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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)

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

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04 Nov 2014, 10:29

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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

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)

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

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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GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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

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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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