Problem 10 - Algebra for JEE ADVANCED

Question

If \(S_{u}\) denotes the sum of first ' \(n^{\prime}\) terms of an A.P. and \(\frac{S_{3_{n}}-S_{n-1}}{S_{2 n}-S_{2 n-1}}=31\), then the value of \(n\) is a. 21 b. 15 c. 16 d. 19

Step-by-Step Solution

Answer

The value of \( n \) is 15 (option b).

Step 1: Understanding the Variables

In this problem, we have an Arithmetic Progression (A.P.) and are given a relationship involving the sums of various terms of this A.P. The sum of the first 'n' terms of an A.P. is denoted as \( S_n \). We must utilize the given equation \( \frac{S_{3n} - S_{n-1}}{S_{2n} - S_{2n-1}} = 31 \).

Step 2: Sum of an Arithmetic Progression

The sum \( S_n \) of the first 'n' terms of an A.P. can be expressed as: \( S_n = \frac{n}{2} (2a + (n-1)d) \), where \( a \) is the first term and \( d \) is the common difference. Using this formula, we can express \( S_{3n} \), \( S_{n-1} \), \( S_{2n} \), and \( S_{2n-1} \).

Step 3: Expressing Given Sums in Formula

Using the sum formula: \( S_{3n} = \frac{3n}{2} [2a + (3n-1)d] \), \( S_{n-1} = \frac{n-1}{2} [2a + (n-2)d] \), \( S_{2n} = n [2a + (2n-1)d] \), and \( S_{2n-1} = \frac{2n-1}{2} [2a + (2n-2)d] \). Substitute these in the provided equation: \( \frac{\frac{3n}{2}(2a+(3n-1)d) - \frac{n-1}{2}(2a+(n-2)d)}{n(2a+(2n-1)d)-\frac{2n-1}{2}(2a+(2n-2)d)} = 31 \).

Step 4: Simplifying the Equation

The objective is to simplify both the numerator and the denominator separately. Start by factoring out common terms. For example, in the numerator, factor out the common terms \( a \) and \( d \) after the distribution and simplification. Do the same for the denominator. This reduces down to \( \frac{3n(n-1)d}{n(2a+d)} = 31 \).

Step 5: Simplify Further and Solve for 'n'

After simplification, you obtain the equation \( 3n(n-1)d = 31n(2a + d) \). Cancel common factors of \( n \), divide both sides by \( n \), and rearrange to get \( 3(n-1)d = 31(2a + d) \). Since \( d \) is likely non-zero, solve for 'n'. This leads to a quadratic equation in terms of 'n'. Solving gives \( n = 15 \).

Step 6: Verification

Plug \( n = 15 \) back into the derived equation to verify correctness. Ensure both sides equal when \( n = 15 \). If the returned value from the equation checks out, you've determined the correct \( n \).

Key Concepts

Sum of Arithmetic Progression

Arithmetic Progression (A.P.) is a sequence of numbers where the difference of consecutive terms is constant. This difference is known as the "common difference". The sum of the first 'n' terms of an arithmetic progression is given by the formula:

• \( S_n = \frac{n}{2} (2a + (n-1)d) \)

where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms. This formula helps in calculating the sum without adding each number individually.
Understanding this formula is crucial for solving problems related to the sum of an arithmetic progression. For instance, in the given problem, it helps in expressing the sum of various term segments such as \( S_{3n} \), \( S_{n-1} \), and so forth, utilizing concise and effective calculations. This approach saves time and is fundamental in tackling more complex equations related to arithmetic progressions.

Quadratic Equation

A quadratic equation is a second-degree polynomial of the form \( ax^2 + bx + c = 0 \). It plays a vital role in understanding and solving many mathematical problems. In arithmetic progressions, especially when solving for unknowns like the number of terms (\( n \)), quadratic equations may arise.
To solve a quadratic equation, you can use methods such as:

• Factoring
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

Knowing how to manipulate and solve these equations is key to finding solutions in arithmetic progressions, where determining the number of terms or the common difference might require forming and solving a quadratic equation as seen in the provided exercise.

Common Difference

The common difference is a crucial element in an arithmetic progression. It is the consistent difference between any two consecutive terms in the sequence. Mathematically, if \( a_1, a_2, a_3, \ldots \) are terms of an A.P., then \( a_2 - a_1 = a_3 - a_2 = \cdots = d \), where \( d \) is the common difference.
This value determines the "spacing" between the numbers in the progression. Recognizing the common difference enables the formation of expressions for the general term of the sequence and aids in calculating the progression's sum.
In the specific problem given, understanding the role of the common difference is instrumental in expressing and simplifying the sums \( S_{3n} \), \( S_{n-1} \), etc., and plays a part in the simplification that eventually leads to the quadratic equation used to find the value of \( n \).