Question

Find the fist 4 terms of a geometric series with first term of 8 and the sum to infinity of 12

Answer 1

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 2

`8,\quad (8)/(3),\quad (8)/(9),\quad(8)/(27)`

Explanation:

Sum to infinity of a geometric series:

`S_\infty=(a)/(1-r) \quad \textsf{for }|r| < 1`

Given:

• 
  `a` = 8
• 
  `S_\infty` = 12

Substitute given values into the formula and solve for
`r`:

`\implies 12=(8)/(1-r)`

`\implies 1-r=(8)/(12)`

`\implies r=1-(8)/(12)`

`\implies r=(1)/(3)`

General form of a geometric sequence:
`a_n=ar^(n-1)`

(where a is the first term and r is the common ratio)

Substitute the found values of
`a` and
`r`:

`\implies a_n=8\left((1)/(3)\right)r^(n-1)`

The first 4 terms:

`\implies a_1=8\left((1)/(3)\right)r^0=8`

`\implies a_2=8\left((1)/(3)\right)r^1=(8)/(3)`

`\implies a_3=8\left((1)/(3)\right)r^2=(8)/(9)`

`\implies a_4=8\left((1)/(3)\right)r^3=(8)/(27)`