Question 1035275
the first term of the infinite geometric series shown below is equal to a.
the common ratio of the infinite geometric series shown below is equal to r.


formula for sum of infinite geometric series is equal to a/(1-r)
if you square r, then the formula becomes a/(1-r^2)


you are given that a/(1-r) = 3
you are given that a/(1-r^2) = 1.8


solve for a in the first equation to get a = 3 * (1-r) = 3 - 3r
solve for a in the second equation to get a = 1.8 * (1 - r^2) = 1.8  - 1.8r^2.


since both expressions on the right side of each equation are equal to a, set them equal to each other to get:
3 - 3r = 1.8 - 1.8r^2


add 1.8r^2 to both sides of the equation and subtract 1.8 from both sides of the equation to get:
3 - 3r + 1.8r^2 - 1.8 = 0


combine like terms and rearrange the terms in descending order of degree to get:
1.8r^2 - 3r + 1.2 = 0


factor this quadratic equation to get:
r = 2/3 or r = 1

since 0 < r < 1, then r can't be equal to 1, so r has to be equal to 2/3.


when r = 2/3, the equation of a = 3 - 3r becomes a = 3 - 3*2/3 which becomes a = 3 - 6/3 which becomes a = 3 - 2 which becomes a = 1.


the first term of the infinite series is 1.
the common ratio is 2/3.


-----


recurring decimal of .524 is equal to 524 / 999.


you multiply .524..... by 1000 to get 524.525.....


1 * .524..... is equal to .524.


subtract 1 from 1000 to get 999.
subtract .524..... from 524.524..... to get 524.


your fraction is 524 / 999.


that can't be simplified any further.