Problem 5 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a_{1}, a_{2}, a_{3}\) (with \(\left.a_{1}>0\right)\) are in G.P. with common ratio \(r\), then the value of \(r\) for which the inequality \(9 a_{1}+5 a_{3}\) \(>14 a_{2}\) holds, cannot be in the interval(A) \(\left[1, \frac{9}{2}\right]\) (B) \((-\infty, 0)\) (C) \(\left[\frac{5}{9}, 1\right]\) (D) \(\left[1, \frac{9}{5}\right]\)

Step-by-Step Solution

Answer

(D) \([1, \frac{9}{5}]\) because \(r > 1.8\) fails it.

Step 1: Understanding the Given Problem

We are given a geometric progression (G.P.) with terms \(a_1, a_2, a_3\). Each term is related by a common ratio \(r\), i.e., \(a_2 = a_1 r\) and \(a_3 = a_1 r^2\). We need to determine the values of \(r\) such that the inequality \(9a_1 + 5a_3 > 14a_2\) does not hold.

Step 2: Expressing Terms Using Common Ratio

Using the properties of a geometric progression, express \(a_2\) and \(a_3\) using the common ratio \(r\):- \(a_2 = a_1 r\)- \(a_3 = a_1 r^2\)Now substitute these into the inequality \(9a_1 + 5a_3 > 14a_2\).

Step 3: Substituting Values and Simplifying the Inequality

Substitute the expressions for \(a_2\) and \(a_3\) into the inequality:\[9a_1 + 5a_1 r^2 > 14a_1 r\]Factor \(a_1\) along the inequality:\[a_1 (9 + 5r^2) > 14a_1 r\]. Since \(a_1 > 0\), it can be canceled out (since it’s positive):\[9 + 5r^2 > 14r\].

Step 4: Rearranging and Solving the Quadratic Inequality

Rearrange the inequality to standard quadratic form:\[5r^2 - 14r + 9 > 0\].To find \(r\), we solve the corresponding quadratic equation \(5r^2 - 14r + 9 = 0\) to determine where the inequality changes sign.

Step 5: Solving the Quadratic Equation

Use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) on \(5r^2 - 14r + 9 = 0\), where \(a = 5\), \(b = -14\), and \(c = 9\):\[r = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 5 \cdot 9}}{10}\]. Calculate the discriminant:\[b^2 - 4ac = 196 - 180 = 16\].Compute the roots:\[r = \frac{14 \pm 4}{10}\]. This yields roots \(r = \frac{18}{10} = 1.8\) and \(r = \frac{10}{10} = 1\).

Step 6: Identifying Intervals for Inequality

The quadratic inequality \(5r^2 - 14r + 9 > 0\) changes sign at the roots \(r = 1\) and \(r = 1.8\) (or \(\frac{9}{5}\)). This inequality holds true outside these roots, i.e., for \(r \in (-\infty, 1) \cup (1.8, \infty)\).

Step 7: Determining the Interval that Cannot Fulfill Inequality

We now examine the provided intervals:- (A) \([1, \frac{9}{2}]\)- (B) \((-finity, 0)\)- (C) \([\frac{5}{9}, 1]\)- (D) \([1, \frac{9}{5}]\)The only interval that does not satisfy the inequality is \((D)\), because we have \(r > 1.8\) for \(\frac{9}{5} = 1.8\).

Key Concepts

Geometric Progression

Geometric Progression (G.P.) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. For example, in the sequence 2, 4, 8, 16, each term is obtained by multiplying the previous term by 2. This pattern continues indefinitely.

Some key characteristics of geometric progressions include:

• The terms grow or shrink exponentially.
• They are defined by just two parameters: the first term and the common ratio.
• This type of sequence is common in finances (compound interest) and physics (exponential growth/decay).

In mathematical terms, if we denote the first term by \(a_1\) and the common ratio by \(r\), the general form of the sequence is \(a_1, a_1r, a_1r^2, a_1r^3, \ldots\).
In this problem, understanding the structure of the geometric progression helps us express connected terms through a common ratio, which is crucial for solving the inequality.

Common Ratio

The common ratio in a geometric progression is the constant factor between consecutive terms. It is key to determining the sequence's behavior. For a positive ratio, terms increase; for values between 0 and 1, they decrease. If \(r < 0\), the terms alternate in sign.

To find the common ratio in a given geometric progression, observe the relationship between terms: if \(a_2 = a_1r\) and \(a_3 = a_1r^2\), then \(r = \frac{a_2}{a_1}\) or \(r = \frac{a_3}{a_2}\).

In our problem, identifying \(r\) allowed us to express all terms of the sequence in terms of \(a_1\) and \(r\). Understanding the role of the common ratio simplifies the task of manipulating and solving equations involving sequences. Identifying how \(r\) behaves gives insights into whether an inequality involving these terms holds.

Quadratic Equation

A quadratic equation is an expression in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. Solving a quadratic equation involves finding the value(s) of \(x\) that satisfy the equation.

The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) provides a way to find these solutions. The part under the square root, \(b^2 - 4ac\), is called the discriminant, which determines the nature of the roots:

• If positive, there are two real roots.
• If zero, there is one real root.
• If negative, the roots are complex.

In the exercise, solving the quadratic equation given by \(5r^2 - 14r + 9 = 0\) helped us find critical points for the inequality problem. These roots provided boundaries where our inequality changes sign, essential for determining the regions of interest for \(r\).

Solving Inequality

Solving an inequality like \(5r^2 - 14r + 9 > 0\) involves finding the range of values for which the inequality holds. First, solve the associated quadratic equation to find critical points, then test intervals defined by these points.

Once the roots \(r=1\) and \(r=1.8\) were found, they provided boundaries for splitting the \(r\)-axis into intervals. Testing points within these intervals tells us where the inequality holds or does not. For instance, choose a test point in each interval:

• For \(r < 1\), say \(r = 0\) will satisfy the inequality.
• For \(1 < r < 1.8\), say \(r = 1.5\) will not satisfy the inequality.
• For \(r > 1.8\), say \(r = 2\) will satisfy the inequality.

In this problem, the task is to find which interval of \(r\) does not satisfy the inequality. Analyzing these intervals shows that the inequality holds for \(-(\infty, 1)\) and \((1.8, \infty)\), but not for \([1, 1.8]\). Hence, we determine which provided interval is not compatible.