Worksheet · Calculus II · Unit 4A

Geometric Series Worksheet with Answers and Printable PDF

Twenty finite, infinite, shifted-index, decimal, modeling, and error-analysis problems.

20 problems60 min estimated timeFoundational to intermediate progression

What is included

Practice geometric sequences and series with complete HTML solutions and separate student and key PDFs.

Skills assessed

common ratio
finite sums
infinite sums
index shifts
applied models

Prerequisites

sequences
sigma notation

Geometric Series Worksheet instructional sequence — Twenty finite, infinite, shifted-index, decimal, modeling, and error-analysis problems. The numbered labels and written sequence preserve meaning without relying on color.

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Find the common ratio of 3, 12, 48, …
Find a_8 when a_1=5 and r=2.
Evaluate $\sum_{k=0}^{5}3(2)^k$ .
Evaluate $\sum_{k=0}^{4}16(1/2)^k$ .
Evaluate $\sum_{k=0}^{\infty}12(1/3)^k$ .
Evaluate $\sum_{k=0}^{\infty}8(-1/2)^k$ .
Decide whether $\sum_{k=0}^{\infty}5(1.02)^k$ converges.
Evaluate $\sum_{n=3}^{\infty}2(1/4)^n$ .
Rewrite $\sum_{n=1}^{\infty}3^{1-n}$ in $ar^k$ form and sum it.
Write 0.333… as a fraction.
Write 0.272727… as a fraction.
Write 0.145145… as a fraction.
A ball rebounds 70% of each previous height after a 10 m drop. Find total vertical distance.
Deposit 100 dollars at the end of each year for four years at 5 percent. Find the value immediately after the fourth deposit.
A square has area 1; each stage shades one fourth of the remaining area. Find total shaded area.
Find S_n for 7+7(0.8)+7(0.8)^2+…
An infinite geometric series has first term 6 and sum 15. Find r.
An infinite geometric series has ratio -1/4 and sum 8. Find its first term.
A student uses a/(1-r) for r=2. Diagnose the error.
Evaluate $\sum_{n=2}^{6}5(3)^{n-2}$ .

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

01

Problem 1: Find the common ratio of 3, 12, 48, …

Answer: $r=4$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Divide either term by its predecessor: $12/3=4$ and $48/12=4$ .
The constant quotient is $r=4$ .

02

Problem 2: Find a_8 when a_1=5 and r=2.

Answer: $a_8=640$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Use $a_n=a_1r^{n-1}$ .
Thus $a_8=5(2)^7=640$ .
Therefore the result is $a_8=640$ .

03

Problem 3: Evaluate $\sum_{k=0}^{5}3(2)^k$ .

Answer: $189$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

There are six terms with first term $3$ and ratio $2$ .
Using $S_6=3(1-2^6)/(1-2)$ gives $189$ .

04

Problem 4: Evaluate $\sum_{k=0}^{4}16(1/2)^k$ .

Answer: $31$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

There are five terms with first term $16$ and ratio $1/2$ .
Using $S_5=16(1-(1/2)^5)/(1-1/2)$ gives $31$ .

05

Problem 5: Evaluate $\sum_{k=0}^{\infty}12(1/3)^k$ .

Answer: $18$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Here $a=12$ , $r=1/3$ , and $|r|<1$ .
Therefore $S=a/(1-r)=12/(1-1/3)=18$ .
Therefore the result is $18$ .

06

Problem 6: Evaluate $\sum_{k=0}^{\infty}8(-1/2)^k$ .

Answer: $16/3$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Here $a=8$ , $r=-1/2$ , and $|r|<1$ .
Therefore $S=8/(1+1/2)=16/3$ .
Therefore the result is $16/3$ .

07

Problem 7: Decide whether $\sum_{k=0}^{\infty}5(1.02)^k$ converges.

Answer: $\text{diverges}$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

The common ratio is $r=1.02$ , so $|r|>1$ .
Its terms do not approach zero, and the infinite geometric series diverges.
Therefore the result is $\text{diverges}$ .

08

Problem 8: Evaluate $\sum_{n=3}^{\infty}2(1/4)^n$ .

Answer: $1/24$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

The first included term is $2(1/4)^3=1/32$ , not $2$ .
With ratio $1/4$ , the sum is $(1/32)/(1-1/4)=1/24$ .
Therefore the result is $1/24$ .

