SATHEE: Sequences and Series 4 Question 6

Sequences and Series 4 Question 6

6.

The sum of first 20 terms of the sequence $0.7, 0.77, 0.777, \dots$ , is

(a) $\frac{7}{81} (179 - 10^{- 20})$

(b) $\frac{7}{9} (99 - 10^{- 20})$

(d) $\frac{7}{9} (99 + 10^{- 20})$

(2013 Main)

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Answer:

Correct Answer: 6. (c)

Solution:

Let $S = 0.7 + 0.77 + 0.777 + \dots$

$= \frac{7}{10} + \frac{77}{10^{2}} + \frac{777}{10^{3}} + \dots$ upto 20 terms

$= 7 [\frac{1}{10} + \frac{11}{10^{2}} + \frac{111}{10^{3}} + \dots$ upto 20 terms]

$= \frac{7}{9} [\frac{9}{10} + \frac{99}{100} + \frac{999}{1000} + \dots$ upto 20 terms]

$= \frac{7}{9} [(1 - \frac{1}{10}) + (1 - \frac{1}{10^{2}}) + (1 - \frac{1}{10^{3}})$ +… +upto 20 terms]

$= \frac{7}{9} [(1 + 1 + \dots + upto 20 terms)$

$- \frac{1}{10} + \frac{1}{10^{2}} + \frac{1}{10^{3}} + \dots + upto 20 terms]$

$= \frac{7}{9} [20 - \frac{\frac{1}{10} 1 - \frac{1}{10}}{1 - \frac{1}{10}}]$

$[\sum_{i = 1}^{20} = 20 and sum of n terms of GP , S_{n} = \frac{a (1 - r^{n})}{1 - r} when (r < 1)]$

$= \frac{7}{9} [\frac{179}{9} + \frac{1}{9} (\frac{1}{10})^{20}]$

$= \frac{7}{81} [179 + (10)^{- 20}]$

Sequences and Series 4 Question 6

6.

Answer:

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Sequences and Series 4 Question 6

6.

Answer:

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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