Find the sum of the nth terms of a G.P 5+15+45+...........What is the smallest number of the terms which will give a total of more than 10^8.
a = 5
r = 3
Sn = a(1-r^n)/(1-r) = 5(3^n - 1)/2
So, we want
5(3^n-1)/2 > 10^8
3^n-1 > 4*10^7
3^n > 4*10^7 - 1
n > log(4*10^7 - 1)/log3
Now, using base 10 logs, and ignoring the useless -1,
n > (
n > 7+log4)/log3
n > 15.9
So the first 16 terms will sum to more than 10^8
As a sanity check, 5*3^15 = 7*10^7 so I figure just the last two or three terms will produce the desired amount, and the first 13 terms are just noise.
To find the sum of the nth terms of a geometric progression (G.P.), we use the formula:
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.
In this case, the given G.P. is 5, 15, 45, ...
We can see that a = 5 (the first term), and to find r, we can divide any term by its previous term. Let's divide the second term by the first term:
r = 15 / 5 = 3.
Now, we have the value of r.
The formula for finding the smallest number of terms to get a sum greater than a certain value is:
n = ceil(log((S * (r - 1) / a) + 1) / log(r)),
where:
- S is the desired sum,
- ceil is the ceiling function that rounds up to the nearest whole number.
In this case, the desired sum is 10^8.
Let's calculate the smallest number of terms required:
n = ceil(log((10^8 * (3 - 1) / 5) + 1) / log(3)).
Using a calculator, we can simplify and solve this equation.
n ≈ ceil(log(6 * 10^8 / 5 + 1) / log(3)),
n ≈ ceil(log(1.2 * 10^8 + 1) / log(3)),
n ≈ ceil(log(1.2 * 10^8 + 1) / log(3)),
n ≈ ceil(log(1,200,000,001) / log(3)).
Evaluating the above expression using a calculator gives:
n ≈ ceil(11.6962 / 0.4771),
n ≈ ceil(24.4744),
n ≈ 25.
Therefore, the smallest number of terms required to get a total sum greater than 10^8 is 25 terms.