Question

How do you find `S_n` for the geometric series `a_1=3`, `a_n=46,875`, r=-5?

Answer 1

`S_n=39063`

Explanation:

`a_1=3 , r =-5 , a_n= 46875`. We know `n` th term of a

geometric series is `a_n=a_1*r^(n-1) or 45875= 3* (-5)^(n-1)` or

`(-5)^(n-1) =46875/3=15625` or

`((-1)*5)^(n-1) =15625` or

`(-1)^(n-1)* 5^(n-1) =15625`

`5^(n-1) =15625` Taking log on both sides we get

`(n-1) log 5 = log 15625 or (n-1) = log 15625/log5` or

`(n-1)=6 :. n=7 ; (-1)^(n-1)=(-1)^6=1`

Number of terms is `n=7`. We know sum of`n` terms of a

geometric series is `S_n=a_1*(r^n-1)/(r-1)` or

`S_n=(3*((-5)^7-1))/(-5-1)=(3*(-78125-1))/(-6)` or

`S_n=(3*(-78126))/(-6)=39063` [Ans]

Answer 2

`39063`

Explanation:

For a geometric series:

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

Where:

`bba` is the first term, `bbr` is the common ratio and `bbn` is the nth term.

We know the first term `a_1=3` and common ratio is `-5`

We need to find the value of `n`.

If we find the sum of the first `n` terms and then subtract the sum of the first `n-1` terms, this will give us the `a_n`th term.

`(3(1+5^(n-1)))/(1+5)=(1+(5)^(n-1))/2`

`(1+(5)^(n))/2-(1+(5)^(n-1))/2=a_n=46875`

`(1+(5)^(n)-1-(5)^(n-1))/2=46875`

`((5)^(n)-(5)^(n-1))/2=46875`

`(5)^(n)-(5)^(n-1)=93750`

`5^n=5^(n-1+1)`

`5^(n-1+1)=5^1*5^(n-1)`

`5^1*5^(n-1)-5^(n-1)=93750`

`5^(n-1)(5^1-1)=93750`

`5^(n-1)*4=93750`

`5^(n-1)=93750/4=23437.5`

Taking logs:

`(n-1)ln(5)=ln(23437.5)`

`n=ln(23437.5)/ln(5)+1~~7.251929636`

`n` must be an integer so `n=7`

`S_n=S_7=(3(1+5^7))/6=234378/6=39063`