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Find the sum of the first 15 terms of the series whose nth

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Find the sum of the first 15 terms of the series whose nth [#permalink] New post 14 Oct 2013, 08:34
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84% (02:17) correct 16% (01:41) wrong based on 39 sessions
Find the sum of the first 15 terms of the series whose nth term is (4n+1).

A. 485
B. 495
C. 505
D. 630
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Oct 2013, 01:54, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Find the sum of the first 15 terms of the series whose nth [#permalink] New post 14 Oct 2013, 08:42
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jrymbei wrote:
Find the sum of the first 15 terms of the series whose nth term is (4n+1).

A. 485
B. 495
C. 505
D. 630


Note - I am clueless about this question.


First term : 4*1+1 = 5.

15th term : 4*15+1= 61

Sum of 15 terms : \frac{No of terms*(first term + last term)}{2} = \frac{15*(61+5)}{2} = 495.

B.
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Re: Find the sum of the first 15 terms of the series whose nth [#permalink] New post 14 Oct 2013, 11:07
Thanks for the explanation...it was so simple!!!
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Re: Find the sum of the first 15 terms of the series whose nth [#permalink] New post 30 Oct 2013, 01:53
jrymbei wrote:
Find the sum of the first 15 terms of the series whose nth term is (4n+1).

A. 485
B. 495
C. 505
D. 630


Note - I am clueless about this question.


Formula used: Sum of n terms = Average (First and Last term) * Number of terms.

First term : n=1, (4*1+1) = 5
Last term : n=15, (4*15 + 1) = 61

Sum = (5+61)/2 * 15 = 495. Answer B.
Re: Find the sum of the first 15 terms of the series whose nth   [#permalink] 30 Oct 2013, 01:53
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