SME
Grade 11 Math · Problem type
Infinite geometric series
Use this for “Find the sum of 12 + 3 + 3/4 + ….”
Chart
Terms: 12, 3, 0.75, 0.1875, 0.04688, 0.01172 (multiply by r = 0.25 each time).
a₆ = 12 × 0.25⁵ = 0.01172
S₆
=
12
×
1 −
0.25
6
1 − 0.25
=
15.9961
S
=
12
1 − 0.25
=
16
S₆ is 0.003906 from S; more terms close the gap.
This is an example. Type your own values to start a new problem.
S
=
a₁
1
−
n
Terms added so far
example
Sₙ
Sum of the first n terms
worked out (not typed)
Formulas
Sₙ
=
a₁
×
1 −
r
n
1 −
r
15.9961
=
12
×
(1
−
(
1
4
)
6
)
÷
(1
−
1
4
)
S
=
a₁
1 −
r
16
=
12
1 −
1
4
Assumptions
•
With r between −1 and 1, rⁿ shrinks toward 0 as n grows.
•
The partial sums Sₙ = a₁(1 − rⁿ) ÷ (1 − r) close in on S = a₁ ÷ (1 − r).
•
With |r| ≥ 1 the terms do not shrink, and the sum has no limit.
Step-by-step
WE KNOW
a₁ = 12, r = 1/4, n = 6
FIND
sum of the first n terms (Sₙ), sum of the series (S)
Step 1 · Find sum of the first n terms (Sₙ)
Use Sₙ = a₁ × (1 − rⁿ) ÷ (1 − r)
The
first
n
terms
of
the
geometric
series.
Sₙ
=
12
×
(1
−
(
1
4
)
6
)
÷
(1
−
1
4
)
Sₙ
=
12
×
1 −
1
4096
1 −
1
4
Sₙ
=
12
×
4095
4096
3
4
Sₙ = 15.9961
Step 2 · Find sum of the series (S)
Use S = a₁ ÷ (1 − r)
As
n
grows,
rⁿ
goes
to
0,
leaving
a₁
1 −
r
.
S
=
12
1 −
1
4
S
=
12
3
4
S
=
16
Check
Put the numbers back in. Both sides should match.
15.9961
≈
12
×
(1
−
(
1
4
)
6
)
÷
(1
−
1
4
)
✓
16
=
12
1 −
1
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.