# 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

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

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.

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lemjay | High School Teacher | (Level 2) Senior Educator

Posted on

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.

jsqueek12 | eNotes Newbie

Posted on

Thank you everyone you been so help full