Question

How do you find the sum of the infinite geometric series given `5/3-10/9+20/27-...`?

Answer 1

`S=1`

Explanation:

We first find the factor :

`r=(-10/9)/(5/3)=-2/3`

The sum of an infinite geometric series is given by the formula :

`S=a_1/(1-r)` where `a_1` is the fisrt term and `|r|<=1` (which in our case it is).

So the sum is :

`S=a_1/(1-r)=(5/3)/(1-(-2/3))=(5/3)/(1+2/3)=(5/3)/(5/3)=1`

Answer 2

Sum of the infinite geometric series is `1`

Explanation:

In the series `5/3-10/9+20/27-.....`

whilr first term `a_1=5/3`, common ratio is `(-10/9)/(5/3)=(20/27)/(-10/9)=-2/3`

As for common ratio `r=-2/3`, `|r|=|-2/3|<1`,

sum of the infinite geometric series is

`a_1/(1-r)=(5/3)/(1-(-2/3))=(5/3)/(5/3)=1`

Answer 3

`S_oo=1`

Explanation:

`"for a geometric sequence the sum of n terms is"`

`S_n=(a(1-r^n))/(1-r);(r!=1)`

`"where a is the first term and r, the common ratio"`

`"as " ntooo,r^nto0" and " S_n" can be expressed as"`

`S_oo=a/(1-r);(|r|<1)`

`rArrr=(-10/9)/(5/3)=-2/3rarr|r|<1`

`rArrS_oo=(5/3)/(1+2/3)=1`