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The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  30 Jan 2014, 00:50
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

The sequence \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) is such that \(a_n= a_{n-1} + 5\) for \(2\leq n \leq 5\). If \(a_5 = 31\), what is the value of \(a_1\) ?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 21

Problem Solving
Question: 67
Category: Algebra Sequences
Page: 70
Difficulty: 600


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The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  30 Jan 2014, 00:50
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SOLUTION

The sequence \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) is such that \(a_n= a_{n-1} + 5\) for \(2\leq n \leq 5\). If \(a_5 = 31\), what is the value of \(a_1\) ?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 21

\(a_5 = 31\);
\(a_4 =31-5=26\);
\(a_3 =26-5=21\);
\(a_2 =21-5=16\);
\(a_1 =16-5=11\).

Answer: C.
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  30 Jan 2014, 02:10
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The sequence \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) is such that \(a_n= a_{n-1} + 5\) for \(2\leq n \leq 5\). If \(a_5 = 31\), what is the value of \(a_1\) ?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 21


Sol: Given a5=31....and also a5=a4+5 ---> a4=26-----> a3=21----->a2=16 ---->a1=11.
Ans C

600 level is okay
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  30 Jan 2014, 02:31
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Ans.C
This is nothing but an AP with common diff=5
a5=31,
a4=26
a3=21,
a2=16,
a1=11

a5=a1+4d
31=a1+4*5
31-20=a1=11
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  30 Jan 2014, 09:42
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This answer can be easily be solve by A.P.
The sequence is an A.P. with each term at an increment of 5 from the previous term.
So a5 is nothing but a1+20.

a5=a1+20=31
=> a1=11.

Answer is C
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  31 Jan 2014, 05:13
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The sequence \(a_1, a_2, a_3, a_4, a_5\) is such that \(a_n= a_{n-1} + 5\) for \(2\leq n \leq 5\). If \(a_5 = 31\), what is the value of \(a_1\) ?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 21

Given sequence is A.P. with common difference of 5.
Last Term - First Term = (Number of terms -1)* common difference

\(a_5-a_1 = (5-1)*\) \(common\) \(difference\)

Or, \(31-a_1=(5-1)*5\)

Or, \(a_1=31-20=11\)

Answer: (C)
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  18 Mar 2014, 22:12
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Bunuel wrote:
SOLUTION

The sequence \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) is such that \(a_n= a_{n-1} + 5\) for \(2\leq n \leq 5\). If \(a_5 = 31\), what is the value of \(a_1\) ?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 21

\(a_5 = 31\);
\(a_4 =31-5=26\);
\(a_3 =26-5=21\);
\(a_2 =21-5=16\);
\(a_1 =16-5=1\).

Answer: C.



But 2<= n <= 5 so how is this applicable to a1 i.e. n=1?
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  19 Mar 2014, 00:59
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havoc7860 wrote:
Bunuel wrote:
SOLUTION

The sequence \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) is such that \(a_n= a_{n-1} + 5\) for \(2\leq n \leq 5\). If \(a_5 = 31\), what is the value of \(a_1\) ?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 21

\(a_5 = 31\);
\(a_4 =31-5=26\);
\(a_3 =26-5=21\);
\(a_2 =21-5=16\);
\(a_1 =16-5=1\).

Answer: C.



But 2<= n <= 5 so how is this applicable to a1 i.e. n=1?


\(a_n= a_{n-1} + 5\) --> for n=2, we get: \(a_2= a_{1} + 5\).

Hope it helps.
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  14 May 2015, 14:25
-5 is constant here so --> a5 - 4*5 = 11 (4*5 because we have 4 numbers between a5 and a1 ... a4,3,2,1 )
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  04 May 2016, 02:40
a5 = 31
a4 = 31-5 = 26
a3 = 26 - 5 = 21
a2 = 21 - 5 = 16
a1 = 16 - 5 = 11
correct answer - C
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  26 Mar 2018, 16:34
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The sequence \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) is such that \(a_n= a_{n-1} + 5\) for \(2\leq n \leq 5\). If \(a_5 = 31\), what is the value of \(a_1\) ?

(A) 1
(B) 6
(C) 11
(D) 16
(E) 21


a(5) = 31, so:

a(5) = a(4) + 5

31 = a(4) + 5

26 = a(4)

The pattern is to subtract 5 to obtain the value of the previous term in this recursively-defined sequence.

Thus, a(3) = 21, a(2) =16, and a(1) = 11.

Answer: C
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  20 Oct 2018, 03:11
a5 = a4 + 5
a4 = a3 + 5
a3 = a2 + 5
a2 = a1 + 5

so a1 = a5 - 4*5 = 31 - 20 = 11

Answer choice C
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  09 Feb 2019, 03:40
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Hi,

I'm still not sure how 2<= n <= 5 is applicable to a1 i.e. n=1?
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  12 Apr 2020, 11:32
an = a(n-1) + 5
a5 = a(5-1) + 5 = a(4) + 5
a5 = 31
So, 31= a(4) + 5
=> 31-5 = a(4)
=>26 = a4

a4 = a3 + 5, => 26 = a3 + 5, => a3 = 21
a3 = a2 + 5, => 21 = a2 + 5, => a2 = 16
a2 = a1 + 5, => 16 = a1 + 5 => a1 = 11
a1 = 11

Answer is 11

Option C is the correct answer
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  05 May 2020, 08:29
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?

an=an−1+5 for 2≤n≤5
or,an−1 = an-5
now,a5=31
so,a4=31−5=26
a3=26−5=21
a2=21−5=16
a1=16−5=11.

correct answer C
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for [#permalink] New post  10 Jul 2020, 10:22
an = an-1 + 5

a5 = a4 + 5 --> a5 - a4 = 5 --> a4 - a3 = 5 ..... and so on

Taking a sum of these two: a5 - a4 + a4 - a3 = 5 + 5

Similarly, taking the sum till a1, we will reach: (a5 - a4) + (a4 - a3) + (a3 - a2) + (a2 - a1) = 5 +5+5+5
--> a5 - a1 = 20 --> a1 = 11 (Since a5 = 31)

No need to do all this. Just by spotting the pattern, we can see that one term is cancelling out if we take the sum of successive terms.
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