Step-by-step explanation:
the sum of an infinite geriatric series with |r| < 1 is
s = a1/ (1 - r)
in our case we have
a1 = 8
12 = 8/(1 - r)
12(1 - r) = 8
1 - r = 8/12 = 2/3
-r = -1/3
r = 1/3
so, the first 4 terms (I added a5 too, just in case your teacher meant the NEXT 4 terms) are
a1 = 8
a2 = a1 × 1/3 = 8/3
a3 = a2 × 1/3 = 8/9
a4 = a3 × 1/3 = 8/27
a5 = a4 × 1/3 = 8/81
...
Answer:
[tex]8,\quad \dfrac{8}{3},\quad \dfrac{8}{9},\quad\dfrac{8}{27}[/tex]
Step-by-step explanation:
Sum to infinity of a geometric series:
[tex]S_\infty=\dfrac{a}{1-r} \quad \textsf{for }|r| < 1[/tex]
Given:
Substitute given values into the formula and solve for [tex]r[/tex]:
[tex]\implies 12=\dfrac{8}{1-r}[/tex]
[tex]\implies 1-r=\dfrac{8}{12}[/tex]
[tex]\implies r=1-\dfrac{8}{12}[/tex]
[tex]\implies r=\dfrac{1}{3}[/tex]
General form of a geometric sequence: [tex]a_n=ar^{n-1}[/tex]
(where a is the first term and r is the common ratio)
Substitute the found values of [tex]a[/tex] and [tex]r[/tex]:
[tex]\implies a_n=8\left(\dfrac{1}{3}\right)r^{n-1}[/tex]
The first 4 terms:
[tex]\implies a_1=8\left(\dfrac{1}{3}\right)r^0=8[/tex]
[tex]\implies a_2=8\left(\dfrac{1}{3}\right)r^1=\dfrac{8}{3}[/tex]
[tex]\implies a_3=8\left(\dfrac{1}{3}\right)r^2=\dfrac{8}{9}[/tex]
[tex]\implies a_4=8\left(\dfrac{1}{3}\right)r^3=\dfrac{8}{27}[/tex]
