# There is a leak in the Rein's roof. This leak must be fixed by the time the 500th tain drop falls through the roof. The leak allows 2 drops to go through the roof on the 1st day, 4 drops through on...

There is a leak in the Rein's roof. This leak must be fixed by the time the 500th tain drop falls through the roof. The leak allows 2 drops to go through the roof on the 1st day, 4 drops through on the second day, 8 drops through on the third day, and so on. On what day will the 500th drop fall through the roof

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### 3 Answers

I agree with the 7.97. I checked my Excel table and saw an error. I attached a correct table here. It shows during the 8th day. Sorry.

In the problem, the number of drops for the first three days are 2, 4, and 8.

Notice that the number of drops forms a sequence. So, our next step is to determine if it is an arithmetic or geometric sequence. Since 2, 4 and 8 are all divisible by 2, then it is possible that this is a geometric sequence. To verify, determine if the consecutive numbers have common ratio.

To do so, apply the formula:

`r =a_n/a_(n-1)`

`r=a_3/a_2 =8/4=2`

`r=a_2/a_1=4/2=2`

Since the value of r's are the same, hence, the number of drops form a geometric sequence.

That means, to determine nth day when 500th rain drops on the leak, apply the formula of sum of geometric sequence.

`S_n=(a_1(1 - r^n))/(1-r)`

So, plug-in `S_n=500` , `a_1=2` and `r=2` .

`500=(2(1-2^n))/(1-2)`

Then, simplify the equation.

`500=(2(1-2^n))/(-1)`

`500=-2(1-2^n)`

`-250=1-2^n`

`-251=-2^n`

`251=2^n`

And, take the natural logarithm of both sides to remove the n in the exponent.

`ln 251=ln2^n`

`ln251=nln2`

`(ln251)/(ln2)=n`

`7.97=n`

Rounding off to the nearest whole number, the value of n becomes:

`8=n`

**Hence, the 500th rain drops on the 8th day.**

Thank you everyone you been so help full