Problem 96 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If the product of three numbers in G.P. is 216 and sum of the products taken in pairs is 156 , find the numbers.

Step-by-Step Solution

Answer

The three numbers in G.P. are 1, 6, and 36.

Step 1: Understand the properties of geometric progression

A geometric progression (G.P.) is a sequence of numbers where each term is found by multiplying the previous term by a constant called the common ratio. Suppose we have three numbers in a G.P., a, ar, and ar^2 where r is the common ratio.

Step 2: Set up the equations based on given information

We are given that product of the three numbers is 216, and the sum of products taken in pairs is 156. Product equation: a * ar * ar^2 = 216 Sum of products taken in pairs equation: (a * ar) + (a * ar^2) + (ar * ar^2) = 156

Step 3: Simplify the product equation

Simplify the product equation: a * a * r * a * r^2 = a^3 * r^3 = 216

Step 4: Simplify the sum of products taken in pairs equation

Simplify the sum of products taken in pairs equation: a^2r + a^2r^2 + a^2r^3 = 156

Step 5: Solve the product equation for a variable

Let's solve the product equation for a. We notice that the expression on the left-hand side is cubic. To simplify it and isolate a, we can take the cube root of both sides: \(a * r = \sqrt[3]{216} = 6\)

Step 6: Substitute the expression for ar

Substitute the expression for ar from Step 5 into the sum of products taken in pairs equation from Step 4: 6a + 6ar + 6a^2r^2 = 156

Step 7: Divide the entire equation by 6

Divide the entire equation by 6 to simplify: a + ar + a^2r^2 = 26

Step 8: Substituting the expression for ar from Step 5

We know, ar = 6 so, a + 6 + a^2 * 6^2 = 26

Step 9: Solve the quadratic equation for a

Following equation becomes, a^2 * 36 + a - 20 = 0 Solve for a: \(a = \frac{-1 \pm\sqrt{1^2- 4 * 36 * (-20)}}{2*36}\) The two possible values for a are 1 and -20/3.

Step 10: Determine the values of r and the numbers in G.P.

Using the value of a in the expression ar = 6, we can find the corresponding values of r and the numbers in G.P. Case 1: a = 1 ar = 6 => r = 6 Numbers in G.P. are 1, 6, and 36. Case 2: a = -20/3 ar = 6 => r = -9/10 Numbers in G.P. are -20/3, 6, and -54/5. We can rule out this case as we cannot have a mixed kind G.P with both positive and negative numbers. Thus we are left with the first case. Therefore, the three numbers in G.P. are 1, 6, and 36.

Key Concepts

Product of Terms in Geometric Progression

In a geometric progression (G.P.), each term can be found by multiplying the previous term by a constant, known as the common ratio. In our exercise, we've identified three terms of a G.P. as \(a, ar,\) and \(ar^2\), where \(r\) is the common ratio. A critical aspect of this exercise is understanding the nature of these terms and their relationship. We are given the product of these terms, which is the multiplication of all three: \(a \times ar \times ar^2\). Simplifying this yields \(a^3 r^3\), a compact form revealing the relationship between the first term \(a\), the common ratio \(r\), and the product of the sequence.

In our specific case, the equation becomes \(a^3 r^3 = 216\). Solving this equation allows us to find the possible values for \(a\) and \(r\), an essential step to finding our numbers in the series. Taking the cube root of both sides provides a simplified form, \(ar = 6\). This result will play a crucial role in unlocking the final answers to the problem.

Sum of Products of Pairs

The sum of the products taken in pairs is an additional condition provided in the problem, meant to guide us to the precise values of the terms in a G.P. Specifically, this involves calculating \((a \times ar) + (a \times ar^2) + (ar \times ar^2)\). These calculations translate to three specific products: \(a^2r, a^2r^2,\) and \(a^2r^3\).

In the context of our exercise, this sum is equal to 156. By substituting \(ar = 6\), the equation reduces to \(6a + 6ar + 6a^2r^2 = 156\). Dividing through by 6 cleans up the equation significantly, simplifying to \(a + ar + a^2r^2 = 26\). The handling of these pair products is crucial to forming the quadratic equation to further solve for \(a\), a vital bridge to derive the correct sequence values.

Solving Quadratic Equations

The final leap in our exercise involves solving a quadratic equation derived from earlier simplifications. After reducing our equation to \(a + 6 + 36a^2 = 26\) using previous results, we rearrange to get the quadratic equation: \(36a^2 + a - 20 = 0\). This form is standard for applying the quadratic formula, given by \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Substitute the coefficients from our equation: \(a = 36, b = 1,\) and \(c = -20\). Calculating the discriminant \(\Delta = b^2 - 4ac = 1 + 2880\) reveals two solutions. Implementing these into the formula, we find \(a = 1\) or \(a = -\frac{20}{3}\). After verifying the consistency with our G.P., positively validated by the requirement for non-negated terms, we confirm our initial hypothesis of \(a = 1\). This culmination illustrates the interconnectedness of solving quadratic equations to confirm values within sequences.