## College Algebra 7th Edition

The first five terms are: $a_1 = -2 \\a_2 = 4 \\a_3 = -8 \\a_4= 16 \\a_5= -32$ The sequence is geometric with $r=-2$. The $n^{th}$ term is: $a_n = -2 \cdot(-2)^{n-1}$
To find the first five terms, substitute 1, 2, 3, 4, and 5 to the given formula to obtain: $a_1 = (-1)^1(2^1) = -1(2) = -2 \\a_2 = (-1)^2(2^2) = 1(4) = 4 \\a_3 = (-1)^3(2^3) = -1(8) = -8 \\a_4=(-1)^4(2^4) = 1(16) = 16 \\a_5=(-1)^5(2^5) = -1(32)=-32$ RECALL: A sequence is geometric if there is a common ratio among consecutive terms. Solve for the ratio of each pair of consecutive terms to obtain: $\dfrac{4}{-2} = -2 \\\dfrac{-8}{4} = -2 \\\dfrac{16}{-8} = -2 \\\dfrac{-32}{16}=-2$ The ratio is common therefore the sequence is geometric with $r=-2$. The $n^{th}$ term of a geometric sequence is given by the formula $a_n = ar^{n-1}$ where $a$ = first term and $r$= common ratio. This means that the $n^{th}$ term of the given geometric sequence, whose $a=-2$ and $r=-2$, is: $a_n = -2 \cdot(-2)^{n-1}$