09

Problem 9: Rewrite $\sum_{n=1}^{\infty}3^{1-n}$ in $ar^k$ form and sum it.

Answer: $3/2$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

At $n=1$ the first term is $1$ , and each next term is multiplied by $1/3$ .
Thus the series is $\sum_{k=0}^{\infty}(1/3)^k$ , whose sum is $1/(1-1/3)=3/2$ .
Therefore the result is $3/2$ .

10

Problem 10: Write 0.333… as a fraction.

Answer: $1/3$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Write the decimal as $3/10+3/100+\cdots$ .
This has $a=3/10$ , $r=1/10$ , so its sum is $(3/10)/(9/10)=1/3$ .
Therefore the result is $1/3$ .

11

Problem 11: Write 0.272727… as a fraction.

Answer: $3/11$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Write the decimal as $27/100+27/10000+\cdots$ .
This has $a=27/100$ , $r=1/100$ , so its sum is $27/99=3/11$ .
Therefore the result is $3/11$ .

12

Problem 12: Write 0.145145… as a fraction.

Answer: $145/999$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Write the decimal as $145/1000+145/1000000+\cdots$ .
This has ratio $1/1000$ , so its sum is $145/999$ .

13

Problem 13: A ball rebounds 70% of each previous height after a 10 m drop. Find total vertical distance.

Answer: $170/3\text{ m}$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

The initial drop contributes $10$ m; each rebound height is traveled once up and once down.
Thus $D=10+2(7+7(0.7)+\cdots)=10+14/(1-0.7)=170/3\text{ m}$ .
Therefore the result is $170/3\text{ m}$ .

14

Problem 14: Deposit 100 dollars at the end of each year for four years at 5 percent. Find the value immediately after the fourth deposit.

Answer: $431.01\text{ dollars}$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Immediately after the fourth deposit, the four deposits have grown for three, two, one, and zero years.
The value is $100(1.05^3+1.05^2+1.05+1)=431.0125$ , or $431.01$ dollars to the nearest cent.
Therefore the result is $431.01\text{ dollars}$ .

15

Problem 15: A square has area 1; each stage shades one fourth of the remaining area. Find total shaded area.

Answer: $1$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

The shaded areas are $1/4,(3/4)(1/4),(3/4)^2(1/4),\ldots$ .
Their sum is $(1/4)/(1-3/4)=1$ .
Therefore the result is $1$ .

16

Problem 16: Find S_n for 7+7(0.8)+7(0.8)^2+…

Answer: $35(1-0.8^n)$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

The first term is $7$ , the ratio is $0.8$ , and the first $n$ terms end at exponent $n-1$ .
Therefore $S_n=7(1-0.8^n)/(1-0.8)=35(1-0.8^n)$ .
Therefore the result is $35(1-0.8^n)$ .

17

Problem 17: An infinite geometric series has first term 6 and sum 15. Find r.

Answer: $3/5$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Use $15=6/(1-r)$ .
Then $1-r=2/5$ , so $r=3/5$ , which satisfies $|r|<1$ .
Therefore the result is $3/5$ .

18

Problem 18: An infinite geometric series has ratio -1/4 and sum 8. Find its first term.

Answer: $10$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

Use $8=a/(1-(-1/4))=a/(5/4)$ .
Thus $a=10$ .
Therefore the result is $10$ .

19

Problem 19: A student uses a/(1-r) for r=2. Diagnose the error.

Answer: $|r|\ge1,\text{ so the infinite series diverges}$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

The formula $a/(1-r)$ requires $|r|<1$ .
For $r=2$ , the terms do not approach zero, so the infinite series diverges.
Therefore the result is $|r|\ge1,\text{ so the infinite series diverges}$ .

20

Problem 20: Evaluate $\sum_{n=2}^{6}5(3)^{n-2}$ .

Answer: $605$

Why this method: Geometric-series structure matches the mathematical structure before any algebraic cleanup.

At $n=2$ the first term is $5$ ; there are five terms through $n=6$ , with ratio $3$ .
Thus $S_5=5(1-3^5)/(1-3)=605$ .
Therefore the result is $605$ .

Common errors

Using the infinite formula when |r|≥1.
Mistaking the first listed term for the coefficient a after an index shift.
Forgetting that a repeating-decimal series begins at a decimal place.