Problem 108 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Three numbers from a G.P. If the 3rd term is decreased by 64 , then the numbers thus obtained will constitute an A.P. If the second term of this A.P. is decreased by 8 , a G.P. will be formed again. Determine the numbers.

Step-by-Step Solution

Answer

The three numbers are 4, 8, 64.

Step 1: Definition of a G.P.

Denote the three numbers of G.P. as \(a\), \(ar\), \(ar^2\). This follows the definition of a geometric sequence, where each term is the previous term multiplied by \emph{r}, the common ratio.

Step 2: Transition to an A.P.

From the problem, we know that if the third term of this sequence is decreased by 64, we get an A.P. So the terms are \(a\), \(ar\), \(ar^2 - 64\). In an arithmetic sequence, the difference between consecutive terms is consistent. Therefore, \(ar - a = ar^2 - 64 - ar\). Simplifying this gives \(2ar - a = ar^2 - 64\).

Step 3: Return to a G.P.

It is stated that if the second term of the A.P. is decreased by 8, the sequence forms a G.P. again. Therefore, the terms of this sequence are \(a\), \(ar - 8\), \(ar^2 - 64\). In a geometric sequence, the ratio of consecutive terms is consistent. This gives the equation \((ar - 8) / a = (ar^2 - 64) / (ar - 8)\). Simplifying this gives \(ar^2 - 8ar = ar^2 - 64\). This simplifies to \(8ar = 64\), from which \(ar = 8\).

Step 4: Find all the terms

Substitute \(ar = 8\) into the equation from Step 2 to find \(a = 4\). Substituting \(a\) and \(ar\) into \(ar^2\) gives the third term as \(ar^2 = 64\). Therefore, the three numbers are 4, 8, 64.

Key Concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is often referred to as the common difference, denoted as 'd'. The general form of an arithmetic progression is \(a, a+d, a+2d, a+3d, \ldots\), where \(a\) is the first term, and 'd' is the common difference.

To understand an AP, it's essential to grasp how changes to the terms can affect the progression. In the context of the exercise, a decrease in the third term of a geometric progression (GP) by a known amount leads to the creation of an AP. This information allows us to set up an equation based on the common difference of the AP, which is critical in finding the actual terms of the sequence.

Using this concept, we discern that the uniform increase from one term to the next in an AP is the stepping stone to transitioning among different types of sequences, such as moving from a GP to an AP and back again.

Sequence Transition

A sequence transition refers to the conversion from one type of sequence to another. In certain problems, as exemplified by the original exercise, a sequence can transform from a geometric progression to an arithmetic progression and vice versa. The key is to identify the conditions or modifications to the terms that lead to this transition.

For instance, when dealing with a GP, subtracting a constant from one term may alter the constant ratio that defines the GP, potentially resulting in an AP, where a common difference is instead the defining characteristic. Conversely, a modification to an AP's term may restore the constant ratio characteristic, turning it back into a GP.

Applying the Concept

In the exercise, we observed that by decreasing the third term of the GP by 64, an AP was formed. Further, reducing the second term of this new AP by 8 reverted the sequence to a GP. Understanding how these transitions occur and setting up the resulting equations are pivotal in solving problems that involve changing sequences.

Geometric Sequence Equations

The equations governing a geometric sequence are pivotal to solving problems related to GPs. A geometric sequence is defined by its common ratio, denoted by 'r', which is the factor by which we multiply one term to get the next. The general formula for the n-th term of a GP is \(a_n = a \times r^{(n-1)}\), where \(a\) is the first term and 'n' is the term's position in the sequence.

Diving deeper into the equations, we understand that manipulating a GP's terms can result in various outcomes. In the given exercise, specific values were subtracted from the terms to trigger transitions between sequences. Through setting up equations that reflect these manipulations, we're able to solve for the unknown values in the sequence. These equations are essential for understanding the relationships between consecutive terms and provide a methodical approach to navigating complex sequence transitions.

By mastering how to derive and solve these equations, students can tackle a broad range of problems involving sequences, including those that require identifying and transitioning between arithmetic and geometric progressions.