Question

a geometric progression for which the common ratio is positive has a second term 18 and a fourth term of 8 find
1. the first term and the common ratio progression
2. the sum to infinity of the progression

show how you did

Answer 1

a = first term
r = common ratio

1st term = a
2nd term = ar = 18
3rd term = ar^2
4th term = ar^3 = 8
...

Sum = a/(1-r)

Go!!

Answer 2

substitute in sn=a(1-r)^n/1-r ?

Answer 3

No, that is not quite the right formula for a FINITE geometric series. I provided the right one for an INFINITE geometric series. Your task is to find 'a' and 'r'.

Answer 4

the formula is for geometric progression not sum also i dont any other ways of findin a and r

Answer 5

You must look for clues rather than throwing your hands in the air.

The problem statement provides these clues.

ar = 18
ar^3 = 8

Answer 6

i throw my hand in the air saying ayoo gotta let go
also i was saying a= 18/r and then substitute this in formula in place of a for finding ratio

Answer 7

Very good. What do you get for 'a' and 'r'?

Answer 8

solving
in progress

Answer 9

so far i am @ 18r^2 - 18r = 18^2 - 18r^3
can u tell if i am doin it right ?

Answer 10

What happened to the lovely a = 18/r from the 2nd term??

Substituting this value into the 4th term:

(18/r)r^3 = 8

18r^2 = 8

r^2 = 8/18 = 4/9

Now what?

Answer 11

damn i subbed in formula for progression ill take square root and go over you method again

Answer 12

Careful with "take the square root".

There are TWO solutions to r^2 = 4/9.

Answer 13

its + and - same answer but question says its positive so

Answer 14

Bonus points for referring to the problem statement to resolve that conflict! Excellent work.

Answer 15

So, r = ?????

Answer 16

i got 2/3
but i think its wrong

Answer 17

No, r = 2/3. This, however is not the end.

a = 18/r = ?????

Answer 18

27

Answer 19

is it right ?

Answer 20

Okay, we are ready for the sum to infinity!

Answer 21

\(\dfrac{a}{1-r} = \dfrac{27}{1 - (2/3)} =\;??\)