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Sequence S is such that Sn

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Sequence S is such that Sn [#permalink] New post 02 Mar 2018, 03:32
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65% (01:50) correct 34% (02:23) wrong based on 38 sessions
Sequence S is such that Sn = Sn-1 + 3/2 and S1 = 2

Sequence A is such that An = An-1 - 1.5 and A1 = 18.5

Quantity A
Quantity B
The sum of the terms in S from S1 to S13, inclusive
The sum of the terms from in A from A1 to A13, inclusive


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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[Reveal] Spoiler: OA

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Re: Sequence S is such that Sn [#permalink] New post 02 Mar 2018, 10:03
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Formula given to us:

\(Sn = Sn-1 +1.5; s1 = 2 and
An = An-1 - 1.5; A1 = 18.5\)

For the first sequence
Each subsequent value of \(s\) is greater by \(1.5\) For solving the question we need to calculate the sum of first 13 terms
we have the 1st term as \(2\). There remains \(12\) more terms. We know that there is an increase of \(1.5\) per term hence there is a total increase of \(12 * 1.5 = 18\)
Last term of the sequence = \(2+18 = 20\)

Since the sequence is equally spaced avg = \(\frac{2 +20}{2} = 11\)
There are a total of \(13\) terms hence sum of the sequence = \(11* 13 = 143\)

For the second sequence
Each subsequent value of A is less by \(1.5\) For solving the question we need to calculate the sum of first \(13\) terms
we have the 1st term as \(18.5\) There remains \(12\) more terms. We know that there is a decrease of \(1.5\) per term hence there is a total decrease of \(12 * 1.5 = 18\)
Last term of the sequence = \(18.5 - 18 = 0.5\)
Since the sequence is equally spaced avg = \(\frac{0.5 +18.5}{2} = 9.5\)
There are a total of \(13\) terms hence sum of the sequence = \(9.5* 13 = 123.5\)
Option A
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Re: Sequence S is such that Sn [#permalink] New post 12 Mar 2018, 01:38
option A
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Re: Sequence S is such that Sn [#permalink] New post 15 Mar 2018, 16:11
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Answer: A

A:
Sn = Sn-1 + 3/2 and S1 = 2
Sn = a + (n-1)d
a = S1 (the first element in the sequence)
d = Sn - Sn-1 = 3/2 (distance between two following elements in this kind of sequence which every new element is made by adding a constant value to it’s previous element)

Sn = 2 + (n-1) * 3/2 = 2 + 3/2*n - 3/2 = 1/2 + 3/2*n
Sum of the terms in S from s1 to S13, inclusive = (S1+ S13) *n / 2 = (2+20)*13/2=11*13 = 143
S1 = 2
S13 = 1/2 + 3/2*13 = 20

B:
The sum of the terms in A from A1 to A13, inclusive =

An = An-1 -1/2 and S1 = 18.5
Sn = a + (n-1)d
a = A1 (the first element in the sequence)
d = Sn - Sn-1 = -1.5 (distance between two following elements in this kind of sequence which every new element is made by adding a constant value to it’s previous element)

An = 18.5 + (n-1) * (-1.5) = 18.5 - 3/2*n + 3/2 = 20- 3/2*n

Sum of the terms in A from A1 to A13, inclusive = (A1+A13) *n / 2 = (18.5 + 0.5)*13/2=19*13/2 = 123.5
A1 = 18.5
A13 = 20-3/2*13 = 1/2

So A is bigger.
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Last edited by FatemehAsgarinejad on 28 Nov 2018, 19:32, edited 1 time in total.
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Re: Sequence S is such that Sn [#permalink] New post 25 Oct 2018, 06:06
I didnt get it
Re: Sequence S is such that Sn   [#permalink] 25 Oct 2018, 06:06
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