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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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Sequence S is such that \(S_n = S_{n-1} + \frac{3}{2}\) and \(S_1 = 2\)

Sequence A is such that \(A_n = A_{n-1} - 1.5\) and \(A_1 = 18.5\)

Quantity A
Quantity B
The sum of the terms in \(S\) from \(S_1\) to \(S_{13}\), inclusive
The sum of the terms from in \(A\) from \(A_1\) to \(A_{13}\), 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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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   Updated on: 28 Nov 2018, 19:32
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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.

Originally posted by Fatemeh on 15 Mar 2018, 16:11.
Last edited by Fatemeh 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
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Re: Sequence S is such that Sn [#permalink] New post   14 Nov 2021, 09:42
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Take a general arithmetic series, \(a = a_1, a_2, a_3, ......, a_n\)
where, \(a_n = a_n-1 + d\)

Sum of all terms till \(a_n = \frac{n}{2}(2a_1 + [n-1]d)\)

\(S_n = S_{n-1} + \frac{3}{2}\)

\(S_1 = 2, d_S = 1.5\)

\(A_n = A_{n-1} - \frac{3}{2}\)

\(A_1 = 18.5, d_A = -1.5\)

Quantity A - \( = \frac{n}{2}(2S_1 + [n-1]d_S) = \frac{13}{2}(2(2) + [13-1]1.5) = \) 143

Quantity B - \( = \frac{n}{2}(2A_1 + [n-1]d_A) = \frac{13}{2}(2(18.5) + [13-1](-1.5)) = \) 123.5

Hence, Answer is A
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Re: Sequence S is such that Sn [#permalink] New post   16 Aug 2024, 03:26
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Re: Sequence S is such that Sn [#permalink]
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