Problem 69 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(a, b\) and \(c\) be the \(7^{\text {th }}, 11^{\text {th }}\) and \(13^{\text {th }}\) terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then \(\frac{a}{c}\) is equal to: [Jan. 09,2019 (II)] (a) 2 (b) \(\frac{1}{2}\) (c) \(\frac{7}{13}\) (d) 4

Step-by-Step Solution

Answer

The ratio \(\frac{a}{c}\) is 4.

Step 1: Understand the Definitions

We have two sequences - an arithmetic progression (A.P.) and a geometric progression (G.P.). In an A.P., the difference between consecutive terms is constant, called the common difference, denoted usually by \(d\). In a G.P., the ratio between consecutive terms is constant, called the common ratio, denoted as \(r\).

Step 2: Express Terms of A.P.

In an A.P. with the first term \(a_1\) and common difference \(d\), the \(n^{th}\) term is expressed as \(a_n = a_1 + (n-1)d\). For the 7th, 11th, and 13th terms, we have: \(a = a_1 + 6d\), \(b = a_1 + 10d\), and \(c = a_1 + 12d\).

Step 3: Set up the G.P. Condition

According to the problem, the terms \(a\), \(b\), and \(c\) must also form a G.P. This implies that the ratio between the terms is constant. So, \(\frac{b}{a} = \frac{c}{b}\).

Step 4: Simplify G.P. Condition

Substitute the A.P. terms into the G.P. condition: \(\frac{a_1 + 10d}{a_1 + 6d} = \frac{a_1 + 12d}{a_1 + 10d}\). Cross-multiplying gives: \((a_1 + 10d)^2 = (a_1 + 6d)(a_1 + 12d)\).

Step 5: Expand and Solve Equation

Expand both sides: LHS becomes \((a_1^2 + 20a_1d + 100d^2)\) and RHS becomes \(a_1^2 + 18a_1d + 72d^2\). Set them equal: \(a_1^2 + 20a_1d + 100d^2 = a_1^2 + 18a_1d + 72d^2\).

Step 6: Simplify the Equation Further

Cancel \(a_1^2\) from both sides: \(20a_1d + 100d^2 = 18a_1d + 72d^2\). Rearrange to find: \(2a_1d = -28d^2\). Dividing throughout by \(2d\) (assuming \(d eq 0\)), gives \(a_1 = -14d\).

Step 7: Calculate \(\frac{a}{c}\)

Substitute \(a_1 = -14d\) into \(a = a_1 + 6d\) and \(c = a_1 + 12d\), giving \(a = -14d + 6d = -8d\) and \(c = -14d + 12d = -2d\). Therefore, \(\frac{a}{c} = \frac{-8d}{-2d} = 4\).

Step 8: Conclusion

Our computations show that \(\frac{a}{c} = 4\). This corresponds to option (d).

Key Concepts

Geometric Progression

A geometric progression (G.P.) is a sequence where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number known as the common ratio. For example, in the sequence 2, 4, 8, 16..., the common ratio is 2, since each term is the previous term multiplied by 2.
To express this mathematically, if the first term of the G.P. is denoted by \(a_1\) and the common ratio by \(r\), then the \(n^{th}\) term can be represented as:

• Second term: \(a_2 = a_1 \times r = a_1 \times r\)
• Third term: \(a_3 = a_2 \times r = a_1 \times r^2\)
• \(n^{th}\) term: \(a_n = a_1 \times r^{(n-1)}\)

Understanding G.P. is crucial as it shows how exponential growth can occur, such as in the case of compound interest or population growth. In the problem, the terms \(a, b, c\) form a G.P., meaning their ratios are constant, emphasizing the predictability and regularity of a G.P.

Common Difference

The common difference is a key feature of arithmetic progression (A.P.). It is the difference between any two successive terms of an A.P., and it remains constant throughout the sequence. For example, in the A.P. 3, 6, 9, 12..., the common difference is 3, as each term differs from the previous one by this constant amount.
If we denote the first term by \(a_1\) and the common difference by \(d\), the formula for the \(n^{th}\) term in an A.P. is given by:

• \(n^{th}\) term: \(a_n = a_1 + (n-1)d\)

In the exercise, terms \(a, b, c\) are the 7th, 11th, and 13th terms of an A.P., respectively. This consistent difference forms the "backbone" of any arithmetic sequence, ensuring the increments between terms do not change.
Recognizing this consistent pattern allows us to establish relationships between terms and solve problems involving finding unknown values within the progression.

Common Ratio

The concept of a common ratio is fundamental to understanding a geometric progression (G.P.). In any G.P., the common ratio \(r\) is the factor by which we multiply a term to get the next term in the sequence. For instance, in the sequence 5, 10, 20, 40..., the common ratio is 2. Each term is 2 times the preceding one.
In this sequence:

• First term \(a_1\), second term: \(a_2 = a_1 \times r\)
• Third term: \(a_3 = a_2 \times r = a_1 \times r^2\)
• Similarly, \(n^{th}\) term: \(a_n = a_1 \times r^{(n-1)}\)

In the given problem, the terms \(a, b, c\) must satisfy the condition of a G.P. This means \(\frac{b}{a} = \frac{c}{b}\), illustrating how the common ratio governs the relationship between any three consecutive terms.
This fixed multiplicative rule enables us to solve and predict values at various positions within a G.P., adding an exponential twist to sequences compared to arithmetic progressions, which are additive in nature.