The nth term of the sequence $a_{1}$ , $a_{2}$ , $a_{3}$ , …, an is defined as $a_{n} = - (a_{n - 1} )$. The first term $a_{1} = – 1$.

Quantity A	Quantity B
$a_{5}$	$1$

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for the right answer and solution.

On resolving we would get a sequence where all odd numbers in the series would be -1 and all even numbers would be +1. So a1, a3, a5, ..so on have values -1. a2, a4, a6, ... so on have +1.

Hence B is greater as a5 would have value of -1

can someone please explain this. I do not understand how we solve the terms

It is given that $a_{1}=-1$ and $a_{n}=-a_{n-1}$.

So $a_{2}=-a_{1}=1$.

And $a_{3}=-a_{2}=-1$.

So we can see every odd term is -1 and every even term is +1. So $a_{5]$ must be -1.

Hence option B is correct!

My Answer is B

The sequence: -1 1 -1 1 -1 -> a5 = -1 -> B.

thanks for the explanation

Basically

n_1 = -1
n_2 = (-1)^2
n_3 = (-1)^3

So whenever the term (=exponent) is odd = -1; whenever the term=exponent is even = 1.

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