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Scoretop 6 - Question 5

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Scoretop 6 - Question 5

by akay » Thu Sep 27, 2007 2:28 am
Can anybody help with this progression question...

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
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by camitava » Thu Sep 27, 2007 4:42 am
Akay, ur qs -
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
So a5 = 5 - 4 = 1
a6 = 6 - 4 = 2
:
:
a94 = 94 - 4 = 90
So sum of this numbers = n/2(a + b) = 90 / 2 * (1 + 90) = 4095
where n = No of terms
a = First term
b = Last term
And sum of the first few terms; a1 , a2, a3 and a4 = 3
So total summation is = 4095 + 3 = 4098

But this is not provided in the options given. :cry: :)
Correct me If I am wrong


Regards,

Amitava
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by josephcho77 » Thu Sep 27, 2007 5:36 am
I think the answer is B

4 terms are repeating.
Sum of the first 4 terms is 3 = 2-3+5-1

If you divide 97 by 4 = 24 + 1
Therefore, the sum of the first 97 terms is 3*24 + 2 = 74

97th term is same as first term which is 2 since 4 terms are repeating.
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by kajcha » Thu Sep 27, 2007 5:38 am
I am a little confused

As per the question an= an-4. Is this a(n-4) or (an)-(4)?
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by akay » Thu Sep 27, 2007 6:32 am
kajcha as per the question it is a(n-4) /* n-4 in subscript */
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by kajcha » Thu Sep 27, 2007 6:47 am
That's what I was thinking.. Ans should be B

Same explanation as josephcho77
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by sudeeparies » Thu Sep 27, 2007 11:42 am
Agreed.If an=a(n-4) meaning if n=5, a5=a(5-4)=a1=2 then a1+a2+a3+a4=3 meaning the addition of these 4 terms happens 97/4=24+1 (the "1" meaning a1).so ((24*3)+2)=74.Hence B
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by camitava » Thu Sep 27, 2007 8:05 pm
Sorry guys! I misunderstood the qs and took a wrong approach to solve it. Sorry one again for my misunderstanding!!! :oops: :cry: :wink:
Correct me If I am wrong


Regards,

Amitava
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