. are seldom solved on a mathematical basis...'. Focus on points 2 ('Average of an A.P.') and 3 ('Process for finding the sum of an A.P.'). This reinforces the intuitive method we just discussed.

Example: Find the sum of the first 20 terms of the AP: 3, 7, 11, 15, ...

Given: $a = 3, d = 4, n = 20$

Method 1: Standard Formula (when last term is unknown)
$S_{20} = \frac{20}{2} [2 (3) + (20 - 1) 4]$
$S_{20} = 10 [6 + (19 \times 4)]$
$S_{20} = 10 [6 + 76] = 10 [82] = 820$

Method 2: Average Method

Find the last term ( $t_{20}$ ):
$t_{20} = a + (19) d = 3 + 19 \times 4 = 3 + 76 = 79$
Find the average:
$Average = \frac{First Term + Last Term}{2} = \frac{3 + 79}{2} = \frac{82}{2} = 41$
Calculate the sum:
$Sum = n \times Average = 20 \times 41 = 820$

Test your understanding!

How many terms of the series –12, –9, –6, … must be taken for the sum to be 54? (This is Problem 2.9 from the CETking handbook). Try solving it without looking at the solution.

Show answer

We are given $S_{n} = 54$ , $a = - 12$ , and $d = - 9 - (- 12) = 3$ . We need to find $n$ .
Let's use the standard sum formula:
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$54 = \frac{n}{2} [2 (- 12) + (n - 1) 3]$
$108 = n [- 24 + 3 n - 3]$
$108 = n [3 n - 27]$
$108 = 3 n^{2} - 27 n$
Divide by 3 to simplify:
$36 = n^{2} - 9 n$
$n^{2} - 9 n - 36 = 0$
This is a quadratic equation. We can solve it by factoring. We need two numbers that multiply to -36 and add to -9. These numbers are -12 and 3.
$(n - 12) (n + 3) = 0$
The possible values for $n$ are 12 and -3. Since the number of terms cannot be negative, $n = 12$ .

Alternatively, you could test options in an exam or simply write out the terms and sum them until you reach 54: -12, -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21. Summing these gives 54.

Conclusion

In this lesson, we introduced Arithmetic Progressions and covered the two fundamental tasks: finding a specific term and finding the sum of terms. While formulas are reliable, your speed in the CAT will dramatically improve by adopting intuitive approaches like the "jumps" method for the nth term and the "average" method for the sum.

Key Takeaways:

An Arithmetic Progression (AP) has a constant difference ( $d$ ) between terms.
The nth term is found with $t_{n} = a + (n - 1) d$ . Think of this as the first term plus $(n - 1)$ jumps.
The sum of n terms is $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . For faster calculation, think of it as $S_{n} = Number of terms \times Average$ , where the average is $\frac{first term + last term}{2}$ .

Preview of the Next Lesson:
We will continue our exploration of sequences by looking at the second major type of progression. In the next lesson, we will solve problems on Geometric Progressions (GP), including infinite GP sums. In a GP, terms change by a constant ratio instead of a constant difference.

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