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

Question

If the sum of \(n\) terms of an A.P. is 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

The sum of the squares of the terms is given by option (C) \( \frac{2c^2}{3} n(n+1)(2n+1) \).

Step 1: Identify components of A.P.

If the sum of the first \( n \) terms of an arithmetic progression (A.P.) is given by \( S_n = cn(n-1) \), equate it to the known formula: \( S_n = \frac{n}{2} (2a + (n-1)d) \), where \( a \) is the first term and \( d \) is the common difference.

Step 2: Equate the sum formulas

Equate the given sum formula with the standard sum formula of an A.P. to find \( a \) in terms of \( c \) and \( d \):\[ cn(n-1) = \frac{n}{2} (2a + (n-1)d) \]. Solving for \( 2a + (n-1)d = 2cn - 2c \) leads us to \( 2a = 2c \), implying \( a = c \).

Step 3: Determine common difference

From \( 2a + (n-1)d = 2cn - 2c \), substitute \( a = c \) to get:\[ 2c + (n-1)d = 2cn - 2c \]leads to \( (n-1)d = 2c(n-1) \), which simplifies to \( d = 2c \).

Step 4: Express each term of A.P.

Each term \( T_k \) of the A.P. is given by \( T_k = a + (k-1)d = c + (k-1)(2c) = c(1 + 2k - 2) \) which simplifies to \( T_k = c(2k - 1) \).

Step 5: Find sum of squares of terms

The sum of squares of the terms is given by \( \sum_{k=1}^n (c(2k-1))^2 = c^2 \sum_{k=1}^n (2k-1)^2 \).The sum \( \sum_{k=1}^n (2k-1)^2 \) can be expanded as \[ \sum_{k=1}^n ((4k^2 - 4k + 1)) \].

Step 6: Calculate the expanded sum

Solve for \( \sum_{k=1}^n ((4k^2 - 4k + 1)) \):- \( \sum_{k=1}^n 4k^2 = 4 \frac{n(n+1)(2n+1)}{6} \)- \( \sum_{k=1}^n 4k = 4 \frac{n(n+1)}{2} \)- \( \sum_{k=1}^n 1 = n \).This yields:\[ \sum_{k=1}^n (2k-1)^2 = \frac{4n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n \].

Step 7: Simplify and obtain the solution

Simplify the expression to calculate the sum of squares:\[ = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n = \frac{2n(n+1)(2n+1)}{3} \].Thus, the sum of the squares of A.P. terms is \( c^2 \cdot \frac{2n(n+1)(2n+1)}{3} \) which matches option \( C \).

Key Concepts

Sum of n Terms

In arithmetic progressions (A.P.), understanding the sum of the first \( n \) terms, denoted as \( S_n \), is crucial for solving problems involving sequences. For a general A.P., \( S_n \) is expressed as \( \frac{n}{2} (2a + (n-1)d) \), where \( a \) is the first term and \( d \) is the common difference. This formula is derived from adding up all the terms in the sequence:

• The first term is \( a \).
• The second term is \( a + d \).
• The nth term is \( a + (n-1)d \).

By adding the first and nth terms, the second and (n-1)th terms, and so on, you get \( n/2 \) pairs where each pair sums to \( 2a + (n-1)d \).
In this problem, given that \( S_n = cn(n-1) \), comparing with the standard formula reveals equations that help determine the values of \( a \) and \( d \).

Sum of Squares

The sum of the squares of terms in an A.P. is a more complex calculation than the sum of the terms themselves. However, it reveals interesting properties about the sequence. For each term \( T_k \) in the sequence, you compute \( (T_k)^2 \). In our case, the formula of each term \( T_k = c(2k-1) \) leads to:

• Calculate \( (c(2k-1))^2 = c^2(2k-1)^2 \).
• To find the sum of squares, evaluate \( \sum_{k=1}^n c^2(2k-1)^2 \).

This expression simplifies using the formula: \( \sum_{k=1}^n (2k-1)^2 \). The expansion of this sum involves the simplification of terms using known summation formulas for squares: \( \sum_{k=1}^n k^2 \) and linear terms: \( \sum_{k=1}^n k \).
Through careful calculation, the expanded form sums up to provide the desired result.

Common Difference

In any arithmetic progression, the common difference \( d \) holds the pivotal role of determining the gap between successive terms. If you have the difference correctly, finding any term in the sequence becomes straightforward. For the given problem, using the sum equation \( cn(n-1) = \frac{n}{2} (2a + (n-1)d) \), we find that:

• Express the equation as \( (n-1)d = 2c(n-1) \).
• This gives \( d = 2c \).

The common difference \( d \) is resolved here as \( 2c \), highlighting the constant rate of increase between terms in this specific sequence. Once \( d \) is determined, calculations involving any term in the A.P. or their properties become facilitated.

First Term

The first term \( a \) of an arithmetic progression anchors the starting point of the sequence. It's where everything begins, and its value directly influences the entire sequence. In an A.P., given the first term, every subsequent element is just a step away, determined by adding multiples of the common difference \( d \).

In our problem, using the relationship \( 2a = 2c \) derived from equating sum formulas, it's simple to see:

• \( a = c \).

With this formulation, each term \( T_k \) in the sequence is determined effortlessly as you proceed with the formula \( T_k = a + (k-1)d \), substituting in \( a = c \) gives the explicit expression for the terms. Understanding \( a \) thus serves as a foundational key to mastering arithmetic progressions.