Problem 26 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the sum of \(n\) terms of an A.P. is \(\mathrm{cn}(n-1)\), where \(c \neq 0\), then sum of the squares of these terms is (A) \(c^{2} n^{2}(n+1)^{2}\) (B) \(\frac{2}{3} c^{2} n(n-1)(2 n-1)\) (C) \(\frac{2 c^{2}}{3} n(n+1)(2 n+1)\) (D) None of these

Step-by-Step Solution

Answer

Option (C) \( \frac{2c^2}{3} n(n+1)(2n+1) \) is correct.

Step 1: Understand the Formula for Sum of n Terms of an A.P.

The sum of the first \(n\) terms of an arithmetic progression (A.P.) can be represented by \(S_n = \frac{n}{2} [2a + (n-1)d]\), where \(a\) is the first term and \(d\) is the common difference.

Step 2: Analyze Given Information

We are given that \(S_n = cn(n-1)\). By equating this with the general formula for the sum, we have \[ \frac{n}{2} [2a + (n-1)d] = cn(n-1) \].

Step 3: Simplify the Equation

Simplify the equation by canceling \(n\) from both sides: \[ [2a + (n-1)d] = 2c(n-1) \]. This implies \(2a + (n-1)d = 2cn - 2c\).

Step 4: Find First Term and Common Difference

To find \(a\) and \(d\), assume \(n=1\), which gives \(a = -c\).Then assume \(n=2\), which gives \[2a + d = 2c(2-1)\], leading to \[2(-c) + d = 2c\], so \(d = 4c\).

Step 5: Formulate and Simplify the Expression for the Sum of Squares

The square of the \(k^{th}\) term is \((a + (k-1)d)^2\). The sum of squares of \(n\) terms is \[ \sum_{k=1}^{n} (a + (k-1)d)^2 \].

Step 6: Calculate Individual Square Terms

Substitute \(a = -c\) and \(d = 4c\) into \((a + (k-1)d) = -c + (k-1)(4c) = (4k-5)c\). Then the square is \((4k-5)^2 c^2\).

Step 7: Sum of Squares Formula Setup

The sum of squares is \(c^2 \sum_{k=1}^{n} (4k-5)^2\). We know \( (4k-5)^2 = 16k^2 - 40k + 25 \).

Step 8: Simplify the Sum of Squares

Calculate \[ \sum_{k=1}^{n} (16k^2 - 40k + 25) \] using summation formulas: \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \], and \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \].

Step 9: Calculate the Resulting Expression

Perform the calculations: 1. \( 16 \sum_{k=1}^{n} k^2 = 16 \cdot \frac{n(n+1)(2n+1)}{6} \).2. \( 40 \sum_{k=1}^{n} k = 40 \cdot \frac{n(n+1)}{2} \).3. Sum \(25n\).Combine these results to get \[ c^2 \left( \frac{16n(n+1)(2n+1)}{6} - 40 \cdot \frac{n(n+1)}{2} + 25n \right) \].

Step 10: Identify Correct Answer

Simplify the combined result to match one of the options:This simplifies to\( \frac{2c^2}{3}n(n+1)(2n+1) \), which corresponds to option (C).

Key Concepts

Sum of squares

The sum of squares in an arithmetic progression (A.P) refers to the aggregation of squared terms from the sequence. If you imagine an A.P. as a list of numbers with a constant gap between each number, the sum of squares is simply adding up each of those numbers after they've been squared.
For an A.P., each term squared is \((a + (k-1)d)^2\) where \(a\) is the first term, \(d\) is the common difference, and \(k\) is the term's position. The sum of these squares over \(n\) terms is expressed as:

• \[\sum_{k=1}^{n} (a + (k-1)d)^2\]

Understanding how each square relates to the original sequence helps you grasp why it's useful for detailed analysis, like finding variance or in this case, simply determining the sum of square terms. For arithmetic progressions rooted in analytic or numerical problems, this computation can reveal deeper insights into the behavior of the sequence.

Formula for sum of terms

The formula for the sum of terms in an arithmetic progression is a central concept. It is given by:
\[S_n = \frac{n}{2} [2a + (n-1)d]\]
where \(S_n\) represents the sum of the first \(n\) terms, \(a\) is the first term, and \(d\) is the common difference. This formula is derived based on the idea that an A.P. has a constant difference between its terms, allowing us to pair terms from the beginning and the end to make calculations easier.

• For example, in a series where \(n=5\), each term adds up with the last, next with second last, etc.

It's a systematic way to quickly find the sum without the necessity to add each individual term manually. In practical terms, it's like creating shortcuts for bulk calculations. This knowledge paves the way for solving more complex series-related problems efficiently.

Common difference

The common difference \(d\) in an arithmetic progression is a fundamental characteristic that describes how much each term increases (or decreases) from the previous term. This constant increment \(d\) ensures that the sequence progresses in a uniform manner.
In an A.P., if you know any term and the common difference, you can determine subsequent terms. The formula for calculating any term in an A.P. is:
\[a_k = a + (k-1)d\]
where \(a\) is the first term and \(k\) is the position of the term in the sequence.

• For instance, if \(a=3\) and \(d=2\), the sequence is \(3, 5, 7, 9, \ldots\)

In our problem, understanding the common difference \(d\) was crucial as it allowed us to deduce critical properties of the sequence, such as identifying terms to formulate the sum of squares. This insight is key to tackling arithmetic progression problems effectively.