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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