Problem 52 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The sum of the first three terms of a G.P. is \(S\) and their product is \(27 .\) Then all such \(S\) lie in : \(\quad\) [Sep. 02, \(\mathbf{2 0 2 0}\) (I)] (a) \((-\infty,-9] \cup[3, \infty)\) (b) \([-3, \infty)\) (c) \((-\infty,-3] \cup[9, \infty)\) (d) \((-\infty, 9]\)

Step-by-Step Solution

Answer

The answer is (c) \((- fty,-3] \cup [9, fty)\).

Step 1: Understand the terms of a G.P.

Let the three terms of a geometric progression (G.P.) be denoted as \(a, ar, ar^2\). Here, \(a\) is the first term and \(r\) is the common ratio.

Step 2: Write sum and product equations

According to the problem, the sum of the terms is \(S = a + ar + ar^2\), which factors as \(S = a(1 + r + r^2)\). The product of the terms is given by \(a \cdot ar \cdot ar^2 = 27\), simplifying to \(a^3 r^3 = 27\).

Step 3: Solve for common ratio or first term using product

Express \(a^3 r^3 = 27\) as \((ar)^3 = 27\), giving \(ar = 3\) or \(ar = -3\) since the cubic roots of 27 are 3 and -3.

Step 4: Express the first term in terms of the common ratio

From \(ar = 3\), we have \(a = \frac{3}{r}\). Similarly, if \(ar = -3\), then \(a = \frac{-3}{r}\).

Step 5: Find S in terms of r

Substituting \(a = \frac{3}{r}\) into \(S = a(1 + r + r^2)\), we get \(S = \frac{3}{r}(1 + r + r^2) = 3\left(\frac{1}{r} + 1 + r\right)\). For \(a = \frac{-3}{r}\), we get \(S = -3\left(\frac{1}{r} + 1 + r\right)\).

Step 6: Analyze possible values for S

The expression \(1/r + 1 + r\) has a minimum value of 3 if \(r = 1\), resulting in \(S = 9\) or \(S = -9\). Thus, all values are \(S \geq 9\) or \(S \leq -9\), when \(ar = 3\) or \(ar = -3\), balancing G.P. terms symmetrically on the axis of symmetry (AM-GM principle).

Step 7: Determine the correct interval

Considering the possible values for \(S\), the range \((S \leq -9) \,\text{or}\, (S \geq 9)\) fits into the union of intervals \((-\infty,-9] \cup [9, \infty)\), which matches option (c).

Key Concepts

Sum of GP terms

In a geometric progression (GP), the sum of terms is crucial to evaluate. The sum for the first few terms of a GP can be expressed in simplest form as a series consisting of powers of the common ratio. Suppose we have the first three terms of a GP: \( a, ar, \'and\, ar^2 \). The total sum \( S \) of these terms is given by:

• The formula is: \( S = a + ar + ar^2 \).
• You can factor this equation as: \( S = a(1 + r + r^2) \).

This equation shows how the terms grow depending on the common ratio \( r \). The sum of the three terms plays a pivotal role in understanding the behavior and structure of the GP. With this formula, one can understand how deeply interconnected the terms are and how they sum up to a finite value, which acts as a guiding reference in many mathematical exercises related to GPs.

Product of GP terms

Another important concept in understanding geometric progressions is the product of its terms. In our problem, the product of the first three terms gives us significant information. Consider the GP terms \( a, ar, ar^2 \). Their product can be expressed as follows:

• The formula is: \( a \cdot ar \cdot ar^2 = a^3r^3 \).
• In our case, the product is equal to 27: \( a^3r^3 = 27 \).
• This simplifies to: \( (ar)^3 = 27 \).

By taking the cube root of 27, we find possible values for \( ar \), namely \( ar = 3 \) or \( ar = -3 \). This understanding helps in connecting each term of the GP through their multiplicative relationship. When tasked with finding the individual values of the GP terms or solving for \( a \) and \( r \), this product relationship becomes invaluable.

Common ratio in GP

The common ratio \( r \) is central to defining a geometric progression. It determines how each successive term behaves in relation to the previous one. Understanding its properties can help solve numerous GP problems.

• The common ratio is part of every term’s derivation: \( ar = a \times r \), \( ar^2 = ar \times r \).
• From the product equation \( (ar)^3 = 27 \), we solve: \( ar = 3 \rightarrow a = \frac{3}{r} \) \, and \, \( ar = -3 \rightarrow a = \frac{-3}{r} \).

The common ratio determines not only the growth of the sequence but also affects whether sequential terms alternate or consistently scale by a constant factor. This alternate value occurs when the common difference is negative, changing the direction of growth or decay within the progression. Hence, the common ratio is valuable in predicting sequence behavior within any problem formulation of a geometric series.