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10 Sep 2019, 11:34
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Can the below be solved by taking the sum of the first (m1) and last (m10) number divided by the total numbers (10)?
For every integer m from 1 to 10 inclusive, the mthmth term of a certain sequence is given by (−1)(m+1)∗(12)m(−1)(m+1)∗(12)m. If T is the sum of the first 10 terms in the sequence, then T is:
(A) greater than 2
(B) between 1 and 2
(C) between 0.5 and 1
(D) between 0.25 and 0.5
(E) less than 0.25
There is an article from GMAT Economist that explains (but I can't post the article), can this be applied?
Example:
Applying the rules to find the sum of the sequence
How do we apply these useful rules to this question?
First, calculate the average of the first and last terms.
The first term is the sum of 1, 2 and 3 = 6
The last term is the sum of 99, 100 and 101 = 300
The average of the first and last terms = (6 + 300) / 2 = 306 / 2 = 153
Second, multiply the average by the number of terms.
There are 99 terms
Therefore, the answer to our question is 153 x 99
For every integer m from 1 to 10 inclusive, the mthmth term of a certain sequence is given by (−1)(m+1)∗(12)m(−1)(m+1)∗(12)m. If T is the sum of the first 10 terms in the sequence, then T is:
(A) greater than 2
(B) between 1 and 2
(C) between 0.5 and 1
(D) between 0.25 and 0.5
(E) less than 0.25
There is an article from GMAT Economist that explains (but I can't post the article), can this be applied?
Example:
Applying the rules to find the sum of the sequence
How do we apply these useful rules to this question?
First, calculate the average of the first and last terms.
The first term is the sum of 1, 2 and 3 = 6
The last term is the sum of 99, 100 and 101 = 300
The average of the first and last terms = (6 + 300) / 2 = 306 / 2 = 153
Second, multiply the average by the number of terms.
There are 99 terms
Therefore, the answer to our question is 153 x 99
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10 Sep 2019, 11:37
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Discussed here: for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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