Question

The first term of a geometric progression is 25 and the sum of the first three terms is 61. Find the possible values of the common ratio. Given also that this progression has a sum to infinity, evaluate this sum.

Answer 1

sum of n terms of a GS is a1. r^n - 1
------
r - 1


where a1 = first term and r = common ratio

Answer 2

Sum of three numbers = first term(1-ratio^term number)
----------------------------
1-ratio

Answer 3

Thank you, but i@m stuck here :
25r^3-61r+36=0

Answer 4

so
61 = 25 ( r^3 - 1)
----------
r - 1

Answer 5

yes a cubic equation like that is tricky to solve

Answer 6

Well use trial and error method to figure out one root, then divide this equation by (r-root) and solve the quadratic equation

Answer 7

or use a graphical calculator if you have one!

Answer 8

one solution is r = 1 but that cant be a valid value for r
There are 2 more roots because its a cubic

Answer 9

you could dive the equation by x -1 This would give you a quadratic

Answer 10

- yes that is one way to go

Answer 11

* by r - 1

Answer 12

@welshfella Thats what I said earlier. :)

Answer 13

sorry - so you did!

Answer 14

thank you, i've found the answer however which value of r should i take into consideration, all 3 of them or 2 or 1?

Answer 15

did you get 0.8 and -1.8?

Answer 16

yup

Answer 17

ignore r = 1 as it would not be a GS in that case.

Answer 18

Sum to infinity = a1 / (1 - r)

so for r = 0.8 its 25 / ( 1 - 0.8)

Answer 19

that formula is only valid for -1=< x <= 1

Answer 20

Ahhhhh i see, thanks

Answer 21

What do you think about r = -1.8, Is that a possiblt value for the common ratio?

Answer 22

yes....

Answer 23

It is but it doesn''t converge to a particular value so theres no sum to infinity.

Answer 24

ok, thanks again

Answer 25

yw

Answer 26

its an oscillating series when r = -1.8

first few terms are 25, -45 , 81, -145.8