The amounts of oil pumped from an oil well in each of the years 2001 to 2004 formed a geometric progression with common ratio 0.9. The amount pumped in 2001 was 100000 barrels.
Calculate in which year the amount pumped will fall below 5000.
The amounts of oil pumped from an oil well in each of the years 2001 to 2004 formed a geometric progression with common ratio 0.9. The amount pumped in 2001 was 100000 barrels.
Calculate in which year the amount pumped will fall below 5000
The last term of a geometric series is given by L = ar^(n-1)where a = the first term, r = the common ration and n = the number of terms.
Here, 5000 = 100,000(.9)^(n-1)
.05 = =.9^(n-1)
log(.05) = =log(.9)(n-1)
-.13010299 = -.04575 (n-1)
28.433 = (n-1) making n = 29.433
Therefore, less than 5000 barrels will be pumped in the year 2031 (2001 + 30)
Well, well, well...it seems like the amount of oil being pumped is going down faster than my enthusiasm for doing laundry. According to my calculations, it looks like the amount pumped will fall below 5000 barrels in 2031. That gives us a lot of time to prepare for a world without oil...or maybe just invest in some really good hair products because we won't be able to rely on oil for our lovely locks anymore. Either way, mark your calendars for the year of the great oil shortage!
To calculate in which year the amount pumped will fall below 5000, we can use the formula for the last term of a geometric series:
L = a * r^(n-1)
where
L = the last term,
a = the first term,
r = the common ratio,
and n = the number of terms.
In this case, the first term (a) is 100000 barrels, and the common ratio (r) is 0.9.
So, we need to solve for n when the last term (L) is 5000 barrels:
5000 = 100000 * 0.9^(n-1)
Dividing both sides by 100000, we get:
0.05 = 0.9^(n-1)
Taking the logarithm of both sides (base 0.9):
log(0.05) = (n-1) * log(0.9)
Using a calculator to evaluate these logarithms, we find:
-1.30103 ≈ -0.04575 * (n-1)
Simplifying further:
-1.30103 = -0.04575n + 0.04575
Adding 0.04575n to both sides, and simplifying:
0.04575n ≈ 1.34678
Dividing both sides by 0.04575:
n ≈ 29.433
Therefore, less than 5000 barrels will be pumped in the year 2031 (2001 + 30).
To calculate in which year the amount pumped will fall below 5000, we first need to use the formula for the nth term of a geometric progression, which is L = ar^(n-1), where L is the last term, a is the first term, r is the common ratio, and n is the number of terms.
Given that the amount pumped in 2001 was 100,000 barrels and the common ratio is 0.9, we can substitute these values into the formula as follows:
5000 = 100000 * (0.9)^(n-1)
To solve for n, we can take the logarithm of both sides of the equation. Let's use the natural logarithm (ln) here:
ln(5000) = ln(100000) + (n-1) * ln(0.9)
Simplifying further:
ln(5000) = ln(100000) + n * ln(0.9) - ln(0.9)
Next, we can move the terms involving "n" to one side of the equation:
n * ln(0.9) = ln(5000) - ln(100000) + ln(0.9)
Now, divide both sides by ln(0.9):
n = (ln(5000) - ln(100000) + ln(0.9)) / ln(0.9)
Evaluating this using a calculator, we find that n is approximately 29.433.
Since n represents the number of terms and each term represents a year, we round up to the nearest whole number to find the year in which the amount pumped will fall below 5000. Therefore, less than 5000 barrels will be pumped in the year 2031 (2001 + 30).