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# The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for

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The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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30 Jan 2014, 01:50
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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]

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30 Jan 2014, 01:50
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

$$a_5 = 31$$;
$$a_4 =31-5=26$$;
$$a_3 =26-5=21$$;
$$a_2 =21-5=16$$;
$$a_1 =16-5=11$$.

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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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30 Jan 2014, 03:10
1
2
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]

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30 Jan 2014, 03:31
1
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]

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30 Jan 2014, 10:42
2
1
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.

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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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31 Jan 2014, 06:13
1
2
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, $$a_1=31-20=11$$

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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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18 Mar 2014, 23:12
1
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

$$a_5 = 31$$;
$$a_4 =31-5=26$$;
$$a_3 =26-5=21$$;
$$a_2 =21-5=16$$;
$$a_1 =16-5=1$$.

But 2<= n <= 5 so how is this applicable to a1 i.e. n=1?
Math Expert
Joined: 02 Sep 2009
Posts: 58418
Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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19 Mar 2014, 01:59
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

$$a_5 = 31$$;
$$a_4 =31-5=26$$;
$$a_3 =26-5=21$$;
$$a_2 =21-5=16$$;
$$a_1 =16-5=1$$.

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]

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14 May 2015, 15:25
-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]

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04 May 2016, 03:40
a5 = 31
a4 = 31-5 = 26
a3 = 26 - 5 = 21
a2 = 21 - 5 = 16
a1 = 16 - 5 = 11
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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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26 Mar 2018, 17:34
Quote:

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.

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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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20 Oct 2018, 04:11
a5 = a4 + 5
a4 = a3 + 5
a3 = a2 + 5
a2 = a1 + 5

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

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Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for  [#permalink]

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09 Feb 2019, 04:40
Hi,

I'm still not sure how 2<= n <= 5 is applicable to a1 i.e. n=1?
Re: The sequence a1, a2, a3, a4, a5 is such that an=a(n-1)+5 for   [#permalink] 09 Feb 2019, 04:40
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