Question

If \(a, b, c(a>0)\) are three successive terms of a G.P. with common ratio \(r\), the value of \(r\) for which \(c>4 b-3 a\) holds given by : (a) \(13\) (d) none of these

Short answer

Answer: The inequality c > 4b - 3a holds when (a) 1 < r < 3.

Step-by-step solution

Express b and c in terms of a and r

Since \(a, b, c\) are successive terms of a G.P., we have the following relationships: $$b = ar, \quad c = ar^2$$

Substitute expressions for b and c into the inequality

Substitute the expressions for \(c\), and \(b\) we found in step 1 into the inequality \(c>4b-3a\): $$ar^2 > 4ar - 3a$$

Simplify the inequality

Factor out a common term of \(a\), and then divide both sides by \(a\) (remembering \(a > 0\)): $$r^2 > 4r - 3$$

Rearrange the inequality into a quadratic inequality

Rearrange the inequality to have all the terms on one side: $$r^2 - 4r + 3 < 0$$

Factor the quadratic

Now factor the quadratic: $$(r - 1)(r - 3) < 0$$

Solve the quadratic inequality

The inequality is satisfied when \((r - 1)(r - 3) < 0\). This occurs when \(r\) is between the two roots, 1 and 3. Therefore, the inequality is satisfied for: $$1 < r < 3$$

Compare the solution to the given options

Looking at the options, we see that our solution matches option (a): (a) \(1<r<3\) So the correct answer is (a) \(1<r<3\).

Key concepts

Common Ratio in Geometric Progressions

Understanding the 'common ratio' is essential when studying geometric progressions (GPs). A GP is 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, denoted as 'r'.

For example, in the progression 2, 4, 8, 16, the common ratio is 2 because each term is twice its predecessor. This property allows us to express any term in a geometric sequence using the first term and the common ratio as follows: if 'a' is the first term, then the nth term (an) can be calculated using the formula: \[ a_n = a \times r^{(n-1)} \]
This equation forms the basis for solving a wide range of problems involving GPs, such as finding specific terms, the sum of the first n terms, or in our exercise, understanding relationships between successive terms.

Successive Terms in Geometric Progressions

Successive terms in a geometric progression are related by the common ratio. In any GP, if you select any three successive terms, let's call them 'a', 'b', and 'c', then 'b' will be 'a' multiplied by the common ratio 'r', and 'c' will be 'b' multiplied by the same 'r'. This gives us two important formulas:\[ b = a \times r \]\[ c = b \times r = a \times r^2 \]
Using these formulas allows us to find the value of any term if we know the common ratio and the term before it. As in the exercise provided, we utilized these relationships to rewrite an inequality involving three successive terms in terms of the first term and the common ratio.

Solving Quadratic Inequalities

Quadratic inequalities can sometimes intimidate students, but they are just as manageable as quadratic equations. A quadratic inequality, like the one derived from our GP exercise, is an expression of the form \[ ax^2 + bx + c < 0 \] or \[ ax^2 + bx + c > 0, \]where 'a', 'b', and 'c' are constants. The solution to such inequalities involves the following steps: First, rearrange the inequality so that one side equals zero. Then, factor the quadratic (if possible) and use the sign of the factors to determine the intervals that satisfy the inequality.

The fundamental principle is that a quadratic expression (for example, \[ (r - p)(r - q) \]) will change signs whenever 'r' passes through one of the zeros (p or q). Thus, by factoring the quadratic and analyzing these sign changes, we can determine the range of values for which the inequality holds true. In the end, we obtain intervals for 'r' that satisfy the original inequality, as we did in steps 5 and 6 of the provided solution.