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The sequence s1, s2, s3,.....sn,...is such that Sn= [#permalink]

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26 Apr 2012, 06:14

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I don't know whether these problems have already been posted on the site, since I couldn't find the answers I will post them.

1) The sequence s_1, s_2, s_3, ..., s_n, ... is such that s_n=1/n - 1/n+1 for all integers \(n\geq 1\). If k is a positive Integer, is the sum of the first k terms of the sequence greater than [fraction]{9}{10}[/fraction]? (1) k > 10 (2) k < 19

2) In the sequence \(x_0, x_1, x_2, ..., x_n,\) each term from \(x_1 to x_k\)is 3 greater than the previous term, and each term from \(x_k+1 to x_n\)is less than the previous term, where n and k are positive integers and k< n. If \(x_0 = x_n = 0\) and if \(x_k = 15\), what is the value of n? A) 5 B) 6 C) 9 D) 10 E) 15

Please elaborate these problems as simple as possible! Thank you!
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Last edited by Bunuel on 26 Apr 2012, 06:54, edited 1 time in total.

I don't know whether these problems have already been posted on the site, since I couldn't find the answers I will post them.

1) The sequence \(s_1, s_2, s_3, ..., s_n, .., is such that [m]s_n=1/n - 1/n+1\) for all integers \(n\geq 1\). If k is a positive Integer, is the sum of the first k terms of the sequence greater than [fraction]{9}{10}[/fraction]? (1) k > 10 (2) k < 19

2) In the sequence \(x_0, x_1, x_2, ..., x_n,\) each term from \(x_1 to x_k\)is 3 greater than the previous term, and each term from \(x_k+1 to x_n\)is less than the previous term, where n and k are positive integers and k< n. If \(x_0 = x_n = 0\) and if \(x_k = 15\), what is the value of n? A) 5 B) 6 C) 9 D) 10 E) 15

Please elaborate these problems as simple as possible! Thank you!

The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - (1/(n+1)) for all integers n>=1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So: \(s_1=1-\frac{1}{2}\); \(s_2=\frac{1}{2}-\frac{1}{3}\); \(s_3=\frac{1}{3}-\frac{1}{4}\); ...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?

(1) k > 10. Sufficient. (2) k < 19. Not sufficient.

In the sequence \(x_0, \ x_1, \ x_2, \ ... \ x_n\), each term from \(x_1\) to \(x_k\) is 3 greater than the previous term, and each term from \(x_{k+1}\) to \(x_n\) is 3 less than the previous term, where \(n\) and \(k\) are positive integers and \(k<n\). If \(x_0=x_n=0\) and if \(x_k=15\), what is the value of \(n\)?

A.5 B. 6 C. 9 D. 10 E. 15

Probably the easiest way will be to write down all the terms in the sequence from \(x_0=0\) to \(x_n=0\). Note that each term from from \(x_0=0\) to \(x_k=15\) is 3 greater than the previous and each term from \(x_{k+1}\) to \(x_n\) is 3 less than the previous term:

So we'll have: \(x_0=0\), 3, 6, 9, 12, \(x_k=15\), 12, 9, 6, 3, \(x_n=0\). So we have 11 terms from \(x_0\) to \(x_n\) thus \(n=10\).

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