5.4. Sum of First n Terms of an AP

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?

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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

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.

--