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Posts: 91
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The infinite sequence \(a_1\), \(a_2\),..., \(a_n\), ... is such that \(a_1 = 2\), \(a_2 = -3\), \(a_3 = 5\), \(a_4 = -1\), and \(a_n = a_{n-4}\) 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
A. 72
B. 74
C. 75
D. 78
E. 80
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udaymathapati
Given: \(a_1=2\), \(a_2=-3\), \(a_3=5\) and \(a_4=-1\). Also \(a_n=a_{n-4}\), for \(n>4\).
Since \(a_n=a_{n-4}\) then:
\(a_5=a_1=2\);
\(a_6=a_2=-3\);
\(a_7=a_3=5\);
\(a_8=a_4=-1\);
\(a_9=a_5=a_1=2\);
and so on.
So we have groups of 4: {2, -3, 5, -1}, the sum of each of such group is \(2-3+5-1=3\). 97 terms consist of 24 full groups plus \(a_{97}=a_1=2\), so the sum of first 97 terms of the sequence is \(24*3+2=74\).
Answer: B.
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ashiima
Given: a(n) = a(n-4) i.e. the nth term is same as (n-4)th term e.g. 5th term is same as 1st term. 6th term is same as 2nd term etc
The sequence becomes: 2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1 ...
The sum of each group of 4 terms = 2-3+5-1 = 3
How many such groups of 4 are there in the first 97 terms? When you divide by 4, you get 24 as quotient and 1 as remainder.
This means you get 24 complete groups of 4 terms each and 1 extra term i.e. the 97th term.
Sum of 24 groups of 4 terms each = 24*3 = 72
The 97th term will be 2 since it is the first term of the next group of four terms.
Sum of first 97 terms = 72+2 = 74
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General Discussion
Posts: 109
the sequence is repeating after each 4 terms.
i.e 2,-3,5,-1 ,2,-3,5,-1 ,2,-3,5,-1 ,2,-3,5,-1 ....till 97th term.
Now (2,-3,5,-1 ),(2,-3,5,-1) ,(2,-3,5,-1 ),......................(2,-3,5,-1) ,1
97 = 24 *4 + 1
Means 24 similar like (2,-3,5,-1 ) and one more that is first term of the series
the sum of the first four term is = 3
Hence sum is 24*3 + 2 = 74
Hence answer is B.
Consider KUDOS if it is helpful to u.
i.e 2,-3,5,-1 ,2,-3,5,-1 ,2,-3,5,-1 ,2,-3,5,-1 ....till 97th term.
Now (2,-3,5,-1 ),(2,-3,5,-1) ,(2,-3,5,-1 ),......................(2,-3,5,-1) ,1
97 = 24 *4 + 1
Means 24 similar like (2,-3,5,-1 ) and one more that is first term of the series
the sum of the first four term is = 3
Hence sum is 24*3 + 2 = 74
Hence answer is B.
Consider KUDOS if it is helpful to u.
Nice one, a pattern problem with an addition twist.
a1 = 2, a2 = -3, a3 = 5, a4 = -1
What an = a(n-4) means that from a5 onwards, every number in the sequence will be equal to 4 numbers preceding that number. Therefore, a5 = a1 = 2, a6 = a2 = -3,...
Notice that this is a 4-element set pattern. If we sum a1-a4, we get 3. What is the closet multiple of 4 to 97 - 96! So, there are 24 such complete sets. Added to that is 97th number, which is first in the sequence = 2. So, we get (24 x 3) + 2 = 74 -> B
a1 = 2, a2 = -3, a3 = 5, a4 = -1
What an = a(n-4) means that from a5 onwards, every number in the sequence will be equal to 4 numbers preceding that number. Therefore, a5 = a1 = 2, a6 = a2 = -3,...
