Questions 1-6 : Find the sum of each infinite geometric series.
Question 1 :
3 + 1 + 1/3 +.......... ∞
Question 2 :
-3 - 12 - 48 +.......... ∞
Question 3 :
1 - 3 - 48 +.......... ∞
Question 4 :
1 + 1/2 + 1/4 +.......... ∞
Question 5 :
1 - 0.5 + 0.25 + .......... ∞
Question 6 :
k = 1∑k = ∞ 5(-1/5)k-1
Questions 7 and 8 : Determine the common ratio of each infinite geometric series.
Question 7 :
a1 = 1 and S∞ = 1.25
Question 8 :
a1 = -4 and S∞ = -16/5
Questions 9 and 10 : Determine if each geometric series converges or diverges.
Question 9 :
1 + 0.5 + 0.25 + .......... ∞
Question 10 :
3 - 12 + 48 + .......... ∞
1. Answer :
3 + 1 + 1/3 +.......... ∞
a1 = 3
r = a2/a1 = 1/3
The value of r (= 1/3) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S∞ = a1/(1 - r)
Substitute a1 = 3 and r = 1/3.
S∞ = 3/(1 - 1/3)
= 3/(2/3)
= 3(3/2)
= 9/2
2. Answer :
-3 - 12 - 48 +.......... ∞
a1 = 3
r = a2/a1 = -12/(-3) = 4
Since the value of r (= 4) is not in the interval -1 < r < 1, the sum for the given infinite geometric series does not exist.
3. Answer :
1 - 3 - 48 +.......... ∞
a1 = 1
r = a2/a1 = -3/1 = -3
Since the value of r (= -3) is not in the interval -1 < r < 1, the sum for the given infinite geometric series does not exist.
4. Answer :
1 + 1/2 + 1/4 +.......... ∞
a1 = 1
r = a2/a1 = (1/2)/1 = 1/2
The value of r (= 1/2) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S∞ = a1/(1 - r)
Substitute a1 = 1 and r = 1/2.
S∞ = 1/(1 - 1/2)
= 1/(1/2)
= 1(2/1)
= 2
5. Answer :
1 - 0.5 + 0.25 + .......... ∞
a1 = 1
r = a2/a1 = -0.5/1 = -0.5
The value of r (= -0.5) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S∞ = a1/(1 - r)
Substitute a1 = 1 and r = -0.5.
S∞ = 1/[1 - (-0.5)]
= 1/(1 + 0.5)
= 1/(1.5)
= 1/(3/2)
= 1(2/3)
= 2/3
6. Answer :
k = 1∑k = ∞ 5(-1/5)k-1
Let ak = 5(-1/5)k-1.
When k = 1,
a1 = 5(-1/5)0
= 5(1)
= 5
When k = 2,
a2 = 5(-1/5)2-1
= 5(-1/5)1
= 5(-1/5)
= -1
When k = 3,
a3 = 5(-1/5)3-1
= 5(-1/5)2
= 5(1/25)
= 1/5
k = 1∑k = ∞ 5(-1/5)k-1 = 5 - 1 + 1/5 + .......... ∞
a1 = 5
r = a2/a1 = -1/5
The value of r (= -1/5) is in the interval -1 < r < 1.
So, the sum for the given infinite geometric series exists.
Formula to find the sum of infinite geometric series :
S∞ = a1/(1 - r)
Substitute a1 = 5 and r = -1/5.
S∞ = 5/[1 - (-1/5)]
= 5/(1 + 1/5)
= 5/(6/5)
= 5(5/6)
= 25/6
7. Answer :
a1 = 1 and S∞ = 1.25
S∞ = 1.25
a1/(1 - r) = 1.25
1/(1 - r) = 1.25
Take reciprocal on both sides.
(1 - r)/1 = 1/1.25
1 - r = 0.8
Subtract 1 from both sides.
-r = -0.2
Multiply both sides by -1.
r = 0.2
8. Answer :
a1 = -4 and S∞ = -16/5
S∞ = -16/5
a1/(1 - r) = -16/5
-4/(1 - r) = -16/5
Take reciprocal on both sides.
(1 - r)/(-4) = -5/16
Multiply both sides by -4.
1 - r = 5/4
Subtract 1 from both sides.
-r = 1/4
Multiply both sides by -1.
r = -1/4
9. Answer :
1 + 0.5 + 0.25 + .......... ∞
a1 = 1 and a2 = 0.5
The value of the common ratio 'r' :
r = a2/a1 = 0.5/1 = 0.5
The value of r (= 0.5) is in the interval -1 < r < 1.
So, the given infinite geometric series converges.
10. Answer :
3 - 12 + 48 + .......... ∞
a1 = 3 and a2 = -12
The value of the common ratio 'r' :
r = a2/a1 = -12/3 = -4
Since the value of r (= -4) is not in the interval -1 < r < 1, the given infinite geometric series diverges.
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