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18 Aug 2021, 08:00
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B
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Difficulty:
95%
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Question Stats:
34% (02:18) correct
66%
(02:17)
wrong
based on 293
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The sequence of \(n\) integers is such that \(a_{k+1}=a_k-2\) for k > 1. If \(a_1\) is a prime number, is n a prime number?
(1) \(a_1=17\)
(2) The sum of the terms in the sequence is 77
M36-23
(1) \(a_1=17\)
(2) The sum of the terms in the sequence is 77
M36-23
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11 Oct 2021, 03:18
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Bunuel
Official Solution:
The sequence of \(n\) integers is such that \(a_{k+1}=a_k-2\) for \(k > 1\). If \(a_1\) is a prime number, is \(n\) a prime number?
(1) \(a_1=17\)
The sequence is: 17, 15, 13, 11, ... But we don't know the value of \(n\) (how long the sequence continues). Not sufficient.
(2) The sum of the terms in the sequence is 77
The sequence at hand is an arithmetic progression with the common difference of -2. The sum of the \(n\) terms of an arithmetic progression with the common difference of \(d\) is \(\frac{n}{2}*(2a_1 + (n – 1)d))\).
So, we are told that \(\frac{n}{2}*(2a_1 + (n – 1)(-2)))=77\);
\(n*(a_1 - n + 1))=77\)
77 can be expressed as the product of two positive integers (\(n\) and \((a_1 - n + 1)\) in our case) in the following 4 ways: 1*77, 7*11, 11*7 and 77*1. \(n\) cannot be 1 or 77 because in this case \(a_1\) is NOT a prime number (as we are given in the stem), so \(n\) is 7 or 11. Both are primes. Sufficient.
Answer: B
General Discussion
Updated on: 19 Aug 2021, 11:25
Originally posted by sumitkrocks on 18 Aug 2021, 08:18.
Last edited by sumitkrocks on 19 Aug 2021, 11:25, edited 1 time in total.
Last edited by sumitkrocks on 19 Aug 2021, 11:25, edited 1 time in total.
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Answer is (B)
The sequence of \(n\) integers is such that \(a_{k+1}=a_k-2\) for k > 1. If \(a_1\) is a prime number, is n a prime number?
Statement (1) \(a_1=17\)
No information about n
Not sufficient
Statement (2) The sum of the terms in the sequence is 77
a1 + a2 + a3 + a4...... =77
a1 + a1-2 + a1-4.....a1-2(n-1)
n*a1 - \(\frac{2(n)(n-1)}{2}\) = 77
n(a1+1-n) = 7*11 or 1*77
n=7 ; a1 =17
n=11; a1 =17
etc
In either case n is prime
N can be prime
(B) is the answer
The sequence of \(n\) integers is such that \(a_{k+1}=a_k-2\) for k > 1. If \(a_1\) is a prime number, is n a prime number?
Statement (1) \(a_1=17\)
No information about n
Not sufficient
Statement (2) The sum of the terms in the sequence is 77
a1 + a2 + a3 + a4...... =77
a1 + a1-2 + a1-4.....a1-2(n-1)
n*a1 - \(\frac{2(n)(n-1)}{2}\) = 77
n(a1+1-n) = 7*11 or 1*77
n=7 ; a1 =17
n=11; a1 =17
etc
In either case n is prime
N can be prime
(B) is the answer
