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Sequences

Problem Solving — algebra and arithmetic
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by Eternalhope » Wed Feb 29, 2012 11:38 am
The sequence a1, a2, a3, a4, a5 is such that an = an-1 + 5. For 2<=n <=5. If a5 = 31, what is the value of a1?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 2
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by krusta80 » Wed Feb 29, 2012 2:32 pm
Eternalhope wrote:The sequence a1, a2, a3, a4, a5 is such that an = an-1 + 5. For 2<=n <=5. If a5 = 31, what is the value of a1?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 2
Each number in the sequence is 5 more than the previous, so we can quickly work backwards to determine the answer...

a5=31
a4=26
a3=21
a2=16
a1=11

C
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by Jim@StratusPrep » Fri Mar 02, 2012 7:41 am
You are going back 4 steps of 5 or 20 (4*5).

31-20 = 11

C
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by IshanGhose » Sat Apr 21, 2012 9:15 am
31 - 4x5 (since 5 is the difference between 2 successive numbers, and 4 is the number of numbers between a5 and a1) = 31-20 = 11.

Therefore C) 11
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