conty911 wrote:

If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n

(2) a^n > a^(3n)

Please provide me pointers to any similar questions, also if possible

Can some one explain steps to tackle such problems with least amount of time.For eg. how to select the values for plugging etc.

THanks

From the conditions in both statements, we can deduce that \(a\) cannot be neither \(0\) nor \(1.\)

A little algebraic manipulation can help in understanding the inequalities and also in choosing values for \(a\).

Usually it is easier to compare a product of numbers to \(0\), so it is worth rearranging the inequalities such that there is a comparison to \(0\) involved.

(1) \(a^{n-1}>a^n\) can be rewritten as \(a^{n-1}-a^n>0\) or \(a^{n-1}(1-a)>0\).

If \(a>1\), doesn't matter if \(n\) is even or odd, the given inequality cannot hold.

If \(a<0,\) then necessarily \(n\) must be odd.

Sufficient.

(2)\(a^n>a^{3n}\) can be rewritten as \(a^n-a^{3n}\) or \(a^n(1-a^{2n})>0\).

Since \(a\) cannot be neither \(0\) nor \(1,\) \(\,\,1-a^{2n}\) is certainly negative. (Now \(a\) cannot be \(-1\) either.)

Then \(a^n\) is negative if \(a\) is negative and \(n\) is odd, and this is the only way the given inequality can hold.

If \(a\) is positive, doesn't matter the value of \(n\), the inequality cannot hold.

Sufficient.

Answer D.

hello, can you explain more about statement 2, why a is positive, the inequality never holds? if 0<a<1, it still holds right?