Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 670

If \(\mathrm{a}, 4, \mathrm{~b}\) are in \(\mathrm{A} . \mathrm{P}\). and \(\mathrm{a}, 2, \mathrm{~b}\) are in G. P. then \((1 / \mathrm{a}), 1\), \((1 / \mathrm{b})\) are in (A) G. P. (B) A. P. (C) H. P. (D) A. G. P.

Short Answer

Expert verified
(A) A.P.

Step by step solution

01

Use the A.P. condition on a, 4, b

Since a, 4, and b are in A.P., the average of a and b is equal to 4. We can write this as: \[(a + b)/2 = 4\]
02

Use the G.P. condition on a, 2, b

Since a, 2, and b are in G.P., the square of the middle term (2) is equal to the product of a and b. We can write this as: \[2^2 = a \cdot b\] Which simplifies to, \[4 = a \cdot b\]
03

Solve for a and b

We now have two equations: 1. \((a + b)/2 = 4\) 2. \(4 = a \cdot b\) Solve equation 1 for a: \[a = 8 - b\] Substitute the value of a in equation 2: \[4 = (8 - b) \cdot b\] This is a quadratic equation, so we can rewrite it as: \[b^2 - 8b + 4 = 0\] Solve for b: \(b = 4 \pm 2\sqrt{3}\) But as b is in G.P. with a, its value must be positive and |a| is larger in G.P.: \(b = 4 + 2\sqrt{3}\) Then we find the value of \(a = 8 - (4 + 2\sqrt{3})\) Which simplifies to, \(a = 4 - 2\sqrt{3}\)
04

Examine the sequence (1/a, 1, 1/b)

Now let's examine the sequence: \[\left(\frac{1}{a}, 1, \frac{1}{b}\right)\] Generating the new sequence: \[\left(\frac{1}{4 - 2\sqrt{3}}, 1, \frac{1}{4 + 2\sqrt{3}}\right)\] Now we can examine this sequence. If they are in A.P., G.P., or H.P., we can express the sequence relationship as one of the following: - A.P.: \((1 / a) + (1 / b) = 2\) - G.P.: \[(1 / a) \cdot (1 / b) = 1^2\] - H.P.: \[((1 / a) + (1 / b)) / 2 = 1 / 1\] For A.P., we have: \[\frac{1}{4 - 2\sqrt{3}} + \frac{1}{4 + 2\sqrt{3}} = 2\] This simplifies to: \[\frac{(4 + 2\sqrt{3}) + (4 - 2\sqrt{3})}{16 - 12} = 2\] Which is true. So, \((1 / a), 1, \)and \( (1 / b)\) are in A.P. Therefore, the correct answer is: (A) A.P.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\tan ^{-1}(1 / 3)+\tan ^{-1}(1 / 7)+\tan ^{-1}(1 / 13)+\ldots+\tan ^{-1}[1 /(9703)]\) (A) \(\begin{array}{lll}\text { (B) }(\pi / 6) & \text { (C) }(\pi / 3) & \text { (D) } \tan ^{-1}(0.98)\end{array}\)

If the \(\mathrm{H}\). M. of a and \(\mathrm{c}\) is \(\mathrm{b}\), G. \(\mathrm{M}\). of \(\mathrm{b}\) and \(\mathrm{d}\) is \(\mathrm{c}\) and \(\mathrm{A} . \mathrm{M} .\) of \(c\) and e is \(\mathrm{d}\), then the G. M. of a and e is (A) \(b\) (B) \(c\) (C) \(\mathrm{d}\) (D) ae

If \(\left[\left(a^{n+1}+b^{n+1}\right) /\left(a^{n}+b^{n}\right)\right]\) is \(H .\) M. of \(a\) and \(b\) then \(n=\) \(\left(a, b \in R^{+} \quad a \neq b\right)\) (A) 0 (B) - 1 (C) \(-(1 / 2)\) (D) \(-2\)

If a set \(\mathrm{A}=\\{3,7,11, \ldots, 407\\}\) and a set \(\mathrm{B}=\\{2,9,16, \ldots, 709\\}\) then \(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\) (A) 13 (B) 14 (C) 15 (D) 16

If A is the A. M. between a and b, then \([(A-2 b) /(A-a)]+[(A-2 a) /(A-b)]=\) (A) \(-8\) (B) 2 (C) 4 (D) \(-4\)
See all solutions

Recommended explanations on English Textbooks

Essay Plan

Read Explanation

Single Paragraph Essay

Read Explanation

Multiple Choice Questions

Read Explanation

Morphology

Read Explanation

Summary Text

Read Explanation

Sociolinguistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free