Derive the formula for the sum of n terms of a geometric progression given the first term as a and the common ratio as r.
Let's denote the sum of the first n terms of a geometric progression as Sₙ.
We know that the first term is a, so the sum of the first term is Sn = a.
The second term in the geometric progression will be a * r, the third term will be a * r^2, and so on.
So, the sum of the first two terms is S₂ = a + (a * r).
The sum of the first three terms is S₃ = a + (a * r) + (a * r^2).
And so on.
To derive a formula for the sum of the first n terms, we can observe the pattern:
S₂ = a + (a * r)
S₃ = a + (a * r) + (a * r^2)
S₄ = a + (a * r) + (a * r^2) + (a * r^3)
...
Notice that each term is the sum of the previous terms plus an additional term in the series, which is the previous term multiplied by the common ratio, r.
So, we can say that for any value of n:
Sₙ = a + (a * r) + (a * r^2) + (a * r^3) + ... + (a * r^(n-2)) + (a * r^(n-1))
Factoring out a from each term, we get:
Sₙ = a(1 + r + r^2 + r^3 + ... + r^(n-2) + r^(n-1))
This is a sum of the geometric progression, where the common ratio is r and the first term is 1.
Using the formula for the sum of a geometric series, we can rewrite the sum as:
Sₙ = a * ((1 - r^n) / (1 - r))
Thus, the formula for the sum of n terms of a geometric progression when the first term is a and the common ratio is r is:
Sₙ = a * ((1 - r^n) / (1 - r))
To derive the formula for the sum of n terms of a geometric progression, we'll start by considering the following geometric series:
S = a + ar + ar^2 + ar^3 + ... + ar^(n-1)
Now, let's multiply both sides of the equation by the common ratio 'r':
rS = ar + ar^2 + ar^3 + ... + ar^n
Now, subtracting the two equations mentioned above:
S - rS = a - ar^n
Factoring out 'S' on the left side and 'a' on the right side:
S(1 - r) = a(1 - r^n)
Dividing both sides by (1 - r):
S = (a(1 - r^n))/(1 - r)
Therefore, the formula for the sum of n terms of a geometric progression is:
S = a(1 - r^n)/(1 - r)