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If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

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by BTGmoderatorDC » Thu Dec 03, 2020 5:51 pm

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If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

(A) 315
(B) 330
(C) 345
(D) 360
(E) 375



OA B

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Re: If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

by Scott@TargetTestPrep » Sun Dec 06, 2020 8:51 am
BTGmoderatorDC wrote: ↑
Thu Dec 03, 2020 5:51 pm
 If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

(A) 315
(B) 330
(C) 345
(D) 360
(E) 375



OA B

Solution:

We are given that 255 is the lowest of a set of 23 consecutive multiples of 15. That is, the subsequent multiples of 15 after 255 are 255 + 15(1), 255 + 15(2), etc. This pattern
Indicates that we have an arithmetic sequence, with first term a_1 = 255 and common difference d = 15. We need to determine the n = 23rd term in the sequence.

Using the formula for the nth term of an arithmetic sequence a_n = a_1 + (n -1)d, we have:

255 + 15(23 - 1) = 255 + 15(22)

The range of a set is the difference between the highest and lowest numbers in the set. Therefore, the range is [255 + 15(22)] - 255 = 15(22) = 330.

Answer: B

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