5.3. nth Term of an AP

Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of 8000, with an annual increment of 500. What would be her monthly salary for the fifth year?

The key idea here is to calculate how Reena's salary increases over the years. Starting with an initial salary of 8000, and increasing by 500 each year means we will add 500 for each year that passes. For instance, in her first year, she earns 8000, in the second year, it is 8500 (which is 8000 + `500). This process continues, establishing a pattern where each year's salary builds on the last. Hence, we derive a formula for calculating any year's salary efficiently.

Think of this situation like planting a tree where every year the tree grows taller by a fixed amount. If you measure the tree's height in the first year and continue adding that same height every year, at the end of each year, you'll be able to predict how tall the tree will be in the future without having to measure it each time.

Salary for the 15th year = [8000 + (15 – 1) × 500] = 15000.

We observe that by simplifying the salary for any year, moving from 1 to n can be achieved with the formula: Salary(n) = First Salary + (n - 1) * Increment. Hence, when we rearrange our observations mathematically, we create a formula that allows for quick feedback rather than recalculating each year.

Imagine if you are juggling balls, and each time you catch one, you add another ball to the pile you’re juggling. The first year, you catch one ball, then each year you catch one more. If someone asks you how many you’d be juggling after n years, rather than counting each one, you can simply say the initial count plus a ball for each additional year.

So, the nth term a_n of the AP with first term a and common difference d is given by a_n = a + (n – 1) d.

This chunk introduces the standardized formula for the nth term of an arithmetic progression. The first term is labeled 'a', and the difference between each term 'd'. To find any term in the sequence, we calculate it by taking the first term and adding the product of the increment and how far along we are in the sequence minus one.

Consider baking cookies—if the first batch yields 12 cookies (first term), and every subsequent batch adds 4 more cookies (the increment), the number of cookies you’ll have after 'n' batches is easily calculable using this formula without needing to bake batch by batch each time.

Example 3: Find the 10th term of the AP: 2, 7, 12, . . . Solution: Here, a = 2, d = 7 - 2 = 5 and n = 10. We have a_n = 2 + (10 - 1) × 5 = 2 + 45 = 47. Therefore, the 10th term of the given AP is 47.

In this example, we first identify the first term 'a' and the common difference 'd'. Using the nth term formula, we substitute the values for 'n', 'a' and 'd'. Spotting how the components fit together helps us find our target term, the 10th term, efficiently.

Similar to filling a jar with marbles where the first addition is 2 marbles and thereafter 5 marbles per time. By counting how many intervals (n) of additions you've completed using our formula, you can quickly know how many marbles you have without directly counting each time.

Example 4: Which term of the AP: 21, 18, 15, . . . is –81? Also, is any term 0? Give reason for your answer.

In this scenario, we need to find out which term matches with -81. By using the nth term formula, we can set -81 equal to a_n, and solve for 'n' to find its position in the sequence. Similarly, checking if 0 appears requires checking if there exists an integer 'n' that satisfies the nth term equation leading to 0.

Think of this like finding the day of the month that falls below a certain threshold. In this case, we are reversing the logic of counting days, instead counting back to a point of relevance, which allows for efficient solving.

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

This example illustrates how to work backwards from known terms to derive 'a' and 'd'. By setting up equations based on the nth term definitions for the 3rd and 7th terms, we can solve them simultaneously to find the sequence's first term and common difference.

This is like piecing together a jigsaw puzzle where we use the pieces we know (specific terms) to find the edges (first term and common difference) of the puzzle, leading us to the complete picture of the AP.

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