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A, B, C, D, E, F, G, and H are all integers, listed in order of increa

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A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]

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Tough and Tricky questions: Sequences.



A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Sequences.



A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)

Kudos for a correct solution.



Ans : B

Assume that the numbers appear as shown below on the number line

A-----B-----C-----D-----E-----F-----G-----H
| |
(5^12) (5^13)

As the values for G and H are given , we can calculate the difference between any two terms of the series .

Common Difference ,d = (5^13) - (5^12)
= (5^12) *[ 5-1 ]
= (5^12)*(4)

Also F + d = G as the terms are in equidistant and in increasing order.

So F + (5^12)*(4) = (5^12).

That is , F = (5^12) - (5^12)*(4)
= (5^12)[ 1-4]
= (5^12) (-3)

Similarly , E = F - d
= (5^12)[-3-4]
= (5^12)*(-7)

You can see a -4 getting added to the non-exponent part of the values . That is , according to the pattern , D SHOULD BE (5^12)*(-7-4)= (5^12)*(-11)
Following this pattern , A = (5^12)*(-23)

Ans : B

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A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]

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\(H = 5^{13}\)

\(G = 5^{12}\)

\(H - G = 5^{13} - 5^{12} = 4 * 5^{12}\)

A ......1....... B ........2....... C ........3......... D ..........4........... E ......5....... F ........6......... G ................... H

From point A, \(4 * 5^{12}\) has to be added 6 times to reach point \(G = 5^{12}\)

\(A + 6* 4 * 5^{12} = 5^{12}\)

\(A = -23 * 5^{12}\)

Answer = B
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A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Sequences.



A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)


H = 5^13
G = 5^12

Distance between H and G = 5^13 - 5^12
= 5^12(5^1 - 1)
= (5^12)(4)
= (4)(5^12)

So, F = 5^12 - (4)(5^12)
E = 5^12 - (4)(5^12) - (4)(5^12)
D = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12)
C = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12)
B = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12)
A = 5^12 - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12) - (4)(5^12)

So, A = 5^12 - [(4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12)]
= 5^12 - (24)(5^12)
= (1)(5^12) - (24)(5^12)
= (-23)(5^12)
= Answer choice B

Cheers,
Brent
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Sequences.



A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?

A. -24(5^12)
B. -23(5^12)
C. -24(5^6)
D. 23(5^12)
E. 24(5^12)


We are given that A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size, and that the distance between any two consecutive numbers, which we can denote by the variable n, is constant.

Since G and H are equal to 5^12 and 5^13, respectively, then the distance between them is:

n = 5^13 - 5^12

n = 5^12(5 - 1)

n = 5^12(4)

Since A is 6 numbers ahead of G, A is 6n less than G, and therefore:

A = G - 6n

A = 5^12 - 6[5^12(4)]

A = 5^12[1 - 6(4)]

A = 5^12[-23] = -23(5^12)

Answer: B
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]

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New post 31 Aug 2017, 05:39
How I solved it is:
A, B, C, D, E, F, G, H (Total of 8 numbers = n)
Since its a sequence and each term is a certain constant more than the other.
H - G = 5^13 - 5^12 = (5^12) (5-1) = 4 x (5^12) = Our common difference, d.

Also unrelated,
G = A + (n-1) d
5^12 = A + (7-1) 4 (5^12)
As n=7, because its the 7th term in the sequence

Therefore,
A = (5^12) - 24 (5^12)
= (5^12) (1-24) Taking 5^12 common
= - 23 (5^12)
= Answer choice B

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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa   [#permalink] 31 Aug 2017, 05:39
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