In the sum of first n terms of an A.P. is cn², then the sum of squares of these n terms is (A) n(4n²

Q: Wait, why do we use $T_n = S_n - S_{n-1}$? Can't we just assume the A.P. starts at $c$?

That is a dangerous assumption! While $S_n = cn^2$ implies an A.P., we must derive $T_n$ rigorously to ensure we capture the common difference correctly. $S_n - S_{n-1}$ is the universal key for any sequence sum, not just A.P.s. It is the most robust way to define the $n$-th term.

Q: Why does the $6n$ term cancel out so perfectly at the end?

That is the elegance of mathematics! When you expand the expression $2(n+1)(2n+1) - 6(n+1) + 3$, the linear terms in $n$ are designed by the structure of the problem to vanish. If they didn't, it would suggest a calculation error. Always look for these 'mathematical symmetries'—they are the universe's way of telling you that you are on the right path.

Q: Is there a faster way to do this without expanding the summation?

For this specific problem, expansion is the most reliable path. While there are properties of A.P.s you could use, the summation method is foolproof. In JEE Advanced, speed is important, but accuracy is paramount. Mastering this expansion technique ensures you won't make silly errors under pressure.

Question

In the sum of first $n$ terms of an A.P. is $cn^2$, then the sum of squares of these $n$ terms is

Options:

(A) $\frac{n(4n^2 - 1)c^2}{6}$

(B) $\frac{n(4n^2 + 1)c^2}{3}$

(C) $\frac{n(4n^2 - 1)c^2}{3}$

(D) $\frac{n(4n^2 + 1)c^2}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are performing a mathematical autopsy on a sequence. We are given the sum of the first $n$ terms of an Arithmetic Progression as $S_n = cn^2$.

Our mission is to find the sum of the squares of these terms. Think of $S_n$ as the total weight of a stack of bricks; we need to isolate the weight of each individual brick to square them.

We use the fundamental bridge: $T_n = S_n - S_{n-1}$. By subtracting the sum of the first $n-1$ terms from the total sum of $n$ terms, we isolate the very last term, $T_n$.

### The Transformation: From Sum to Term

Let us perform the algebra. We have $S_n = cn^2$, which implies $S_{n-1} = c(n-1)^2$.

Substituting these into our bridge equation, we get:

$$T_n = cn^2 - c(n-1)^2$$

Expanding the square, we have $T_n = cn^2 - c(n^2 - 2n + 1)$. The $cn^2$ terms cancel out, leaving us with:

$$T_n = 2cn - c = c(2n - 1)$$

This linear expression is our general term and holds the key to everything that follows.

### The Summation: Setting the Stage

Now, we need the sum of the squares of these terms, denoted as $\sum_{r=1}^n T_r^2$. Substituting our general term, this becomes:

$$\sum_{r=1}^n [c(2r-1)]^2$$

We can pull the constant $c^2$ outside the summation:

$$c^2 \sum_{r=1}^n (4r^2 - 4r + 1)$$

We distribute the summation operator across three distinct, manageable parts:

$$c^2 \left[ 4 \sum_{r=1}^n r^2 - 4 \sum_{r=1}^n r + \sum_{r=1}^n 1 \right]$$

### The Algebraic Dance: The Grand Finale

Now, we invoke our standard summation identities:
$\sum r^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum r = \frac{n(n+1)}{2}$, and $\sum 1 = n$.

Substituting these into our expression, we get:

$$c^2 \left[ 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n \right]$$

Simplifying the fractions, we have:

$$c^2 \left[ \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n \right]$$

To make this cleaner, we factor out $\frac{n}{3}$:

$$\frac{c^2n}{3} [2(n+1)(2n+1) - 6(n+1) + 3]$$

Expanding the terms inside the bracket gives $2(2n^2 + 3n + 1) - 6n - 6 + 3$, which simplifies to $4n^2 + 6n + 2 - 6n - 3$. The $6n$ and $-6n$ terms cancel out, leaving us with $4n^2 - 1$.

The final result is:

$$\mathbf{\frac{n(4n^2 - 1)c^2}{3}}$$

You have just conquered a classic JEE Advanced problem. Remember, the complexity is just a mask; the underlying structure is always elegant.

Animated Solution for Mathematics - Sequence and Series: In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms is

Select Answer:

Visualized Solution

Understanding the Given Sum $S_{n}$

Finding the General Term $T_{n}$

Substituting the Values

Simplifying $T_{n}$

Setting up the Sum of Squares

Expanding the Square

Applying Summation Formulas

Substituting the Formulas

Taking Common Factors

Expanding Inside the Bracket

Final Algebraic Manipulation

The Way Forward

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Transformation

The Summation

The Algebraic Dance

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is