How do you find the value of R in a geometric series with first term A and and common ratio R, where |R|<1.

Question

Sum of Infinity is 8.
Sum of infinity when adding up all the odd numbered terms(1st term+3rd term+5th term+.....) is 6.

Answer 1

The formula for an infinite series is:

S = A/(1 - R)

Therefore, since 8 is the infinite sum,

8 = A/(1 - R)

It is also given that adding up all the odd numbered terms is 6.

Note that normally, a geometric sequence goes as follows:

(a, ar, ar^2, ar^3, ar^4, ar^5, ar^6, .....)

Now, the odd numbered terms would go

(a, ar^2, ar^4, ar^6, ar^8, ar^10, ....)

And note that in the sequence of odd numbered terms, the ratio is r^2. Therefore, the sum of the sequence of odd numbered terms would be

S = A / (1 - R^2)

But the sum of the odd numbered terms is 6, so

6 = A / (1 - R^2)

That makes our two equations and two knowns to be:

8 = A/(1 - R)
6 = A / (1 - R^2)

Let's solve this by substitution.

For one thing, if 8 = A/(1 - R), multiplying both sides by (1 - R) gives

8(1 - R) = A

Now, all we have to do is substitute A into the second equation,
6 = A / (1 - R^2).

6 = 8(1 - R) / (1 - R^2)

But note that 1 - R^2 factors as a difference of squares; therefore,

6 = 8(1 - R) / [(1 - R)(1 + R)]

It is given that |R| < 1, so we can safely cancel out the (1 - R) on the top and bottom.

6 = 8 / (1 + R)

Now, we just solve for R using conventional algebraic methods.

6(1 + R) = 8
1 + R = 8/6
R = 8/6 - 1
R = 8/6 - 6/6
R = 2/6 = 1/3

Answer 2

Let
S = sum of series
T = sum of odd numbered terms

S = a + ar + ar^2 + ar^3 + ...
S(1 - r) = a
S = a/(1 - r) = 8
a = 8(1 - r)

T = a + ar^2 + ar^4 + ...
T(1 - r^2) = a
T = a/(1 - r^2) = 6
a = 6(1 - r^2)

Setting the two equations for a equal.

8(1 - r) = 6(1 - r^2) = 6(1 - r)(1 + r)
8 = 6(1 + r)
8/6 = 1 + r
4/3 - 1 = r
1/3 = r
r = 1/3

Answer 3

The formulas are: S_n = [a(1-r^n)]/(1-r) and S_infinity = a/(1-r) S_2 = 7.2 = [a(1-r^2)]/(1-r) and S_infinity = 20 = a/(1-r) Rearranging we get: a = 20-20r, inputting this into S_2 we get: 7.2 = [(20-20r)(1-r^2)]/(1-r) Rearranging we get: 4.47214 r = 3.57771, so r = 0.80000 to 5sf. Inputting this into one of the equations we can find a, and we get that a = 20-20(0.8), so a = 4

Answer 4

8 = A/(1-R)
6 = A/(1-R^2)
6 - 6R^2 = 8 - 8R
6R^2 - 8R + 2 = 0
R = (8 ± √(64 - 48))/12
R = (8 ± √(16))/12
R = 1, 1/3
Reject 1 as it causes division by 0.
R = 1/3
A = 16/3

Check:
6 =?(16/3)/(1 - 1/9)
6 =?(16/3)/(8/9)
6 =?(16/3)(9/8)
6 = 2*3

Answer 5

Sum to infinity, S1= A/(1-R)=8
T1=A
T3=AR^2
T5=AR^4 .............
therefore A+AR^2+AR^4+.... = 6
A(1+R^2+R^4+..........) =6
A/(1-R^2)=6 ---------------- (sum to infinity formula)
A=6(1-R^2)
sub this into S1 to get R