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In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k

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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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New post 03 Oct 2017, 01:01
The equation given is applicable for values of n between 2-15.
In statement 1 we are given the value of n=1 which is not applicable to find the value of K hence insufficient.
on the other hand, the value of n in the second statement is given 8, which lies in the range of 2-15.
hence, we can find out the value of k and other Values in series.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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New post 06 May 2018, 00:49
what a lovely question!
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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New post 14 May 2018, 01:01
What is stopping A1 being -10 and K is -10. Then wouldn't it just oscillate between -10 and 10, and the answer is 0?
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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New post 14 May 2018, 01:07
rohit60 wrote:
What is stopping A1 being -10 and K is -10. Then wouldn't it just oscillate between -10 and 10, and the answer is 0?


In this case the sequence would be:

-10, -20, -30, -40, -50, -60, -70, -80, ... So, in this case, a8 would be -80 and not 10 as given in the second statement.
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k  [#permalink]

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New post 14 May 2018, 01:10
Bunuel wrote:
rohit60 wrote:
What is stopping A1 being -10 and K is -10. Then wouldn't it just oscillate between -10 and 10, and the answer is 0?


In this case the sequence would be:

-10, -20, -30, -40, -50, -60, -70, -80, ... So, in this case, a8 would be -80 and not 10 as given in the second statement.


Thanks! Makes sense. I misread it as An-1 - K
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Re: In the sequence shown, a_n=a_(n-1)+k, where 2<=n<=15 and k &nbs [#permalink] 14 May 2018, 01:10

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