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Re: Sequence S is such that Sn [#permalink]
2
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]
I didnt get it
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Re: Sequence S is such that Sn [#permalink]
1
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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