Question

the sum of the first eight terms of a geometric series is seventeen times the sum, of its first four terms. Find the common ratio

Answer 1

Use the formula\[\sum_{k=1}^{n}ar^{n-1} = a \frac{r^n - 1}{r - 1}\]

Answer 2

Plug in n = 8 and n = 4. Then make use of the relationship to give you an equation in terms of r. Solve for r.

Answer 3

can you show me how to solve completely?

Answer 4

Try plugging in n = 8. What do you get on the right-hand side?

Answer 5

you get - a(r^8-1)/(r-1)

Answer 6

what next

Answer 7

Plug in n = 4 for the same equation.

Answer 8

a(r^4-1)/(r-1)

Answer 9

what next

Answer 10

You do know that the sum of the first eight terms of a geometric series is seventeen times the sum, of its first four terms.

Hence, 17a(r^4-1)/(r-1) = a(r^8-1)/(r-1) => (r^8-1) = 17(r^4-1). Am I right?

Answer 11

yes

Answer 12

ohh....

Answer 13

can you just step me through how to simplify this, just so i can be sure?

Answer 14

\[(r^8-1) = 17(r^4-1)\]
\[r^8-1 = 17r^4-17\]
\[r^8 - 17r^4 + 16 = 0\]
Solve for r^4 as a quadratic equation.

Answer 15

(r^4 - 16)(r^4 - 1) = 0
r^4 = 1 (NA since it is not a GP) or r^4 = 16
r = ±sqrt(2)

Answer 16

thanks