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11 Mar 2020, 02:38
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:31) correct
58%
(02:37)
wrong
based on 149
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What is the sum of the first 20 term of the sequence 0.7, 0.77, 0.777, … ?
A. \(\frac{7}{81} (179 − 10^{− 20} )\)
B. \(\frac{7}{9} (99 − 10^{− 20} )\)
C. \(\frac{7}{81} (179 + 10^{− 20})\)
D. \(\frac{7}{9} (99 + 10^{− 20})\)
E. \(\frac{7}{9} (99 + 10^{− 20})\)
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{7}{81} (179 − 10^{− 20} )\)
B. \(\frac{7}{9} (99 − 10^{− 20} )\)
C. \(\frac{7}{81} (179 + 10^{− 20})\)
D. \(\frac{7}{9} (99 + 10^{− 20})\)
E. \(\frac{7}{9} (99 + 10^{− 20})\)
Are You Up For the Challenge: 700 Level Questions
11 Mar 2020, 03:17
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0.7+0.77+0.777.....
= 7[(1/10)+(11/100)+(111/1000).......20 terms]
= 7/9[(9/10)+(99/100)+(999/1000)....20 terms]
= 7/9[(1-1/10)+(1-1/100)+(1-1/1000)......20 terms]
= 7/9[20- {(1/10)+(1/100)+(1/1000)+.....20 terms]
=7/9[20- {(1/10)(1-(1/10)^20)/9/10}
= 7/9[20- (1/9) +(1/9)*10^{-20}]
= 7/9[(179/9) + (1/9)*10^{-20}]
= 7/81[179+10^{-20}]
= 7[(1/10)+(11/100)+(111/1000).......20 terms]
= 7/9[(9/10)+(99/100)+(999/1000)....20 terms]
= 7/9[(1-1/10)+(1-1/100)+(1-1/1000)......20 terms]
= 7/9[20- {(1/10)+(1/100)+(1/1000)+.....20 terms]
=7/9[20- {(1/10)(1-(1/10)^20)/9/10}
= 7/9[20- (1/9) +(1/9)*10^{-20}]
= 7/9[(179/9) + (1/9)*10^{-20}]
= 7/81[179+10^{-20}]
Bunuel
11 Mar 2020, 03:41
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What is the sum of the first \(20\) term of the sequence \(0.7\), \(0.77\), \(0.777\), … ?
\(\frac{7}{10}+ \frac{77}{100}+ \frac{777}{1000}+..=\)
\(\frac{7}{9}*(\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+.. )=\)
\(\frac{7}{9}*(1- \frac{1}{10}+ 1-\frac{1}{100} + 1- \frac{1}{1000}+... )\)
\(\frac{7}{9}*( 20 - (\frac{1}{10}+\frac{1}{100}+ \frac{1}{1000}+...\frac{1}{10^{20}}))\)
--> \(\frac{1}{10}+\frac{1}{100}+ \frac{1}{1000}+...\frac{1}{10^{20}}\) is a geometric sequence.
Sum = \(\frac{(\frac{1}{10})*( 1- \frac{1}{10^{20}})}{(1-\frac{1}{10})}\) = \(\frac{(\frac{1}{10})*( 1- \frac{1}{10^{20}})}{\frac{9}{10}}\) = \(\frac{( 1- \frac{1}{10^{20}})}{9}\),
--> \(\frac{7}{9}*( 20 - \frac{( (1- \frac{1}{10^{20}})}{9})\),
= \((\frac{7}{9})*\frac{(180 -1+ \frac{1}{10^{20}})}{9}\)
\(\frac{7}{81}*( 179 + \frac{1}{10^{20}})\)
The answer is C.
\(\frac{7}{10}+ \frac{77}{100}+ \frac{777}{1000}+..=\)
\(\frac{7}{9}*(\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+.. )=\)
\(\frac{7}{9}*(1- \frac{1}{10}+ 1-\frac{1}{100} + 1- \frac{1}{1000}+... )\)
\(\frac{7}{9}*( 20 - (\frac{1}{10}+\frac{1}{100}+ \frac{1}{1000}+...\frac{1}{10^{20}}))\)
--> \(\frac{1}{10}+\frac{1}{100}+ \frac{1}{1000}+...\frac{1}{10^{20}}\) is a geometric sequence.
Sum = \(\frac{(\frac{1}{10})*( 1- \frac{1}{10^{20}})}{(1-\frac{1}{10})}\) = \(\frac{(\frac{1}{10})*( 1- \frac{1}{10^{20}})}{\frac{9}{10}}\) = \(\frac{( 1- \frac{1}{10^{20}})}{9}\),
--> \(\frac{7}{9}*( 20 - \frac{( (1- \frac{1}{10^{20}})}{9})\),
= \((\frac{7}{9})*\frac{(180 -1+ \frac{1}{10^{20}})}{9}\)
\(\frac{7}{81}*( 179 + \frac{1}{10^{20}})\)
The answer is C.
