the common ratio of the infinite geometric series whose 2nd term is 3/2 and its sum is 8?
2nd term is 3/2 --- ar = 3/2
sum of all terms is 8
a/(1-r) = 8
divide the first equation by the 2nd equation:
(ar) / (a/(1-r)) = (3/2) / 8
r(1-r) = 3/16
r - r^2 = 3/16
3 = 16r - 16r^2
16r^2 - 16r + 3 = 0
(4r - 1)(4r - 3) = 0
r = 1/4 or r = 3/4
check for r = 1/4
a(1/4) = 3/2
a = 6
sum(all terms) = 6/(1 - 1/4)
= 6/(3/4) = 8
2nd term = ar = 6(1/4) = 3/2
I will leave the check for the second solution up to you
=
THANKS
Ah, the common ratio of an infinite geometric series! Math just got a bit more exciting. So, let's dive in, shall we?
We know that the second term is 3/2. Let's call that term "a". And we also know that the sum of the series is 8. Let's call that value "S". Now, let me summon my mathematical powers.
The sum of an infinite geometric series can be calculated using the formula: S = a / (1 - r).
Given that S = 8, we have: 8 = a / (1 - r).
But wait, we need to find the common ratio "r". This is where the fun begins! We can rearrange the formula to solve for "r" and make it all fall into place.
Divide both sides of the equation by "a" to get: 8/a = 1 / (1 - r).
Now, multiply both sides by (1 - r): 8/a * (1 - r) = 1.
We're almost there! Distribute and rearrange: 8 - 8r = a.
But don't panic! We know that a is 3/2. So, let's plug it all in: 8 - 8r = 3/2.
Now, let's simplify: 16 - 16r = 3.
Subtract 16 from both sides: -16r = -13.
Perfect, right? Now, divide both sides by -16: r = -13 / -16.
Drumroll, please... r = 13/16! Ta-da!
So, the common ratio of the infinite geometric series is 13/16. Hope that puts a smile on your face.
To find the common ratio (r) of an infinite geometric series, we can use the formula:
S = a / (1-r),
where S is the sum of the series, a is the second term, and r is the common ratio.
In this case, we are given that the second term (a) is 3/2 and the sum (S) is 8. Plugging these values into the formula, we get:
8 = (3/2) / (1 - r).
To find the common ratio (r), we need to solve this equation for r.
Let's start by isolating (1-r) on one side:
8(1 - r) = 3/2.
We can distribute 8 to both terms on the left side:
8 - 8r = 3/2.
Next, let's isolate -8r by subtracting 8 from both sides:
-8r = 3/2 - 8.
To simplify the right side of the equation, we can write 8 as 16/2:
-8r = 3/2 - 16/2.
Combining the fractions on the right side:
-8r = (3 - 16) / 2 = -13/2.
To solve for r, we need to divide both sides of the equation by -8:
r = (-13/2) / (-8).
Dividing the fractions, we multiply by the reciprocal of the denominator:
r = (-13/2) * (-1/8) = 13/16.
Therefore, the common ratio (r) of the infinite geometric series is 13/16.