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Chapter 6: Problem 13

If \(2^{\text {nd }}, 5^{\text {th }}\), and \(9^{\text {th }}\) terms of an A.P. are in G.P., then the sum of all possible ratios of the first term to the common difference of A.P. (1) \(\frac{9}{8}\) (2) 9 (3) 8 (4) 1

Short Answer

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Step by step solution

01

Identify the terms in the A.P.

In an Arithmetic Progression (A.P.), the general form of the nth term is given by: \[ a_n = a + (n-1)d \]where \(a\) is the first term and \(d\) is the common difference. The given terms are the 2nd, 5th, and 9th terms of the A.P. Thus, we have:\[ a_2 = a + d, \,a_5 = a + 4d, \,a_9 = a + 8d \]
02

Set up the condition for G.P.

Since the 2nd, 5th, and 9th terms of the A.P. are in Geometric Progression (G.P.), the condition for G.P. is:\[ a_5^2 = a_2 \, a_9 \]Substituting the values:\[ (a + 4d)^2 = (a + d)(a + 8d) \]
03

Expand and simplify

Expand both sides of the equation:\[ (a + 4d)^2 = a^2 + 8ad + 16d^2 \]and\[ (a + d)(a + 8d) = a^2 + 8ad + ad + 8d^2 = a^2 + 9ad + 8d^2 \]Set the equations equal to solve for \(a\) and \(d\):\[ a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2 \]
04

Solve for the ratio \( \frac{a}{d} \)

Subtract \( a^2 \) from both sides:\[ 8ad + 16d^2 = 9ad + 8d^2 \]Rearrange to isolate terms involving \(a\) and \(d\):\[ 16d^2 - 8d^2 = 9ad - 8ad \]Simplify:\[ 8d^2 = ad \]Solve for ratio \( \frac{a}{d} \):\[ a = 8d \text{ or equivalently } \frac{a}{d} = 8 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

arithmetic progression
An Arithmetic Progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the common difference, denoted by \(d\). The formula to find the nth term of an A.P. is given by: \( a_n = a + (n-1)d \).
Here, \(a\) represents the first term of the sequence. For example, in the sequence 2, 5, 8, 11,...
  • The first term \(a\) is 2.
  • The common difference \(d\) is 3 because each term is 3 more than the previous term.
  • The nth term can be calculated using \( a_n = 2 + (n-1) \times 3 \).
In the given exercise, we are provided with the 2nd, 5th, and 9th terms of an A.P., which are expressed as:
\( a_2 = a + d, \)
\( a_5 = a + 4d, \)
\( a_9 = a + 8d \).
Understanding this structure helps us set up and solve problems involving arithmetic progressions.
geometric progression
A Geometric Progression (G.P.) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio, denoted by \(r\). The general term of a G.P. can be written as:
\(a_n = ar^{n-1}\).
Here, \(a\) is the first term. For example, in the sequence 3, 6, 12, 24,...
  • The first term \(a\) is 3.
  • The common ratio \(r\) is 2.
  • The nth term can be calculated using \( a_n = 3 \times 2^{(n-1)} \).
In the given problem, the 2nd, 5th, and 9th terms of the A.P. are in G.P. Therefore, the relationship is:
\(a_5^2 = a_2 \times a_9\).
Substituting the terms of the A.P., we have:
\((a + 4d)^2 = (a + d) \times (a + 8d)\).
Solving this equation provides insight into how G.P. relates to the terms of an A.P.
ratio of terms
The ratio of terms, specifically the ratio of the first term to the common difference in an arithmetic progression, can be a crucial factor in solving equations involving both A.P. and G.P. From the given exercise, we arrive at the equation:
\( (a + 4d)^2 = (a + d)(a + 8d) \).
By expanding and simplifying this equation:
\( (a + 4d)^2 = a^2 + 8ad + 16d^2 \) and
\( (a + d)(a + 8d) = a^2 + 9ad + 8d^2 \),
we equate the two:
\( a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2 \).
By solving for \(a\) and \(d\), we simplify to:
\(8d^2 = ad\),
resulting in the ratio:
\( \frac{a}{d} = 8 \).
This ratio indicates that the first term \(a\) is 8 times the common difference \(d\). It's important to grasp this concept to understand how different terms in progressions relate to each other.

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