5.4 - Sum of First n Terms of an AP
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Introduction to the Sum of First n Terms
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Today, we’re diving into how to find the sum of the first n terms of an arithmetic progression, or AP. Can anyone remind me what an AP is?
An AP is a sequence of numbers where each term is obtained by adding a constant value to the previous term.
Great! And that constant value is called the common difference, denoted as 'd'. If we want to sum the first n terms, we need to know the first term, 'a', and the common difference. Does anyone know the formula for this sum?
Is it something like S equals n over 2 times the sum of the first and last terms?
Exactly! The formula is \( S_n = \frac{n}{2} (a + l) \). We can derive it from the properties of the sequence. Let's explore how.
Derivation of the Sum Formula
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To derive the sum formula, let’s write down the first two expressions for the sum of terms in an arithmetic sequence. We have S_n = a + (a + d) + (a + 2d) + ... + [a + (n-1)d]. Now, if we write it in reverse order, what do we get?
We get l + (l - d) + (l - 2d) + ... + a.
Right! If we add these two equations, each pair sums to the same value, resulting in \( 2S = n(2a + (n - 1)d) \). Does anyone remember how we finalize this?
We divide by 2 to get \( S_n = \frac{n}{2} [2a + (n - 1)d] \)!
Correct! This formula allows us to calculate the sum of the first n terms easily.
Practical Example of the Sum Formula
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Let’s apply this formula to a practical scenario—Shakila's contributions to her daughter's money box over 21 birthdays. Can anyone tell me the first term and common difference here?
The first term, a, is 100, and the common difference, d, is 50, since she increases the amount by that each birthday.
Exactly! Now, applying the sum formula, what would be the total amount by her 21st birthday?
Using \( S_n = \frac{n}{2} [2a + (n - 1)d] \), we calculate \( S_{21} = \frac{21}{2} [2(100) + (21 - 1)(50)] = \frac{21}{2} [200 + 1000] = \frac{21}{2} [1200] = 12600 \).
Well done! This method saves us a lot of time compared to adding each number individually.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
The section introduces the concept of finding the sum of the first n terms of an arithmetic progression (AP) using formulas derived from the properties of APs. It explains the significance of the first term, the common difference, and employs historical examples to illustrate the concept.
Detailed
In this section, we explore the process of calculating the sum of the first n terms of an arithmetic progression (AP). An AP is defined as a sequence of numbers in which the difference between consecutive terms is constant, known as the common difference (d). The sum of the first n terms can be represented by two formulas: \( S_n = \frac{n}{2} [2a + (n - 1)d] \) and \( S_n = \frac{n}{2} [a + l] \), where 'a' is the first term, 'l' is the last term, and 'n' is the number of terms. To illustrate the application of these formulas, the section presents a practical scenario involving the collection of money over the years and how to compute the total amount efficiently. Historical anecdotes about mathematicians like Gauss are shared to motivate students to appreciate the formulas intuitively. Additionally, examples and exercises are provided to reinforce the understanding of these formulas in diverse contexts.
Example 1: If the sum of the first 15 terms of the AP: 10, 7, 4, ... is 180, find the first term.
Solution: Here, \( a = 10, \ d = 7 - 10 = -3, \ n = 15 \).
We know that
\[ S_n = \frac{n}{2}[2a + (n-1)d] \]
Therefore,
\[ 180 = \frac{15}{2}[2(10) + (15-1)(-3)] \]
Thus, we have
\[ 180 = \frac{15}{2}[20 - 42] \]
So, the first term of the AP is 10.
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Understanding the Problem
Chapter 1 of 6
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Chapter Content
Let us consider the situation again given in Section 5.1 in which Shakila put 100 into her daughter’s money box when she was one year old, 150 on her second birthday, ` 200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old?
Detailed Explanation
In this scenario, Shakila deposits varying amounts of money into a money box for her daughter on each birthday. The deposits are made in an increasing sequence: 100, 150, 200, and so on, increasing by 50 each year. We need to find the total amount of money in the box after 21 years.
Examples & Analogies
Imagine a piggy bank where every birthday, you put in a little more money than the year before. If on your first birthday, you got 100 rupees, on your second birthday you got 150, by your 21st birthday you would have saved a lot! This represents how small increases can grow over time.
Finding the Total Amount
Chapter 2 of 6
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Chapter Content
Here, the amount of money (in `) put in the money box on her first, second, third, fourth . . . birthday were respectively 100, 150, 200, 250, . . . till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum.
Detailed Explanation
The mentioned deposits form an arithmetic progression (AP). The sequence increases by a common difference of `50. To find the total sum of these deposits for 21 terms, we can utilize the formula for the sum of an AP:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Here, \( a \) is the first term (100), \( d \) is the common difference (50), and \( n \) is the number of terms (21). This formula allows us to compute the total amount quickly without writing each term individually.
