In a sequence of integers, [m]a_1 = 40[/m] and [m]a_n

Question

In a sequence of integers,  a_1 = 40  and  a_n < a_{n-1}  for all integers  n > 1 .

From which of the following can it be concluded that  a_{20}  is positive?

I.  a_{25} = \frac{a_{24}}{2} 
II. The sum of the first 42 terms is positive.
III.  a_{29} > a_{27} - a_{28}

(A) I only
(B) II only
(C) III only
(D) I and III
(E) I, II, and III

Marcelo7 · Answer

I would like to know why this is not the correct answer:

I. a25 = a24/2 => a1+24r = (a1+23r)/2 => 2a1+48r = a1+23r => 25r= -a1 (a1 = 40) => r = -40/25 = -8/5
Since we want to know if a20 > 0? (using a21 which is lower than a20 for easier caculation) a21 = a1+20r = 40 +20*-8/5 = 40-32=8 then a20 is positive.

II. Sn = (a1+an)n/2 => (a1+a42)*42/2 > 0 => a1+a42 > 0 => 2a1+41r >0 => r > -80/41 if this we know that r > -2 (-80/41 aprox. -2) a20 = 40 -2*19 = 40 - 38 = 2 so a20 must be positive since r must be lower than -2 to a20 to be negative.

III. a29>a27-a28 => a29> a1+26r -(a1+27r) => a29 > -r since we know that an<a(n-1) than r must be negative so a29> -r is positive so a20 is also positive.

Thanks for reading!

Edited:

Answer: the exercise never says the sequence is a arithmetic progression!

willykuo · Answer

I. Assume a20 <= 0
then: 0 => a20 > a24 > a25 > 2*a25
then: a24 / 2 > a25 contradiction.... so a20 > 0.
=> True

II. let a1 = 40, a2 = -0.1, a3 = -0.01, a4 = -0.001, a5 = -0.0001, ..... then we just construct a sequence that a20 < 0 but sum of a1~a42 > 0
=> False

III. a20 > a29 > a27 - a28 > 0
=> True

SwethaReddyL · Answer

Bunuel  /  KarishmaB  do you have the solution to this please?

KarishmaB · Answer

a_n < a_{n-1} for all integers n > 1 . This means every term is smaller than the previous term. The first term is 40, a positive number. The moment a term goes negative, all subsequent terms will also go negative. From which of the following can it be concluded that a_{20} is positive? I. a_{25} = \frac{a_{24}}{2} We know that a_{25} < a_{24} . If a_{25} = \frac{a_{24}}{2} , then a_{24} must be positive. Had it been negative, then its half cannot be a smaller value. e.g. If a_{24} = 10 , then \frac{a_{24}}{2} = \frac{10}{2} = 5 (smaller value) But if a_{24} = -10 , then \frac{a_{24}}{2} = \frac{-10}{2} = -5 (greater value) Hence a_{24} must be positive. Which means that a_{20} must be positive. Had a_{20} been negative, every subsequent value including a_{24} would have been negative. Hence, a_{20} is positive. II. The sum of the first 42 terms is positive. No idea about a_{20} . It could be positive or negative. There needn't be any relation between subsequent terms of a sequence. The terms could be 40, 39, 30, 29, 20.5, ... a_{20}=2 , 0, -1, - 1.1, -1.2, -1.3 ... or 40, 39, 38, 37, -1, ... a_{20}=-2 , -2.1, -2.2, -2.3 ... III. a_{29} > a_{27} - a_{28} We know that a_{28} < a_{27} This means a_{27} - a_{28} is positive which means a_{29} is positive. So a_{20} must be positive too. Had a_{20} been negative, every subsequent value including a_{29} would have been negative. Hence, a_{20} is positive Answer (D)

In a sequence of integers, [m]a_1 = 40[/m] and [m]a_n < a_{n

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