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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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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.
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Find the sum of the first 15 terms of the series whose nth [#permalink] New post 26 Mar 2015, 21:30
Its probably obvious.. but can someone explain why the formula for a arithmetic series is used here ... Sum of series n = (n/2)(A1 - An) ... as opposed to the formula for a geometric series ... Sum of series n = [A1*(1-r^n)]/(1-r) where r is the common ratio and n is the nth term.

I understand the solution but am confused by the formula thinking that An=4n+1 is a geometric sequence given that you have to multiply by 4 to get the next term in the sequence. Appreciate your reply!!

THANKS!
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Re: Find the sum of the first 15 terms of the series whose nth [#permalink] New post 27 Mar 2015, 03:31
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sisorayi01 wrote:
Its probably obvious.. but can someone explain why the formula for a arithmetic series is used here ... Sum of series n = (n/2)(A1 - An) ... as opposed to the formula for a geometric series ... Sum of series n = [A1*(1-r^n)]/(1-r) where r is the common ratio and n is the nth term.

I understand the solution but am confused by the formula thinking that An=4n+1 is a geometric sequence given that you have to multiply by 4 to get the next term in the sequence. Appreciate your reply!!

THANKS!


The sequence is define by \(a_n=4n+1\), thus:

\(a_1=4*1+1=5\);
\(a_2=4*2+1=9\);
\(a_3=4*3+1=13\);
\(a_4=4*4+1=17\);
...

As you can see the sequence we have (5, 9, 13, 17, ...) is an arithmetic progression, not geometric.
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Re: Find the sum of the first 15 terms of the series whose nth [#permalink] New post 27 Mar 2015, 04:24
sisorayi01 wrote:
Its probably obvious.. but can someone explain why the formula for a arithmetic series is used here ... Sum of series n = (n/2)(A1 - An) ... as opposed to the formula for a geometric series ... Sum of series n = [A1*(1-r^n)]/(1-r) where r is the common ratio and n is the nth term.

I understand the solution but am confused by the formula thinking that An=4n+1 is a geometric sequence given that you have to multiply by 4 to get the next term in the sequence. Appreciate your reply!!

THANKS!


Adding to what Bunuel just posted, a geometric series would look like following
An = 4^n
And the series would look something like 4, 16, 64, 256, 1024, ...
Re: Find the sum of the first 15 terms of the series whose nth   [#permalink] 27 Mar 2015, 04:24
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