a sum to infinity which is three times the first term. find the first term
a/(1-r) = 3a
divide by a
1/(1-r) = 3
1 = 3 - 3r
3r = 2
r = 2/3
plug in r = 2/3
a/(1 - 2/3) = 3a
a/(1/3) = 3a
3a = 3a , which is a true statement,
so "a" can be anything.
But in a GP, the first term cannot be zero
so the first term can be any real number except zero.
To find the first term of a series for which the sum to infinity is three times the first term, we can use the formula for the sum of an infinite geometric series.
The formula for the sum of an infinite geometric series is:
S = a / (1 - r)
where S is the sum of the series, a is the first term, and r is the common ratio.
In this case, we are given that the sum to infinity is three times the first term, so we have:
S = 3a
We need to find the value of a, the first term.
Substituting the given values into the formula, we get:
3a = a / (1 - r)
Now, we need to solve for a. To do this, we can cross-multiply and simplify the equation:
3a(1 - r) = a
3a - 3ar = a
2a = 3ar
2 = 3r
r = 2/3
Now that we know the ratio (r) between the terms is 2/3, we can find the value of the first term (a) by substituting the value of r into the original equation:
3a = a / (1 - 2/3)
Simplifying further, we get:
3a = a / (1/3)
3a = 3a
a = 1
Therefore, the first term (a) of the series is 1.