A geometric progression has a third term of 20 band sum to infinity which is three times the first term. find the first term.
B3=20
The sum=B1/(1-q)=3*B1 =>q=2/3
B3=B1*q^2
20=B1*4/9 =>B1=45
=90
To find the first term of the geometric progression, let's use the formula for the sum of an infinite geometric series.
The formula for the sum (S) of an infinite geometric series is given by:
S = a / (1 - r),
where:
- a is the first term of the series,
- r is the common ratio of the series.
In the given problem, the sum to infinity is three times the first term:
S = 3a.
We are also given that the third term is 20. In a geometric series, we can express the third term (t3) in terms of the first term (a) and the common ratio (r) as:
t3 = a * r^2.
Substituting the given values into the equation, we have:
20 = a * r^2.
Now let's express the sum S in terms of the third term t3:
S = t3 / (1 - r).
Since S is three times the first term, we can substitute 3a for S:
3a = t3 / (1 - r).
Substituting the value of t3 from the earlier equation, we have:
3a = (a * r^2) / (1 - r).
Next, let's solve this equation to find the value of r:
3a - a * r^2 = 0.
Factoring out 'a,' we have:
a * (3 - r^2) = 0.
Since the first term (a) cannot be zero, we can write:
3 - r^2 = 0.
Solving for r^2:
r^2 = 3.
Taking the square root of both sides, we get two values for r:
r = +sqrt(3) or r = -sqrt(3).
Now, let's substitute these values of r back into the equation 20 = a * r^2 to get the possible values for the first term (a).
When r = +sqrt(3):
20 = a * (sqrt(3))^2,
20 = a * 3,
a = 20 / 3.
When r = -sqrt(3):
20 = a * (-sqrt(3))^2,
20 = a * 3,
a = 20 / 3.
Therefore, the first term of the geometric progression can be either 20/3 or -20/3.