The distance the post is driven each time forms a geometric series

`8,8r,8r^2,8r^3,8r^4,...`

where `r=9/10`

The formula for the sum of a geometric series is given by

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

Here, `a=8`

We want to find the minimum integer value of`n`that gives `S_n > 70`

`implies ` `(8(1-r^n))/(1-r) > 70`

`implies` `8(1-r^n) > 70(1-r)`

`implies` `8r^n < 8 -70(1-r)`

`implies` `r^n < 1-(70/8)(1-r)`

`implies` `n > (log(1-(70/8)(1-r)))/logr` (remember log(r) is negative because r<1)

`implies` `n > log(1/8)/(log(9/10))`

`implies` `n > 19.7`

The minimum number of blows is therefore 20

The limiting sum of the series (the sum as `n ->oo`) is given by the formula

`S = a/(1-r)`

This is because `r^n -> 0` as `n -> oo` so `(1-r^n) -> 1`

Here

`S = 8/(1/10) = 80`cm

**a) minimum number of blows = 20****b) 80cm is the limiting sum of the geometric series**

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