Find how many terms there are in this geometric sequence:
$-1, 2, -4, 8, ..., -16777216$
My attempt:
$a_k=a.r^{k-1}$
And in this sequence:
$a=-1$, $r=-2$
So
$a_k=(-1){(-2)}^{k-1}$
$-16777216=(-1){(-2)}^{k-1}$
$16777216={(-2)}^{k-1}$
$log(16777216)=log({(-2)}^{k-1})$
$log(16777216)=(k-1)log{(-2)}$
$k-1={{log(16777216)} \over {log{(-2)}}}$
But $-2$ is negative, and logarithm not defined for negative numbers?, So what can I do ?
Thanks
$\endgroup$ 14 Answers
$\begingroup$Hint: Since the sign is causing you trouble, get rid of it. The number of terms in this sequence is the same as the number of terms in the sequence $$1,2,4,8,...,16777216 $$
$\endgroup$ 1 $\begingroup$It is: $$-1,2,-4,8,...,(-1)\cdot 2^{24}$$
$\endgroup$ 1 $\begingroup$$-16777216=(-1){(-2)}^{k-1}={(-1)}^k2^{k-1}=-2^{k-1}$ (exponent can't be negative so minus has to come from -1)
$\endgroup$ 1 $\begingroup$To use logarithms, take the absolute values of the terms. Then you have
$|-1|×|-2|^{k-1}=|-16777216|$
$1×2^{k-1}=16777216$
where all numbers are positive and the logarithms can be manipulated without trouble. When you find $k$ you must check against the original equation with the negative signs included to verify that the proposed solution is valid.
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