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
Calculus, AP?
caluculus, fouth, ration, are not words normally used by Calculus students. Choosing words, and checking the spelling, reflects on you personally. If you desire success, work on these types of details. You can do better than this.
Given that a3 is 16, and a4 is 8, the ratio must be r=.5
a3=a0*r^(3-1)
16=a0*(1/2)^(3-1)or a0=16*4
check: 16*4, 16*2, 16*1, 16*1/2, ...
sum:
Sum= ao/(1-r)=16*4/.5=16*8
check my thinking.
this is not arithmetic progression = =
third term = ar^2 and ar^2 = 16
fourth term = ar^3
ar^2 + ar^3 = 8
16 + ar^3 = 8
ar^3 = -8
ar^3/(ar^2) = -8/16
r = -1/2
then a = 64
terms are: 64 -32 16 -8 4 ...
check: is term3 = 16 ? , YES
is 16 - 8 = 8 ? , yes
sum∞ = a/(1-r) = 64/(3/2) = 128/3
Thanks, R. I mis-read the statement, I some how read into the question the fourth term was 8. Duh on me.
To solve this problem, we'll use the formulas of a geometric progression.
(a) Finding the first term and the common ratio:
Let's assume the first term of the geometric progression is "a" and the common ratio is "r".
Given that the third term is 16, we can write the equation:
a * r^2 = 16 ---- (Equation 1)
Also, the sum of the third term and the fourth term is 8, so we have:
a * r^2 + a * r^3 = 8 ---- (Equation 2)
Now, we need to solve these two equations simultaneously to find the values of "a" and "r".
We can rewrite Equation 2 by factoring out common terms:
a * (r^2 + r^3) = 8
a * r^2 * (1 + r) = 8
Substituting Equation 1 into this equation, we get:
16 * (1 + r) = 8
16 + 16r = 8
16r = 8 - 16
16r = -8
r = -8/16
r = -1/2
Substituting the value of "r" into Equation 1, we can find the value of "a":
a * (-1/2)^2 = 16
a * 1/4 = 16
a = 16 * 4
a = 64
Therefore, the first term is 64 and the common ratio is -1/2.
(b) Finding the sum of the infinite progression:
The sum of an infinite geometric progression can be found using the formula: S = a / (1 - r), where "S" is the sum, "a" is the first term, and "r" is the common ratio.
Substituting the values we found, we have:
S = 64 / (1 - (-1/2))
S = 64 / (1 + 1/2)
S = 64 / (3/2)
S = 64 * (2/3)
S = 128/3
Therefore, the sum of the infinite geometric progression is 128/3.