1.what is the least number of terms of the GP.2,4, 8.....that will give a sum greater than 1500000 2.Aball allowed to drop from heigth of 3m on to a horizontal ground rebounds to3/4 of its previous heigth.Find to the nearest metre,the total distance the ball will have traveled when it hits the ground for the eight time
1.
In your GP, a = 2, r = 2
sum(n) < 1500000
a(r^n - 1)/(r-1) > 1500000
2(2^n - 1)/(2-1) > 1500000
2^n - 1 > 750000
2^n > 750001
take log of both sides , and use log rules
n log2 > log 750001
n > log 750001/log2
n > 19.5
at n = 19 we won't be there, so we need 20 terms
check: sum(20) = 2(2^20 - 1)(1) = 2097150
sum(19) = 2(2^19 - 1)/1 = 1048574 which is < 1500000
2. make a sketch of the first few bounces.
Notice that the first bounce is only a one-way down, but the second , third, fourth, etc, have an up and a down and they are a GP
good luck
1. "Well, let's see... 2, 4, 8... that's starting to look like a pattern. It's like the numbers are going on a doubling spree. So, if we keep doubling, we can calculate how many terms we need to get a sum greater than 1500000. But remember, no pressure on the terms, we don't want them to feel multiplied! Ahem. Anyways, after a quick calculation, it seems like we need a whopping 18 terms to reach a sum greater than 1500000. That's quite a growth spurt!"
2. "Ah, the bouncing ball conundrum! So, if the ball rebounds to 3/4 of its previous height every time it hits the ground, it's like a never-ending trampoline party. To find the total distance traveled when it hits the ground for the eight time, we need to consider the distance covered during each bounce. By taking into account the ascending and descending journeys, it turns out that the ball will have traveled approximately 27 meters when it hits the ground for the eighth time. Quite the bouncing marathon, huh?"
1. To find the least number of terms of the geometric progression (GP) 2, 4, 8, ... that will give a sum greater than 1500000, we can use the formula for the sum of a geometric series.
The general formula for the sum of a geometric series is:
Sn = a * (r^n - 1) / (r - 1)
where Sn is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, the first term (a) is 2 and the common ratio (r) is 4/2 = 2, since each term is obtained by multiplying the previous term by 2.
We need to find the least value of n for which Sn > 1500000.
Substituting these values into the formula, we have:
1500000 < 2 * (2^n - 1) / (2 - 1)
1500000 < 2^n - 1
Adding 1 to both sides, we get:
1500000 + 1 < 2^n
1500001 < 2^n
Taking the logarithm of both sides with base 2, we get:
log2(1500001) < log2(2^n)
n > log2(1500001)
Using a calculator, we find that log2(1500001) is approximately 20.87.
So, the least number of terms (n) for the GP 2, 4, 8, ... that will give a sum greater than 1500000 is 21.
2. To find the total distance the ball will have traveled when it hits the ground for the eighth time, we can use the concept of geometric sequences.
The ball rebounds to 3/4 of its previous height, which means the height decreases by a factor of 3/4 each time it rebounds.
Let's represent the initial height as H.
The sequence of heights for each rebound can be written as follows:
H, (3/4)H, (3/4)^2 * H, (3/4)^3 * H, ...
This is a geometric sequence with a first term of H and a common ratio of 3/4.
The sum of the first eight terms of a geometric sequence can be found using the formula:
Sn = a * (1 - r^n) / (1 - r)
where Sn is the sum of the first n terms, a is the first term, r is the common ratio, and n is the number of terms.
In this case, a = H, r = 3/4, and n = 8.
Substituting these values into the formula, we have:
Sn = H * (1 - (3/4)^8) / (1 - 3/4)
Simplifying further:
Sn = H * (1 - 6561/65536) / (1/4)
Sn = 4H * (65536 - 6561) / 65536
Sn = 4H * 58975 / 65536
The total distance traveled by the ball is twice the sum of the sequence, since it travels the same distance on the way up and on the way down.
Therefore, the total distance traveled by the ball when it hits the ground for the eighth time is:
2 * (4H * 58975 / 65536) = 8H * 58975 / 65536
To the nearest metre, we can approximate the value of H as 3 since it falls from a height of 3 meters.
So, the total distance traveled by the ball is approximately:
8 * 3 * 58975 / 65536 = 6.806 meters (rounded to the nearest meter).
Thus, the total distance the ball will have traveled when it hits the ground for the eighth time is approximately 7 meters.
1. To find the least number of terms of a geometric progression (GP) that will give a sum greater than 1,500,000, we need to use the formula for the sum of a GP:
Sn = a * (r^n - 1) / (r - 1),
where Sn is the sum of the first n terms, a is the first term, r is the common ratio, and n is the number of terms.
In this case, the first term (a) is 2 and the common ratio (r) is 4/2 = 2, because each term is twice the previous term. We want to find the least value of n such that Sn > 1,500,000.
Substituting the given values into the equation:
1,500,000 < 2 * (2^n - 1) / (2 - 1).
Simplifying, we get:
1,500,000 < 2^n - 1.
Adding 1 to both sides:
1,500,001 < 2^n.
Taking the logarithm base 2 of both sides:
log2(1,500,001) < log2(2^n).
Using the logarithm identity log_b(x^n) = n * log_b(x):
log2(1,500,001) < n * log2(2).
Calculating the values:
n > log2(1,500,001) / log2(2).
Using a calculator, we find:
n > 19.87.
Since n must be a whole number, we round up to the nearest whole number:
n = 20.
Therefore, the least number of terms of the GP 2, 4, 8... that will give a sum greater than 1,500,000 is 20.
2. To find the total distance the ball will have traveled when it hits the ground for the eighth time, we can use the concept of rebounding height.
Given that the ball rebounds to 3/4 of its previous height, we can determine the heights at each rebound:
First drop: 3m
First rebound: 3m * (3/4) = 2.25m
Second drop: 2.25m
Second rebound: 2.25m * (3/4) = 1.6875m
And so on...
We can observe that the pattern continues with each subsequent drop and rebound.
To find the total distance traveled, we can calculate the sum of the drops and rebounds for eight times:
Total distance = 3m + 2.25m + 2.25m + 1.6875m + 1.6875m + ...
To calculate this sum, we notice that the distances form a geometric progression with a first term (a) of 3m and a common ratio (r) of (3/4).
Using the formula for the sum of an infinite GP:
Sum = a / (1 - r),
Substituting the values:
Sum = 3 / (1 - 3/4).
Simplifying:
Sum = 3 / (1/4).
Sum = 12m.
Therefore, when the ball hits the ground for the eighth time, the total distance traveled will be approximately 12 meters.