1. Good post? |

infinite sequence

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

2. Good post? |
a5=5-4=1
a6=6-4=2
a7=7-4=3
a97=97-4=93

Sum=n(n+1)/2=93*94/2=4371
Also adding the first four numbers os series=4371+2-3+5-1=4374

What is the source of the question

3. Good post? |
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?

an=an-4 for n>4...... means that the sequence repeats itself
2,-3,5,-1
a5 = 2
a6 = -3
a7 = 5
a8 = -1

2 + 5 -3 -1 = 3

97/4 =24 remainder 1

24 * 3 =72

therefore sum = 72 + a97
72 + 2
= 74

4. Good post? |
Originally Posted by Kumaril
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?

an=an-4 for n>4...... means that the sequence repeats itself
2,-3,5,-1
a5 = 2
a6 = -3
a7 = 5
a8 = -1

2 + 5 -3 -1 = 3

97/4 =24 remainder 1

24 * 3 =72

therefore sum = 72 + a97
72 + 2
= 74

I did this the same way that mydreamgmat did. How did you get a5=2? Shouldn't a5=1?

5. Good post? |
you misunderstand what an = an-4 means

a5 = a(5-4)
Therefore a5 = a1

a6 = a(6-4)
therefore a6 = a2
and so on

HTH

6. Good post? |
Originally Posted by solarisgirl
97/4 =24 remainder 1

24 * 3 =72

therefore sum = 72 + a97
72 + 2
= 74
How do you get to the highlighted part?

7. Good post? |
a1 =2
a2 =-3 a3 =5 and a4 =-1

this gives sum of first 4 terms as 3.
an=an-4 gives a5=a1= 2

so a5=2 a6=-3 a7= 5 and a8 =-1 and so on.

So every 4 terms have a sum of 3. So till 96th term we have sum = 3*24=72.
97th term is term a1, so a97=2.

Sum = 72 +2 =74. Hope this clarifies.

8. Good post? |
Kumaril... good one dude.

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