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by mandeepak » Fri Oct 26, 2007 5:25 pm
The infinite sequence a , a ,…, a ,… is such that a = 2, a = -3, a = 5, a = -1, and a =
1 2 n 1 2 3 4 n
a for n > 4. What is the sum of the first 97 terms of the sequence?
n-4

A. 72
B. 74
C. 75
D. 78
E. 80
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by jangojess » Fri Oct 26, 2007 9:40 pm
can u pls repost the Q....Q seems vague for me
Trying hard!!!
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by mandeepak » Sat Oct 27, 2007 3:37 am
I have added the question as a img file
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by ssy » Sat Oct 27, 2007 4:09 am
IMO it's B, 74

Any nth term larger than four equals the value of the term n-4. Therefore, the pattern is self-repeating 2, -3, 5, -1, 2, -3, 5, -1,2,-3,5,-1...and so forth

For example: a5=a5-4=a1=2
a6=a6-4=a2=-3
a7=a7-4=a3=5
a8=a8-4=a4=-1
a9=a9-4=a5=2...

The sum of each set of four is 3 (2+-3+5+-1)
So, 97/4=24 r1
First 96 terms = 24*3=72
97th term=2 (97th term will equal first term in the set)

Sum- 72+2=74.
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by ri2007 » Mon Nov 19, 2007 8:20 am
Hi

Can any one help me understand what I am doing wrong. My exam is approaching real fast so would really apprecaite ur help.

Here is what I did

The 5th term in the series is 1 (i.e 5-4)
the 6th term is 2 (i.e 6-4)
The 97th terms is 93

So essentially from the 5th term in the series to the 97th term we have a series of consequitive integers from 1 to 93

So the sum of all consequitive integers from 1 to 93 inclusive =

(94/2) * (94-1+1) = 4418

Now the sum of the first 4 numbers in the sequence is (2 + (-3) + 5 + (-1) ) = 3

So sum of first 97 terms is 4418 + 3 = 4421

This is obviously wrong, but I dont know why. Please Help.

Thanks in advance
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