The third term of a geometric progression is 16. The sum of the third term and the fouth term is 8. Find
(a)the first term and the common ration of the progression
(b)the sum of infinity of the progression
What is Caluculus ?
I thought you would have at least corrected your spelling from the last post after it was pointed out to you.
see post in
http://www.jiskha.com/display.cgi?id=1322746626
To solve this problem, we will use the formula for the nth term of a geometric progression:
Tn = ar^(n-1)
where Tn is the nth term, a is the first term, r is the common ratio, and n is the term number.
(a) Finding the first term and the common ratio:
Given that the third term of the geometric progression is 16, we can substitute n = 3 and Tn = 16 into the formula:
16 = ar^(3-1)
16 = ar^2
Similarly, we know that the sum of the third term and the fourth term is 8:
T3 + T4 = 8
Substituting the formula for Tn, we get:
ar^2 + ar^3 = 8
To solve these two equations, we can divide the second equation by the first equation:
(ar^2 + ar^3) / (ar^2) = 8 / 16
1 + r = 0.5
r = 0.5 - 1
r = -0.5
Substituting the value of r back into either equation, we can solve for a. Let's use the first equation:
16 = a(-0.5)^2
16 = a * 0.25
a = 16 / 0.25
a = 64
Therefore, the first term (a) is 64, and the common ratio (r) is -0.5.
(b) Finding the sum of infinity of the progression:
The sum of an infinite geometric progression can be found using the formula:
S∞ = a / (1 - r)
where S∞ is the sum of infinity, a is the first term, and r is the common ratio.
Substituting the values we found earlier, we get:
S∞ = 64 / (1 - (-0.5))
S∞ = 64 / (1 + 0.5)
S∞ = 64 / 1.5
S∞ = 42.67
Therefore, the sum of infinity of the geometric progression is approximately 42.67.