Notice that this is a 4-element set pattern. If we sum a1-a4, we get 3. What is the closet multiple of 4 to 97 - 96! So, there are 24 such complete sets. Added to that is 97th number, which is first in the sequence = 2. So, we get (24 x 3) + 2 = 74 -> B
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\(a_1 = 2\)
\(a_2 = -3\)
\(a_3 = 5\)
\(a_4 = -1\)
\(a_5 = a_{(5-4)} = a_1 = 2\)
\(a_6 = a_{(6-4)} = a_2 = -3\)
\((a_1 + a_2 + a_3 + a_4) = (a_5 + a_6 + a_7 + a_8) = (a_9 + a_{10} .....)\) and so on
Addition upto \(a_{96}\) is repeated 24 times
\((a_1 + a_2 + a_3 + a_4) = 2 - 3 + 5 - 1 = 3\)
\(a_1 + a_2 + a_3 + ........ + a_{96} = 3 * 24 = 72\)
\((a_1+a_2+a_3+ ........ + a_{96}) + a_{97} = 72 + 2 = 74\)
Answer = B
\(a_2 = -3\)
\(a_3 = 5\)
\(a_4 = -1\)
\(a_5 = a_{(5-4)} = a_1 = 2\)
\(a_6 = a_{(6-4)} = a_2 = -3\)
\((a_1 + a_2 + a_3 + a_4) = (a_5 + a_6 + a_7 + a_8) = (a_9 + a_{10} .....)\) and so on
Addition upto \(a_{96}\) is repeated 24 times
\((a_1 + a_2 + a_3 + a_4) = 2 - 3 + 5 - 1 = 3\)
\(a_1 + a_2 + a_3 + ........ + a_{96} = 3 * 24 = 72\)
\((a_1+a_2+a_3+ ........ + a_{96}) + a_{97} = 72 + 2 = 74\)
Answer = B
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Expert reply
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udaymathapati
We are given that a(1) = 2, a(2) = -3, a(3) = 5, and a(4) = -1.
Since a(n) = a(n-4), we see that:
a(5) = a(5-4) = a(1) = 2,
a(6) = a(6-4) = a(2) = -3,
a(7) = a(7-4) = a(3) = 5,
and a(8) = a(8-4) = a(4) = -1
As we can see, the terms repeat themselves in a cycle of 4.
Since the sum of the terms a(1) to a(4) inclusive is [2 + (-3) + 5 + (-1)] = 3, the sum of a(5) to a(8) will also be 3, as will the sum of a(9) to a(12), etc.
We see from each “grouping” of 4 numbers, we have a sum of 3, and from 1 to 96 inclusive there are:
(96 - 4)/4 + 1 = 92/4 + 1 = 24 such groupings. Thus, the sum of those 24 groupings is 24 x 3 = 72.
We must also add the value of a(97), which equals 2.
Thus, the sum of the first 97 terms is: 24 x 3 + 2 = 72 + 2 = 74.
Answer: B
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udaymathapati
Can anyone provide me with a quality information(website, youtube etc) on GMAT sequences. I cant seem to decode such sort of questions
dave13
12. Sequences
For ALL other subjects check Ultimate GMAT Quantitative Megathread
(I pointed you to this topic multiple times)Hope it helps.
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Bunuel
Hi can you help understand why can't we use the formula here N(A1+AN)/2 ? Thanks
Sri07
The formula you quote, is for evenly spaced sets (aka arithmetical progression), and cannot be applied to all sequences. The sequence at hand is 2, -3, , 5, -1, ... As you can see this is not an arithmetic progression, so you cannot use that formula.
Check links below for more.
12. Sequences
- Theory
Sequences Made Easy - All in One Topic!
Math : Sequences & Progressions
Time to Tackle Geometric Progressions!
For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
Hope it helps.
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Posts: 5
Hello,
So this progression is neither an AP or GP correct? and thus the equations for the sum do not apply
So this progression is neither an AP or GP correct? and thus the equations for the sum do not apply
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Technically it is 4 APs with d=0 after every 4 terms.
nicolasnemtala
Posts: 161
Here's my video solution:
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