Finding the 20th Term of an Arithmetic Progression Given a Specific Sum Condition

In an arithmetic progression (A.P.), the relationship between its terms and the sum of its terms follows a specific pattern. This article will delve into the steps to find the 20th term of an A.P. given that the sum of the first five terms is one-fourth of the sum of the next five terms. Let's explore the detailed solution step-by-step.

Understanding the Problem

Given an A.P. with the first term (a 2) and common difference (d), the problem requires us to find the 20th term. We know that the sum of the first five terms is one-fourth of the sum of the next five terms.

Step-by-Step Solution

1. Expressing the First Five Terms:

The first five terms of the A.P. can be written as:

First term: (a 2) Second term: (a d 2 d) Third term: (a 2d 2 2d) Fourth term: (a 3d 2 3d) Fifth term: (a 4d 2 4d)

The sum of these terms, (S_5), can be calculated using the formula for the sum of an A.P.:

[S_5 frac{n}{2} times [2a (n-1)d] frac{5}{2} times [2 times 2 4d] frac{5}{2} times (4 4d) 10 10d]

2. Expressing the Next Five Terms:

The next five terms are:

Sixth term: (a 5d 2 5d) Seventh term: (a 6d 2 6d) Eighth term: (a 7d 2 7d) Ninth term: (a 8d 2 8d) Tenth term: (a 9d 2 9d)

The sum of these terms, (S_{10}), is:

[S_{10} frac{n}{2} times [2a (n-1)d] frac{10}{2} times [2 times 2 9d] 10 times (4 9d) 10 45d]

3. Setting Up the Equation:

According to the problem, (S_5 frac{1}{4} times S_{10-S_5}). Substituting the sums, we get:

[10 10d frac{1}{4} times (10 45d - (10 10d)) frac{1}{4} times (35d) frac{35d}{4}]

Multiplying both sides by 4 to eliminate the fraction:

[40 40d 35d]

Re-arranging for (d):

[40d - 35d -40 implies 5d -40 implies d -8]

4. Finding the 20th Term:

The general term, (T_n a (n-1)d), will be used to find the 20th term:

[T_{20} 2 (20-1)(-8) 2 - 19 times 8 2 - 152 -150]

The 20th term is:

[T_{20} boxed{-150}]

Conclusion

This problem demonstrates the application of the sum formula for an A.P. and how to derive the common difference and specific terms from given conditions. Understanding these steps is crucial for solving similar problems related to arithmetic progressions in mathematics.