Problem 97 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Three numbers are in G.P. whose sum is 70 . If the extremes be each multiplied by 4 and the mean by 5 , they will be in A.P. Find the numbers.

Step-by-Step Solution

Answer

The three numbers in the geometric progression are -8, 2, and 26.

Step 1: Assign Variables

Let the three numbers be a, b and c (a being the first term, c being the third term, and b being the middle term).

Step 2: Write the Relations for G.P.

Since the three numbers are in G.P., their middle term is the geometric mean of the other two. So, b = sqrt(ab) or a * c = b^2. Also, a + b + c = 70.

Step 3: Write the Relations for A.P.

According to the problem, if we multiply the first and third numbers by 4 (4a and 4c) and the second number by 5 (5b), they will form an A.P. Hence, for the numbers in A.P.: 2 * (5b) = 4a + 4c

Step 4: Express a in terms of c

From the A.P. relation, we can write a in terms of c: 10b = 4(a+c) So, a = 2.5b - 0.5c

Step 5: Substitute the Expression for a in G.P. Relation

Substitute the expression for a from the previous step into the G.P. relation: (2.5b - 0.5c) * c = b^2

Step 6: Solve for c

Rewrite the G.P. relation equation in terms of c and solve it as a quadratic equation: 2.5c^2 - b*c - 2b^2 = 0 Now, substitute the sum of numbers (a+b+c = 70) in the equation: c = 70 - a - b Plug in the expression for a: c = 70 - (2.5b - 0.5c) - b Solve for c: c = 26

Step 7: Find a and b

Now that we have the value of c, we can find the values of a and b using the previously derived equations: a = 2.5b - 0.5c = 2.5b - 13 b = sqrt(a * c) = sqrt((2.5b - 13) * 26)

Step 8: Solve the Equation for b

Solve the equation for b to find its value: b^2 = (2.5b - 13) * 26 b^2 = 65b - 338 Solve to get the value of b: b = 2

Step 9: Find the Value of a

Now find the value of a using the previously obtained expression for a: a = 2.5b - 13 = 2.5 * 2 - 13 a = 5 - 13 = -8 The three numbers are -8, 2, and 26.

Key Concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, known as the common difference, to the previous term. This property makes AP a linear sequence. For example, if you start with the number 2 and add 3 each time, your AP will be 2, 5, 8, 11, and so on. The nth term of an AP can be found using the formula:
\( a_n = a_1 + (n-1)d \),
where \(a_n\) is the nth term, \(a_1\) is the first term, and \(d\) is the common difference. This formula is extremely helpful when dealing with problems where you need to find specific terms or the sum of an AP.
In the given problem, we transformed the terms of a geometric progression (GP) into an AP by multiplying them with constants. This resulted in solving for the variables to find the original terms of the GP.

Geometric Mean

The geometric mean is a type of mean or average, which indicates the central tendency or typical value of a set of numbers by using the product of their values. It differs from the arithmetic mean as it takes the root of the product of the numbers. Specifically, the geometric mean of two numbers, \(a\) and \(b\), is given by the square root of their product:
\( GM = \sqrt{a \cdot b} \).
In the context of geometric progressions, the middle term is often the geometric mean of the two extremes. This property is crucial as we used it in the problem to establish a relationship between the terms of the GP. This relationship helps transform the problem into solvable equations, linking geometric progressions with quadratic equations through the power of algebra.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, which means the highest exponent of the variable is 2. The standard form of a quadratic equation is:
\( ax^2 + bx + c = 0 \),
where \(a\), \(b\), and \(c\) are coefficients, and \(x\) represents the unknown variable. Quadratic equations are known for having zero, one, or two real solutions. These solutions can be found by factoring, completing the square, or using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
In the given exercise, solving for one of the terms of the GP resulted in a quadratic equation in terms of \(c\). By rearranging and using quadratic solving techniques, we determined the value of \(c\), and subsequently, the values of the other terms of the GP. Understanding how to manipulate and solve quadratic equations is essential in algebra and is frequently used in conjunction with series and progressions.