Question

8 + 40 + 200 + ... + 15625000
Sum of a finite geometric series:
Sn=a1-a1r / 1-r

Answer 1

19,531,248

Explanation:

General form of a geometric sequence:
`a_n=ar^(n-1)`

(where
`a` is the first term and
`r` is the common difference)

Given geometric series: 8 + 40 + 200 + ... + 15625000

`\implies a_1=8`

`\implies a_2=40`

`\implies a_3=200`

To find the common ratio
`r`, divide consecutive terms:

`\implies r=(a_2)/(a_1)=(40)/(8)=5`

Therefore:

`\implies a_n=8(5)^(n-1)`

To find n when
`a_n=15625000` :

`\implies 8(5)^(n-1)=15625000`

`\implies (5)^(n-1)=1953125`

`\implies \ln (5)^(n-1)=\ln1953125`

`\implies (n-1)\ln 5=\ln1953125`

`\implies n=(\ln1953125)/(\ln5)+1`

`\implies n=10`

Sum of the first n terms of a geometric series:

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

Therefore, sum of the first 10 terms:

`\implies S_(10)=(8(1-5^(10)))/(1-5)`

`\implies S_(10)=19,531,248`