BTGmoderatorDC wrote: ↑

Fri Apr 17, 2020 8:48 pm

Each term of a certain sequence is 3 less than the previous term. The first term of this sequence is 19. If the sum of the first n terms of the sequence is n, what is the value of positive integer n?

A 1
B 13
C 15
D 19
E 47

OA B

Source: Veritas Prep

Note that the formulae for the sum (S) of the first n terms, whose constant difference is d and the first term is a, is given by S = n/2[2a + (n – 1)d]

=> n = n/2[2*19 + (n – 1)*–3]

n = 13.

The correct answer: B

Hope this helps!

-Jay
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BTGmoderatorDC wrote: ↑

Fri Apr 17, 2020 8:48 pm

Each term of a certain sequence is 3 less than the previous term. The first term of this sequence is 19. If the sum of the first n terms of the sequence is n, what is the value of positive integer n?

A 1
B 13
C 15
D 19
E 47


OA B

Source: Veritas Prep

We have an arithmetic sequence, which has a general formula for the nth term as: a(n) = a(1) + d*(n - 1), where a(1) is the first term and d is the common difference. In this question, a(1) = 19 and d = -3.

We can let a(1) = 19, a(2) = 16, a(3) = 13 and so on. Therefore, the nth term, a(n) = 19 - 3*(n - 1) = 22 - 3n. Since the sequence is evenly spaced, we can use the formula: sum = average x quantity. Since the average of a finite evenly spaced sequence of n terms is [a(1) + a(n)] / 2, we have:

n = ([a(1) + a(n)] / 2) x n

n = ([19 + 22 - 3n] / 2) x n

1 = (41 - 3n) / 2

2 = 41 - 3n

3n = 39

n = 13

Answer: B