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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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21 Apr 2018, 04:36
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In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k
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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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21 Apr 2018, 06:09
sachinpoovanna
In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k
Show more

\(a_{n}=a_{(n−1)}+(−1)^{n}(n)(k)\)

\(a_{2}=a_{1}+(−1)^{2}(2)(k)=k+2k=3k\)

\(a_{3}=a_{2)}+(−1)^{3}(3)(k)=0\)

\(a_{4}=a_{3)}+(−1)^{4}(4)(k)=4k\)

So we have our series as k,3k,0,4k.....

In this series even terms of the series are, 3k, 4k,.......\(a_{100}\)

\(a_{100}\), will be the 50th term of this even series and can be found out by using the AP formula -

\(t_n=a+(n-1)*d=>a_{100}=3k+(50-1)*k=52k\)

Option \(B\)
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chetan2u
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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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21 Apr 2018, 06:11
sachinpoovanna
In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k
Show more

look at the series..
\(a_2=k+2k=3k\)
\(a_3=3k-3k=0\)
\(a_4=0+4k=4k\)
\(a_5=4k-5k=-k\)
\(a_6=-k+6k=5k\) and so on..

even number sequence = \(3k, 4k, 5k......(\frac{n}{2}+2)k\)
thus \(a_{100}=(\frac{100}{2}+2)K=52k\)

B

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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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05 Apr 2019, 08:50
GMATSkilled
In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k
Show more

a2 = 3k,
a4 = 4k = a2 + k
a6 = 5k = a2 + 2k
a8 = 6k = a2 + 3k
...
a(n) = a2 + (n-2)/2 *k => a100 = 3k + (100 - 2)/2 * k = 3k + 49k = 52k. Hence B

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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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21 Apr 2020, 02:57
Hi,
I encountered this question and some similar questions on Veritas tests.
How to solve this question in less than 2 mins?
Unable to see a pattern until I solve till a6. During tests I had to take a wrong guess inorder to complete test in time, however during revisions I was able to solve it correctly in about 4 mins.
Its hard to complete tests in time with such question.

Please help with a relatively faster approach to such questions.
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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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14 Jul 2020, 00:43
Given: an=a(n−1)+(−1)n(n)(k) for all n>1 and a1=k

a1: k
a2: k + 2k = 3k
a3: 3k - 3k = 0
a4: 0 + 4k = 4k
a5: 4k - 5k = -k
a6: -k + 6k = 5k
a7: 5k - 7k = -2k
a8: -2k + 8k = 6k
a9: 6k - 9k = -3k

Thus,
even numbered terms are increasing by k
a2=3k, a4=4k, a6=5k, a8=6k....
odd numbered terms are decreasing by k
a1=k, a3=0, a5=-k, a7=-2k, a9=-3k....

To calculate a100 we need to use AP
first term: 3k common difference is: k and a100 will be the 50th even number
a100 = 3k+(50-1)k = 3k+49k = 52k
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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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12 Dec 2020, 21:33
chetansharma
sachinpoovanna
In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k

look at the series..
\(a_2=k+k=2k........a_3=2k-3k=-k........a_4=-k+4k=3k.....a_5=3k-5k=-2k.......a_6=-2k+6k=4k\) and so on..
so even numbers, say a_x, is \(a_x=xk-\frac{(x-4)k}{2}\)..
thus \(a_{100}=100k-\frac{(100-4)k}{2}=100k-48k=52k\)

B
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Hi , chetansharma i think highlighted part is wrong. I think A3 should be zero. Could you please check ?
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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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12 Dec 2020, 23:02
ShankSouljaBoi
chetansharma
sachinpoovanna
In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k

look at the series..
\(a_2=k+k=2k........a_3=2k-3k=-k........a_4=-k+4k=3k.....a_5=3k-5k=-2k.......a_6=-2k+6k=4k\) and so on..
so even numbers, say a_x, is \(a_x=xk-\frac{(x-4)k}{2}\)..
thus \(a_{100}=100k-\frac{(100-4)k}{2}=100k-48k=52k\)

B

Hi , chetansharma i think highlighted part is wrong. I think A3 should be zero. Could you please check ?
Show more

Hi, I had gone wrong in the very first calculations for a_2 as I missed out putting n in it. So, all values thereafter have also gone wrong.
Edited. Thanks

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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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10 May 2021, 06:05
niks18
sachinpoovanna
In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k

\(a_{n}=a_{(n−1)}+(−1)^{n}(n)(k)\)

\(a_{2}=a_{1}+(−1)^{2}(2)(k)=k+2k=3k\)

\(a_{3}=a_{2)}+(−1)^{3}(3)(k)=0\)

\(a_{4}=a_{3)}+(−1)^{4}(4)(k)=4k\)

So we have our series as k,3k,0,4k.....

