Problem 9 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the \(10^{\text {th }}\) term of an A.P. is \(\frac{1}{20}\) and its \(20^{\text {th }}\) term is \(\frac{1}{10}\), then the sum of its first 200 terms is: \(\quad\) [Jan. 8, \(\mathbf{2 0 2 0}\) (II)] (a) 50 (b) \(50 \frac{1}{4}\) (c) 100 (d) \(100 \frac{1}{2}\)

Step-by-Step Solution

Answer

The sum is \(104.5\) or option (d) \(100 \frac{1}{2}\).

Step 1: Understand the Problem

We need to find the sum of the first 200 terms of an arithmetic progression (A.P.). We know the 10th term and the 20th term of this A.P.

Step 2: Recall the Formula

The general formula for the nth term of an A.P. is \(a_n = a + (n-1) d\), where \(a\) is the first term and \(d\) is the common difference.

Step 3: Write Equations for Known Terms

The 10th term is given by \(a_{10} = a + 9d = \frac{1}{20}\). The 20th term is given by \(a_{20} = a + 19d = \frac{1}{10}\).

Step 4: Solve the System of Equations

We subtract the equation for the 10th term from the 20th term: \((a + 19d) - (a + 9d) = \frac{1}{10} - \frac{1}{20}\). This simplifies to \(10d = \frac{1}{20}\), giving \(d = \frac{1}{200}\).

Step 5: Substitute to Find the First Term

Substitute \(d = \frac{1}{200}\) back into the equation for the 10th term: \(a + 9 \left(\frac{1}{200}\right) = \frac{1}{20}\). This simplifies to \(a + \frac{9}{200} = \frac{1}{20}\). Solving for \(a\), we get \(a = \frac{1}{40}\).

Step 6: Sum of First 200 Terms Formula

The sum of the first \(n\) terms of an A.P. is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

Step 7: Calculate the Sum

Substitute \(n = 200\), \(a = \frac{1}{40}\), and \(d = \frac{1}{200}\) into the formula: \(S_{200} = \frac{200}{2} (2 \times \frac{1}{40} + 199 \times \frac{1}{200})\). Simplifying: \(S_{200} = 100 \left(\frac{1}{20} + \frac{199}{200}\right)\).

Step 8: Final Simplification

Convert \(\frac{1}{20}\) to \(\frac{10}{200}\). Adding the fractions gives \(\frac{209}{200}\). Thus, \(S_{200} = 100 \times \frac{209}{200} = 104.5\).

Key Concepts

Sum of A.P. terms

In an arithmetic progression (A.P.), finding the sum of the terms can help in understanding the entire series. The formula for calculating the sum of the first \(n\) terms of an A.P. is given by:\[S_n = \frac{n}{2} \left(2a + (n-1)d\right)\]where:

• \(S_n\) is the sum of the first \(n\) terms.
• \(n\) is the number of terms to be added.
• \(a\) is the first term of the series.
• \(d\) is the common difference between consecutive terms.

This formula is derived from the expression of the series in terms of \(a\) and \(d\), allowing us to add terms efficiently without calculating each one individually.
When applied to our specific problem, knowing the values of \(a\) and \(d\) allows us to plug these into the sum formula, simplifying our calculation of the sum of the first 200 terms.

Common difference in A.P.

The common difference in an arithmetic progression (A.P.) is a key feature that defines the uniform step between consecutive terms. The common difference, denoted by \(d\), can be easily found using the formula:\[ d = a_{n+1} - a_n \]For our problem, this difference was calculated from the 10th and 20th terms provided:

• The 10th term: \(a + 9d = \frac{1}{20}\)
• The 20th term: \(a + 19d = \frac{1}{10}\)

By subtracting the equation of the 10th term from the 20th, we found:\[ 10d = \frac{1}{10} - \frac{1}{20} \]Simplifying, this results in:\[ d = \frac{1}{200} \]The role of \(d\) ensures that each term in the sequence can be predicted based on the previous terms, making it a valuable component of the series.

nth term of an A.P.

The formula for the nth term of an arithmetic progression (A.P.) is instrumental in defining each term in the series. It's expressed as:\[ a_n = a + (n-1)d \]where:

• \(a_n\) represents the value of the nth term.
• \(a\) is the first term of the series.
• \(d\) is the common difference.
• \(n\) is the term number.

Applying this to our exercise, we utilized the formula to express the 10th and 20th terms in terms of \(a\) and \(d\):

• The 10th term, \(a + 9d = \frac{1}{20}\)
• The 20th term, \(a + 19d = \frac{1}{10}\)

These expressions allowed us to set up a system of equations to solve for both \(a\) and \(d\). Understanding how to use this formula is crucial in analyzing any arithmetic sequence, as it provides a direct connection between the sequence parameters and its individual terms.