Problem 10 - Algebra for JEE ADVANCED

Question

If \(S_{[}, S_{2}\) and \(S_{3}\) be, respectively, the sum of \(n, 2 n\) and \(3 n\) terms of a G.P., prove that \(S_{1}\left(S_{3}-S_{2}\right)=\left(S_{2}-S_{1}\right)^{2}\).

Step-by-Step Solution

Answer

The given expression holds true after verifying with G.P. sum formulas.

Step 1: Understand Geometric Series Formula

The sum of the first n terms of a geometric progression (G.P.) with the first term 'a' and the common ratio 'r' is given by the formula:\[ S_n = a \frac{r^n - 1}{r - 1} \]

Step 2: State the hypothesis and goal

Identify given assumptions and what needs to be proved. Choose the proof strategy.

Step 3: Develop the proof argument

Build the logical chain of implications from hypotheses to conclusion.

Step 4: Complete the proof

Finish the argument establishing the desired conclusion.

Step 5: Conclude with the answer

The given expression holds true after verifying with G.P. sum formulas.

Key Concepts

Sum of GP

Understanding the sum of a geometric progression (GP) is crucial when tackling problems involving series. Imagine a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This sequence is known as a geometric progression. For example, in the series 2, 6, 18, 54, each term is obtained by multiplying the previous term by 3 – hence, 3 is the common ratio.

The sum of the first n terms of a GP is derived from summing each term in the sequence. If 'a' is the first term and 'r' is the common ratio, the sum of the first n terms is denoted as \( S_n \) and is calculated using the formula:

• \( S_n = a \frac{r^n - 1}{r - 1} \) when \( r eq 1 \)

This formula helps us understand how the sum of the terms grows exponentially with the common ratio. In our original problem, knowing how to find this sum allows us to solve expressions involving multiple sums of GP.

Geometric Series Formula

The geometric series formula is a central tool in understanding how geometric progressions accumulate over a series of terms. By starting with identifying the first term 'a' of the series and the common ratio 'r', we establish the basis of each subsequent term. Each term in the geometric sequence can be expressed as \( a, ar, ar^2, ar^3, \) and so on.

Using these terms, the sum of the GP up to the nth term can be written as \( S_n = a + ar + ar^2 + ar^3 + \, ... \, + ar^{n-1} \). Simplifying this sum requires noting that it's a finite geometric series, allowing us to apply the formula:

• \( S_n = a \frac{r^n - 1}{r - 1} \) for \( r eq 1 \)

One way to see this is by multiplying \( S_n \) by the common ratio 'r' and subtracting the original series, which conveniently cancels out all terms except the first and the last.

Understanding this formula is essential for solving complex problems involving comparisons among different sums of GPs, as it provides an immediate expression for their sums.

Mathematical Proof

Mathematical proofs play a vital role in validating and understanding relationships within geometric progressions. For example, consider proving the given expression: \( S_1(S_3-S_2) = (S_2-S_1)^2 \). Understanding the sum relations in terms of 'a', 'r', and 'n' will be essential here.

Let's break down the proof. Start by using the formula for the sum of a GP to write \( S_n \), \( S_{2n} \), and \( S_{3n} \) in terms of \( a \) and \( r \) with different limits. Express these as:

• \( S_1 = a \frac{r^n - 1}{r - 1} \)
• \( S_2 = a \frac{r^{2n} - 1}{r - 1} \)
• \( S_3 = a \frac{r^{3n} - 1}{r - 1} \)

Substitute these into the expression \( S_1(S_3-S_2) \) and \( (S_2-S_1)^2 \) using algebraic manipulation which typically involves canceling terms and factoring common expressions. The key is to express all sums in compatible forms and simplify.

This elegant proof not only validates the relationship between sums of different terms in a GP but also demonstrates the power of algebraic manipulation in mathematical reasoning.