In this series even terms of the series are, 3k, 4k,.......\(a_{100}\)

\(a_{100}\), will be the 50th term of this even series and can be found out by using the AP formula -

\(t_n=a+(n-1)*d=>a_{100}=3k+(50-1)*k=52k\)

Option \(B\)
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VeritasKarishma how to figure out that \(a_{100}\), will be the 50th term ?
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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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 Updated on: 10 May 2021, 22:40
Originally posted by KarishmaB on 10 May 2021, 22:35.
Last edited by KarishmaB on 10 May 2021, 22:40, edited 1 time in total.
GMATSkilled
In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k
Show more
Ansh777

\(a_{1} = k\)

\(a_{2}=k+2k \)

\(a_{3}=k+2k - 3k \)

In every subsequent term, nk is either added or subtracted alternately. So

\(a_{100}\)= k + 2k - 3k + 4k - 5k + 6k - 7k ... +98k - 99k + 100k

We get 50 such pairs.

\(a_{100}= 3k + k + k + k...\)

49 pairs give us k and the first pair gives us 3k.

So \(a_{100} = 3k + 49k = 52k\)

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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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10 May 2021, 22:39
dave13
niks18
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In a sequence for which \(a_{n}\)=\(a_{(n−1)}\)+\((−1)^{n}\)(n)(k) for all values n>1, where k is a constant and \(a_{1}\)=k, what is the value of \(a_{100}\)?

A. 53k

B. 52k

C. 48k

D. 47k

E. 23k

\(a_{n}=a_{(n−1)}+(−1)^{n}(n)(k)\)

\(a_{2}=a_{1}+(−1)^{2}(2)(k)=k+2k=3k\)

\(a_{3}=a_{2)}+(−1)^{3}(3)(k)=0\)

\(a_{4}=a_{3)}+(−1)^{4}(4)(k)=4k\)

So we have our series as k,3k,0,4k.....

In this series even terms of the series are, 3k, 4k,.......\(a_{100}\)

\(a_{100}\), will be the 50th term of this even series and can be found out by using the AP formula -

\(t_n=a+(n-1)*d=>a_{100}=3k+(50-1)*k=52k\)

Option \(B\)

VeritasKarishma how to figure out that \(a_{100}\), will be the 50th term ?
Show more


a100 is the 100th term in the given series. Every nth term is gives as \(a_{n}\)

If you consider only the even terms of our given series i.e. the second term, fourth term, sixth term etc, then in this even term series, 100th term will be the 50th term because all odd terms have been ignored. So half the terms are ignored and only half are kept.

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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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10 May 2021, 23:44
Follow the pattern for the first 5 to 6 values:

A1——K
A2——3k
A3——0
A4——-4K
A5——-K
A6——5K
A7——-2K
A8—— 6K

The terms that are even integers starting from A2 = 3K have an increase of (+1K) for each consecutive even integer.

How many consecutive even integers from 2 (2nd term) to 100(100th term)

50 consecutive even integers

A2 = 3k

.....



Add another 49 (+K) to A2 to get to the 50th consecutive even integer term ———> Term A-100


A-100 = (3K) + (49K) = 52K

Answer: 52K

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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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21 Mar 2024, 22:37
Let k=1

a100-a99=100
a99-a98=-99
a98-a97=98
..
..
a4-a3=4
a3-a2=-3
a2-a1=2

--------------
=> a100-a1=2-3+4-5+6-7+....+98-99+100
=(-1)*49+100=51
a100=51+a1=51+1=52
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In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh
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06 Apr 2025, 19:36
Given an=a(n-1)+((-1)^n)(n)(k) for all values n>1 and a1=k

From given condition
a1=k
a2=3k
a3=0
a4=4k
a5=-k
a6=5k

we can see a pattern for even terms in series which are continuous set of numbers starting from 3k
With that a100 should be equal to 52 k Option B
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