The first term of a g.p is twice it's common ratio, find the sum of the first two term of the progression, if it's sum to infinity is 8
a = 2r, so r = a/2
a/(1-r) = 8
a/(1-a/2) = 8
a = 8-4a
a = 8/5
So, the sequence is
8/5, 32/25, 128/125, ...
To find the sum of the first two terms of a geometric progression (g.p), we need to know the first term and the common ratio.
Let's assume the first term of the g.p is "a" and the common ratio is "r". According to the given information, the first term is twice the common ratio, so we have:
a = 2r
The sum of the first two terms of a g.p can be found using the formula:
S2 = a + ar = a(1 + r)
Since we are given that the sum of the progression to infinity is 8, we can use the formula for the sum of an infinite g.p:
S∞ = a / (1 - r)
Given S∞ = 8, we can substitute the value of a in terms of r into the formula:
8 = (2r) / (1 - r)
To solve this equation for r, we can multiply both sides by (1 - r):
8(1 - r) = 2r
Expanding:
8 - 8r = 2r
Rearranging:
10r = 8
Dividing both sides by 10:
r = 8/10
Simplifying:
r = 4/5
Now that we have the value of r, we can substitute it back into the equation for a:
a = 2r = 2(4/5) = 8/5
Therefore, the first term (a) is 8/5 and the common ratio (r) is 4/5.
Now, let's find the sum of the first two terms (S2):
S2 = a(1 + r) = (8/5)(1 + 4/5)
Expanding and simplifying:
S2 = (8/5)(9/5) = 72/25
So, the sum of the first two terms of the geometric progression is 72/25.