In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n1 , wh

Question

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

niks18 · Accepted Answer

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

chetan2u · Answer

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

Skyline393 · Answer

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

Ansh777 · Answer

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.

ProfChaos · Answer

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

ShankSouljaBoi · Answer

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

chetan2u · Answer

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

dave13 · Answer

VeritasKarishma  how to figure out that  a_{100} , will be the 50th term ?

KarishmaB · Answer

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

KarishmaB · Answer

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.

Fdambro294 · Answer

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

Posted from my mobile device

Oppenheimer1945 · Answer

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

rohitz98 · Answer

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

In a sequence for which an=a(n−1)+(−1)n(n)(k) for all values n>1 , wh

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions