Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 643

If the first, second and last terms of an A. P. are a, b and \(3 a\) respectively, the sum of the series is (A) \(\left[\left(4 a^{2}\right) /(b-a)\right]\) (B) \(\left[\left(2 a^{2}+2 a b\right) /(b-a)\right]\) (C) \(\left[\left(2 a b+a^{2}\right) /(b-a)\right]\) (D) \(\left[\left(2 a^{2}-2 a b\right) /(a-b)\right]\)

Short Answer

Expert verified
The short answer is: \(\textbf{(B)} \left[\left(2 a^{2}+2 a b\right) /(b-a)\right]\).

Step by step solution

01

Identify given values and find the common difference

Given: First term, \(a\), Second term, \(b\), and Last term, \(3a\). The common difference 'd' can be found by subtracting the first term from the second term. So, \(d = b - a\).
02

Find the nth term formula with the given last term

Using the formula for the nth term and the given last term: \(T_{n} = a + (n - 1) * d = 3a\). Now, replace \(d\) with the previously found value: \(b - a\). So, \(a + (n - 1) * (b - a) = 3a\).
03

Find the value of n

From the equation in Step 2, \(a + (n - 1) * (b - a) = 3a\), We can now find the value of n by solving for n: \((n - 1) * (b - a) = 2a\). Divide both sides by (b - a): \(n - 1 = \frac{2a}{b - a}\). Therefore, \(n = \frac{2a}{b - a} + 1\).
04

Find the sum of the series

Now we have the values of a, \(T_{n}\) (which is \(3a\)), and n. We can use the formula for the sum of an A.P. series to find the sum: \(S_{n} = \frac{n}{2} * (a + T_{n})\), where \(n = \frac{2a}{b - a} + 1\) and \(T_{n} = 3a\). Substitute these values into the formula: \(S_{n} = \frac{\left(\frac{2a}{b - a} + 1\right)}{2} * (a + 3a)\). Simplify this expression: \(S_{n} = \frac{1}{2} * \left(\frac{2a + b - a}{b - a}\right) * 4a\), \(S_{n} = \frac{(4a^{2} + 4ab - 2a^{2})}{(b - a)}\). Simplify further: \(S_{n} = \frac{2a^{2} + 4ab}{b - a}\). Comparing this expression with the given options, we find that the answer is: (B) \(\left[\left(2 a^{2}+2 a b\right) /(b-a)\right]\).

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

Sum of products of first n natural numbers taken two at a time is (A) \(\left[\left\\{\overline{ \left.n\left(n^{2}-1\right)(3 n+2)\right\\} /(24)}\right]\right.\) (B) \(\left[\left\\{n(n+1)^{2}(3 n+2)\right\\} /(72)\right]\) (C) \(\left[\left\\{n^{2}(n+1)(3 n+2)\right\\} /(48)\right]\) (D) \([\\{n(n+1)(n+2)(3 n+2)\\} /(96)]\)

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) \(-2\)

For an A. P., \(S_{100}=3 S_{50}\). The value of \(S_{150}: S_{50}=\) (A) 8 (B) 3 (C) 6 (D) 10

If the \(1^{\text {st }}\) term and common ratio of a G. P. are 1 and 2 respectively then \(\mathrm{s}_{1}+\mathrm{s}_{3}+\mathrm{s}_{5}+\ldots+\mathrm{s}_{2 \mathrm{n}-1}=\) (A) \((1 / 3)\left(2^{2 n}-5 n+4\right)\) (B) \((1 / 3)\left(2^{2 n+1}-5 n\right)\) (C) \((1 / 3)\left(2^{2 n+1}-3 n-2\right)\) (D) \((1 / 3)\left(2^{2 n+1}-5 n^{2}\right)\)

If the sum of the roots of the equation \(a x^{2}+b x+c=0\) is equal to the sum of the squares of their reciprocals then \(\mathrm{bc}^{2}\), \(\mathrm{ca}^{2}, \mathrm{ab}^{2}\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G. P.
See all solutions

Recommended explanations on English Textbooks

Essay Plan

Read Explanation

Listening and Speaking

Read Explanation

Pragmatics

Read Explanation

Rhetoric

Read Explanation

Lexis and Semantics

Read Explanation

Free Response Essay

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