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by KICKGMATASS123 » Fri Jul 24, 2009 12:46 pm
The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80


OA is B

Can someone explain a general approach for such complicated sequences??


Thanks in advance
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by prindaroy » Fri Jul 24, 2009 1:26 pm
So here is what you should do;

a5 = a1 since an = an-4

a6 = a2

a7 = a3

a8 = a4

a9 = a5 = a1

So it forms a repetitive process;

So we need to know what a1+a2+......+a97 is

so 2+-3+5+-1+2+-2+5+-1.....all the way till 96 where the last number will -1 so, we have (2-3+5-1)*24 = 72+2 = 74
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by KICKGMATASS123 » Fri Jul 24, 2009 2:11 pm
prindaroy wrote:So here is what you should do;

a5 = a1 since an = an-4

a6 = a2

a7 = a3

a8 = a4

a9 = a5 = a1

So it forms a repetitive process;

So we need to know what a1+a2+......+a97 is

so 2+-3+5+-1+2+-2+5+-1.....all the way till 96 where the last number will -1 so, we have (2-3+5-1)*24 = 72+2 = 74
did u pick 96 coz it's divisible by 4? and the sequence repeats every 4 numbers?
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by prindaroy » Fri Jul 24, 2009 3:38 pm
Yes. We need to look for the greatest number smaller than 97 and divisible by 4 because we know 97 isn't divisible by 4
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