Calculating the Sum of the First 20 Terms of an Arithmetic Progression (AP)

Understanding the properties of an arithmetic progression (AP) and how to calculate the sum of its terms is an essential skill in both mathematics and practical applications. In this article, we will delve into an example where the nth term of an AP is given by Tn 7 - 3n, and we will calculate the sum of the first 20 terms of this AP.

Step-by-Step Explanation

Let's start by identifying the first few terms of the given AP to understand its behavior:

T1 7 - 3(1) 4 T2 7 - 3(2) 1 T3 7 - 3(3) -2

From these calculations, we can see that the first term of the AP is a 4 and the common difference d is -3, as each term decreases by 3.

The Sum of an AP

The sum of the first n terms of an AP is given by the formula:

Sn n/2 [2a (n-1)d]

To find the sum of the first 20 terms of our AP, we will use this formula. Let's substitute the values step-by-step:

Step 1: Calculate the First Term (a)

T1 4

Step 2: Calculate the 20th Term (T20)

T20 7 - 3(20) 7 - 60 -53

Step 3: Substitute into the Sum Formula

Using the formula Sn n/2 [2a (n-1)d], we get:

S20 20/2 [2(4) (20-1)(-3)] 10 [8 19(-3)] 10 [8 - 57] 10(-49) -490

Therefore, the sum of the first 20 terms of the given AP is -490.

Additional Examples and Methods

Example 1

Consider another example where the nth term of an AP is given by Tn 7 - 3n. We will find the sum of the first 25 terms of this AP.

Find the first term: T1 7 - 3(1) 4 Find the 25th term: T25 7 - 3(25) -68 Use the sum formula: S25 25/2 [2(4) (25-1)(-3)] 25/2 [8 72] 25/2 [80] 25×40 -800

Thus, the sum of the first 25 terms is -800.

Example 2

Another method involves using the sum formula directly without explicitly finding the first and last terms:

Tn 7 - 3n Substitute n 25: S25 7(25) - 3(25(26)/2) 175 - 75×13 175 - 975 -800

The result is the same, confirming the correctness of our calculations.

In conclusion, understanding the formula and steps for calculating the sum of the first n terms of an AP is crucial for solving similar problems. Whether via direct substitution or stepwise calculations, the process remains consistent and reliable.