Answer:
The first five terms are 2, [tex]-\frac{4}{3}[/tex] , [tex]\frac{8}{9}[/tex] , [tex]-\frac{16}{27}[/tex] , [tex]\frac{32}{81}[/tex]
Step-by-step explanation:
In the geometric sequence, there is a constant ratio between each two consecutive numbers
Examples:
5, 10, 20, 40, 80, ………………………. (×2)
5000, 1000, 200, 40, …………………………(÷5)
General term (nth term) of a Geometric sequence is:
a1 = a, a2 = ar, a3 = ar², a4 = ar³, ..........
[tex]an=ar^{n-1}[/tex], where
a is the first term
r is the constant ratio between each two consecutive terms
Let us solve the question
∵ A sequence has a first term of 2 and a constant ratio of [tex]-\frac{2}{3}[/tex]
∴ This is a geometric sequence, where
→ We need to find the first 5 terms
∴ n = 5
∵ [tex]an=ar^{n-1}[/tex]
∵ At n = 1, First term = a
∴ The first term = 2
∵ At n = 2, [tex]a2=2(-\frac{2}{3})^{2-1}=2(-\frac{2}{3})[/tex]
∴ The second term = [tex]-\frac{4}{3}[/tex]
∵ At n = 3, [tex]a3=2(-\frac{2}{3})^{3-1}=2(-\frac{2}{3})^{2}=2(\frac{4}{9})[/tex]
∴ The third term = [tex]\frac{8}{9}[/tex]
∵ At n = 4, [tex]a4=2(-\frac{2}{3})^{4-1}=2(-\frac{2}{3})^{3}=2(-\frac{8}{27})[/tex]
∴ The fourth term = [tex]-\frac{16}{27}[/tex]
∵ At n = 5, [tex]a5=2(-\frac{2}{3})^{5-1}=2(-\frac{2}{3})^{4}=2(\frac{16}{81})[/tex]
∴ The fourth term = [tex]\frac{32}{81}[/tex]
∴ The first five terms are 2, [tex]-\frac{4}{3}[/tex] , [tex]\frac{8}{9}[/tex] , [tex]-\frac{16}{27}[/tex] , [tex]\frac{32}{81}[/tex]