Examples & Analogies
Think of using a recipe that allows you to prepare a dish much quicker than measuring each ingredient separately for each serving. In this method, you create a formula that gives you the total without tedious calculations.
Applying Gauss's Method
Chapter 3 of 6
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We consider the problem given to Gauss (about whom you read in Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote: S = 1 + 2 + 3 + . . . + 99 + 100 and then, reversed the numbers to write S = 100 + 99 + . . . + 3 + 2 + 1.
Detailed Explanation
Gauss observed that if you add the series from both ends towards the center (1 + 100, 2 + 99, etc.), each pair sums to the same value (101) and there are 50 such pairs. Therefore, the total sum is \[ 50 \times 101 = 5050. \] This method simplifies the addition of consecutive numbers by pairing them, showing how useful patterns in arithmetic can be.
Examples & Analogies
Imagine counting all the candy pieces you have. Instead of counting each piece individually, you pair them up (one in each hand). It’s a faster way to ensure you haven’t missed any, just like Gauss paired numbers to quickly calculate the total.
Deriving the Formula for the Sum of an AP
Chapter 4 of 6
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We will now use the same technique to find the sum of the first n terms of an AP: a, a + d, a + 2d, . . . The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms of the AP. We have S = a + (a + d) + (a + 2d) + . . . + [a + (n – 1) d] (1). Rewriting the terms in reverse order, we have S = [a + (n – 1) d] + [a + (n – 2) d] + . . . + (a + d) + a (2).
Detailed Explanation
By writing the sum of the AP in reverse, adding both forms provides a simplified way to compute S. Each time you add, you get the same value for each position (\[2a + (n-1)d\] for each term). Therefore, total adding gives us \[ 2S = n (2a + (n-1)d) \] leading to\[ S = \frac{n}{2} [2a + (n-1)d]. \] This formula allows calculating the sum of n terms quickly, emphasizing the efficiency in solving mathematical problems.
Examples & Analogies
Think about using an efficient trick to track expenses over time rather than checking every single item on a shopping list. The formula summarizes and simplifies the significant pattern, leading to quicker conclusions.
Practical Application Example
Chapter 5 of 6
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Chapter Content
Now we return to the question that was posed to us in the beginning. The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday, . . ., were 100, 150, 200, 250, . . ., respectively. This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Detailed Explanation
Given the first term as 100 and the number of terms as 21, using the previously derived formula makes it easy. The values plugged in would yield a sum that gives us the total amount in the money box by the time she turns 21.
Examples & Analogies
Consider a building project where each layer or floor added contributes more structures. Using the formula lets you quickly find the total height all the way up without having to measure it step by step, just as calculating the AP lets us find the total money without individually adding all the contributions.
Conclusion and Summary
Chapter 6 of 6
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Chapter Content
We can also write this as S = [a + a + (n – 1) d] / 2, i.e., S = (a + a_n) / 2. This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
Detailed Explanation
This final rearrangement presents an alternative way to compute the sum of an AP when the last term is known. It's especially handy when details about the common difference are lacking, but both the first and last terms can be utilized to find the sum.
Examples & Analogies
This can be likened to averaging the scores of two exams to find a general performance measure, where knowing both high and low scores helps illustrate overall achievement without needing every detail in between.
Key Concepts
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Sum of First n Terms: Formula to calculate the sum of the first n terms of an AP.
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Arithmetic Structure: Understanding the structure and characteristics of an AP.
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First and Last Terms: Roles of the first term and the last term in summation.
Examples & Applications
Finding the sum of the first 21 terms collected by Shakila amounts to a calculation simplifying the tedious addition process.
Using Gauss's technique of summing sequential numbers to derive a formula for calculating sums of terms in sequences.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
To find S, it's nifty, just take a, add l, and then divide by two, real quick it's swell!
Stories
Imagine Shakila saves money for her daughter each birthday. The total grows like branches on a tree, spreading wide with each year. Adding them up is complicated, but using the magical sum formula makes it easy!
Memory Tools
Remember S = n/2 (first term + last term) as 'Silly Ninjas Paint Leafy Trees' to recall the elements: Sum, Number of terms, First and Last.
Acronyms
FORMULA for AP Sum
'A Popular Hero' stands for 'a + l' (first term + last term) since it is central to finding our sum.
Flash Cards
Glossary
- Arithmetic Progression (AP)
A sequence of numbers in which the difference between consecutive terms is constant.
- Common Difference (d)
The fixed amount added to each term in an arithmetic progression.
- First Term (a)
The initial term in an arithmetic progression.
- Nth Term (a_n)
The term located at position n in an arithmetic progression.
- Sum of n Terms (S_n)
The total value when summing the first n terms of an arithmetic progression.
Reference links
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