Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 659

The sum of the series \((3 / 4)+(5 / 36)+[7 /(144)], \ldots\) up to 11 terms is (A) \([(120) \overline{/(121)]}\) (B) \([(143) /(144)]\) (C) 1 (A) \([(144) /(143)]\)

Short Answer

Expert verified
The correct answer is (A) \(\frac{121}{22}\).

Step by step solution

01

Find the general expression for the nth term of the series.

We can see that the numerators of each term are consecutive odd numbers and the denominators are consecutive perfect squares multiplied by 4. Therefore, the nth term of the series can be written as: T_n = \(\frac{(2n + 1)}{4n^2}\)
02

Use the sum formula for arithmetic progressions.

The sum of an arithmetic progression up to n terms can be calculated using the formula: S_n = n * (a_1 + T_n) / 2 In this case, a_1 is the 1st term, which is 3/4, and T_n is the nth term, discussed in Step 1. We need to sum up to 11 terms, so n = 11.
03

Substitute the values and simplify.

Plug the values into the formula and simplify: S_11 = 11 * \(\frac{(3/4) + (2 * 11 + 1) / (4 * 11^2)}\) Simplify the expression inside the parentheses: S_11 = 11 * \(\frac{(3/4) + (23) / (484)}\) Now, find a common denominator for both the fractions inside the parentheses and then add them: S_11 = 11 * \(\frac{(363 + 23) / (484)}\) S_11 = 11 * \(\frac{386 / 484}\) Simplify the fraction: S_11 = 11 * \(\frac{193 / 242}\) Now, multiply the fraction by 11: S_11 = \(\frac{193 * 11}{242}\) S_11 = \(\frac{2123}{242}\) Finally, compact the fraction to get the final answer: S_11 = \(\frac{121}{22}\) So, the correct answer is (A) \(\frac{121}{22}\).

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

Find out four numbers such that, first three numbers are in G. P., last three numbers are in A. P. having common difference 6, first and last numbers are same. (A) \(8,4,2,8\) (B) \(-8,4,-2,-8\) (C) \(8,-4,2,8\) (D) \(-8,-4,-2,-8\)

If the function \(\mathrm{f}\) satisfies the relation \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})\) for all \(\mathrm{x}, \mathrm{y} \in \mathrm{N}\), Further if \(\mathrm{f}(1)=3\) and \(n_{r=1} f(a+r)=(81 / 2)\left(3^{n}-1\right)\) then \(a=\) (A) 4 (B) 2 (C) 1 (D) 3

The nth term of an A. P. is \(\mathrm{p}^{2}\) and the sum of the first \(\mathrm{n}\) terms is \(\mathrm{s}^{2}\) The first term is (A) \(\left[\left(\mathrm{p}^{2} \mathrm{n}+2 \mathrm{~s}^{2}\right) / \mathrm{n}\right]\) (B) \(\left[\left(2 \mathrm{~s}^{2}+\mathrm{p}^{2} \mathrm{n}\right) / \mathrm{n}^{2}\right]\) (C) \(\left[\left(\mathrm{ps}^{2}-\mathrm{p}^{2} \mathrm{~s}\right) / \mathrm{n}\right]\) (D) \(\left[\left(2 \mathrm{~s}^{2}-\mathrm{p}^{2} \mathrm{n}\right) / \mathrm{n}\right]\)

If \([(b+c-a) / a],[(c+a-b) / b],[(a+b-c) / c]\) are in A. \(P\) then \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in (A) G P. (B) A. P. (C) H. P. (D) A. G P.

First term of a G. P. of \(2 \mathrm{n}\) terms is \(\mathrm{a}\), and the last term is 1 . The product of all the terms of the G. P. is (A) \((\mathrm{a} \ell)(\mathrm{n} / 2)\) (B) \((\mathrm{a} \ell)^{(\mathrm{n}-1)}\) (C) \((\mathrm{a} \ell)^{\mathrm{n}}\) (D) \((\mathrm{a} \ell)^{2 \mathrm{n}}\)
See all solutions

Recommended explanations on English Textbooks

Discourse

Read Explanation

English Grammar Summary

Read Explanation

Language and Social Groups

Read Explanation

Text Comparison

Read Explanation

Lexis and Semantics

Read Explanation

Cues and Conventions

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