Problem 52 - A Complete Resource Book in Mathematics for JEE Main

Question

The sum of \(n\) terms of \(m\) A.P.s are \(S_{1}, S_{2}, S_{3}, \ldots, S_{m}\), If the first term and common difference are \(1,2,3, \ldots, m\) respectively, then \(S_{1}+S_{2}+S_{3}+\ldots+S_{m}=\) (A) \(\frac{1}{4} m n(m+1)(n+1)\) (B) \(\frac{1}{2} m n(m+1)(n+1)\) (C) \(m n(m+1)(n+1)\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (A) \( \frac{1}{4} m n(m+1)(n+1) \).

Step 1: Understand the Problem

We need to find the sum of sums of the first \( n \) terms of \( m \) arithmetic progressions (A.P.s). Each A.P. here has a distinct first term and a corresponding common difference.

Step 2: Use the Formula for Sum of an A.P.

Recall that the sum of the first \( n \) terms of an A.P. is given by \( S_n = \frac{n}{2} (2a + (n-1)d) \), where \( a \) is the first term and \( d \) is the common difference.

Step 3: Apply the Formula to Each A.P.

For the first A.P., \( a_1 = 1 \) and \( d_1 = 1 \), so \( S_1 = \frac{n}{2} (2 \times 1 + (n-1) \times 1) = \frac{n}{2} (n+1) \).

Step 4: Compute Each A.P. Sum

For the second A.P., \( a_2 = 2 \) and \( d_2 = 2 \), so \( S_2 = \frac{n}{2} (2 \times 2 + (n-1) \times 2) = n(n+1) \). Similarly, sum of the \( n \) terms for \( m \)-th A.P. becomes \( S_m = \frac{n}{2}(2m + (n-1)m) = \frac{nm(m+1)}{2} \).

Step 5: Sum Up All Sums

Combine the sums from all \( m \) A.P.s: \( S_1 + S_2 + \ldots + S_m = \sum_{i=1}^{m} \frac{n(i)(i+1)}{2} = \frac{n}{2} \sum_{i=1}^{m} i(i+1) \).

Step 6: Calculate the Summation

Calculate \( \sum_{i=1}^{m} i(i+1) \). This simplifies to \( \sum_{i=1}^{m} i^2 + \sum_{i=1}^{m} i \), which further simplifies using known formulas: \( \sum_{i=1}^{m} i^2 = \frac{m(m+1)(2m+1)}{6} \) and \( \sum_{i=1}^{m} i = \frac{m(m+1)}{2} \).

Step 7: Simplify and Finalize

Substituting these into the sum, we get: \( \sum_{i=1}^{m} i(i+1) = \frac{m(m+1)(2m+3)}{3} \). Therefore, the required sum is \( \frac{n}{2} \times \frac{m(m+1)(2m+3)}{3}\), which simplifies to \( \frac{1}{4}mn(m+1)(n+1) \).

Step 8: Choose the Correct Option

After calculation, comparing with the options, the correct answer is (A) \( \frac{1}{4} m n(m+1)(n+1) \).

Key Concepts

Sum of Arithmetic Series

In an arithmetic progression (A.P.), the sum of the first \( n \) terms, often denoted as \( S_n \), can be calculated using a simple formula. The formula is \( S_n = \frac{n}{2} \times (2a + (n-1)d) \). Here, \( a \) represents the first term, and \( d \) is the common difference between consecutive terms.
This formula arises from the idea that the series has both an increasing and a constant difference as it progresses. By averaging the first and the last term while multiplying by the number of terms, this calculation succinctly provides the sum.
For example, if you know the first term is 1 and the common difference is also 1, the sum of the first \( n \) terms becomes \( \frac{n}{2} \times (2 \times 1 + (n-1) \times 1) \), simplifying to \( \frac{n(n+1)}{2} \). This structure of the series makes it predictable and manageable for calculations, which are often required in solving problems related to sequences.

Common Difference

The concept of a common difference \( d \) is fundamental to understanding arithmetic sequences. The common difference is the constant amount added to each term of the sequence to arrive at the next term.
In the exercise provided, we see that each sequence has its corresponding common difference equal to its sequence number. For example, for the sequence identified as the first A.P., the common difference is 1, for the second it is 2, and for the \( m \)-th sequence, it is \( m \). This variability allows each sequence to progressively differ, providing a range of arithmetic series combinable in the problem.
The role of the common difference is crucial as it controls the rate at which the terms of the series increase or decrease. Recognizing this element allows one to predict, compute, and manipulate the entire sequence efficiently. Knowing \( d \) simplifies the task of determining any specific term in the progression using the formula \( a_n = a + (n-1)d \), enabling deeper exploration of these series.

First Term of Series

The first term of a series, denoted by \( a \), serves as the starting point for the arithmetic progression. Understanding the importance of this term is crucial as it influences all subsequent terms in the sequence.
In arithmetic progressions, each subsequent term builds from the first by adding a multiple of the common difference \( d \). For instance, when dealing with multiple sequences like in the exercise, the first term of each series is conveniently the same as its sequence number: for the first A.P., it's 1, for the second it's 2, and so on up to \( m \).
Having a clear grasp on the initial term allows for straightforward calculations of any term in the series. Whether you are computing the third term or the \( n \)-th term, you start from \( a \) and systematically add increments dictated by \( d \). This foundational term empowers computations involving both individual terms and aggregated aspects like the series' sum.