SATHEE: Sequences and Series 2 Question 3

Sequences and Series 2 Question 3

3.

For $x \in R$ , let $[x]$ denote the greatest integer $\leq x$ , then the sum of the series $[- \frac{1}{3}] + [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + \dots + [- \frac{1}{3} - \frac{99}{100}]$ is

(2019 Main, 12 April I)

(a) $- 153$

(b) $- 133$

(d) $- 135$

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Answer:

Correct Answer: 3. (b)

Solution:

Given series is

$[- \frac{1}{3}] + [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + \dots \dots + [- \frac{1}{3} - \frac{99}{100}]$

[where, $[x]$ denotes the greatest integer $\leq x]$

Now,

$[- \frac{1}{3}], [- \frac{1}{3} - \frac{1}{100}], [- \frac{1}{3} - \frac{2}{100}], \dots + [- \frac{1}{3} - \frac{66}{100}]$

all the term have value -1

and $[- \frac{1}{3} - \frac{67}{100}], [- \frac{1}{3} - \frac{68}{100}], \dots, [- \frac{1}{3} - \frac{99}{100}]$ all the term have value -2 .

So, $[- \frac{1}{3}] + [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + \dots + [- \frac{1}{3} - \frac{66}{100}]$

$= - 1 - 1 - 1 - 1 \dots 67$ times.

$= (- 1) \times 67 = - 67$

and $[- \frac{1}{3} - \frac{67}{100}] + [- \frac{1}{3} - \frac{68}{100}] + \dots + [- \frac{1}{3} - \frac{99}{100}]$

$= - 2 - 2 - 2 - 2 \dots 33$ times

$= (- 2) \times 33 = - 66$

$∴ [- \frac{1}{3}] + [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + \dots + [- \frac{1}{3} - \frac{99}{100}]$

$= (- 67) + (- 66) = - 133$ .

Alternate Solution

$∵ [- x] = - [x] - 1$ , if $x \notin$ Integer,

and $[x] + [x + \frac{1}{n}] + [x + \frac{2}{n}] + \dots + [x + \frac{n - 1}{n}] = [n x]$ ,

$n \in N$ .

So given series

$\begin{aligned} [- \frac{1}{3}] + [- \frac{1}{3} - \frac{1}{100}] + [- \frac{1}{3} - \frac{2}{100}] + \dots \dots + [\frac{- 1}{3} - \frac{99}{100}] \\ = [- \frac{1}{3} - 1] + [- \frac{1}{3} + \frac{1}{100} - 1] \\ + [- \frac{1}{3} + \frac{2}{100} - 1] + \dots + [- \frac{1}{3} + \frac{99}{100} - 1] \\ = (- 1) \times 100 - [\frac{1}{3} \times 100] = - 100 - 33 = - 133 . \end{aligned}$

Sequences and Series 2 Question 3

3.

Answer:

Alternate Solution

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Sequences and Series 2 Question 3

3.

Answer:

Alternate Solution

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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