# Eureka Math Algebra 1 Module 3 Lesson 3 Answer Key

## Engage NY Eureka Math Algebra 1 Module 3 Lesson 3 Answer Key

### Eureka Math Algebra 1 Module 3 Lesson 3 Exercise Answer Key

Exercise 2.
Think of a real – world example of an arithmetic or a geometric sequence. Describe it, and write its formula.
Answers will vary. An example of an arithmetic sequence would be a person’s salary that increases by $2,000 each year. A recursive formula would be S(n + 1) = S(n) + 2,000 for n ≥ 1 for some initial salary S(1). An example of a geometric sequence would be a person’s salary that increases by 2% each year. A recursive formula would be S(n + 1) = 1.02S(n) for n ≥ 1 for some initial salary S(1). Exercise 3. If we fold a rectangular piece of paper in half multiple times and count the number of rectangles created, what type of sequence are we creating? Can you write the formula? Answer: We are creating a geometric sequence because each time we fold, we double the number of rectangles. R(n) = 2n, where n is the number of times we have folded the paper. ### Eureka Math Algebra 1 Module 3 Lesson 3 Problem Set Answer Key For Problems 1–4, list the first five terms of each sequence, and identify them as arithmetic or geometric. Question 1. A(n + 1) = A(n) + 4 for n ≥ 1 and A(1) = – 2 Answer: – 2, 2, 6, 10, 14 Arithmetic Question 2. A(n + 1) = $$\frac{1}{4}$$ ⋅ A(n) for n ≥ 1 and A(1) = 8 Answer: 8, 2, $$\frac{1}{2}$$, $$\frac{1}{8}$$, $$\frac{1}{32}$$ Geometric Question 3. A(n + 1) = A(n) – 19 for n ≥ 1 and A(1) = – 6 Answer: – 6, – 25, – 44, – 63, – 82 Arithmetic Question 4. A(n + 1) = $$\frac{2}{3}$$ A(n) for n ≥ 1 and A(1) = 6 Answer: 6, 4, $$\frac{8}{3}$$, $$\frac{16}{9}$$, $$\frac{32}{27}$$ Geometric For Problems 5–8, identify the sequence as arithmetic or geometric, and write a recursive formula for the sequence. Be sure to identify your starting value. Question 5. 14, 21, 28, 35, … Answer: f(n + 1) = f(n) + 7 for n ≥ 1 and f(1) = 14 Arithmetic Question 6. 4, 40, 400, 4000, … Answer: f(n + 1) = 10f(n) for n ≥ 1 and f(1) = 4 Geometric Question 7. 49, 7, 1, $$\frac{1}{7}$$, $$\frac{1}{49}$$, … Answer: f(n + 1) = $$\frac{1}{7}$$f(n) for n ≥ 1 and f(1) = 49 Geometric Question 8. – 101, – 91, – 81, – 71, … Answer: f(n + 1) = f(n) + 10 for n ≥ 1 and f(1) = – 101 Arithmetic Question 9. The local football team won the championship several years ago, and since then, ticket prices have been increasing$20 per year. The year they won the championship, tickets were \$50. Write a recursive formula for a sequence that models ticket prices. Is the sequence arithmetic or geometric?
T(n) = 50 + 20n, where n is the number of years since they won the championship; n ≥ 1 (n ≥ 0 is also acceptable). The sequence is arithmetic.
OR
T(n + 1) = T(n) + 20, where n is the number of years since the year they won the championship; n ≥ 1 and T(1) = 70 (n ≥ 0 and T(0) = 50 is also acceptable). The sequence is arithmetic.

Question 10.
A radioactive substance decreases in the amount of grams by one – third each year. If the starting amount of the substance in a rock is 1,452 g, write a recursive formula for a sequence that models the amount of the substance left after the end of each year. Is the sequence arithmetic or geometric?
A(n + 1) = $$\frac{2}{3}$$ A(n) or A(n + 1) = 2A(n)÷3, where n is the number of years since the measurement started, A(0) = 1,452
The sequence is geometric.
Since the problem asked how much radioactive substance was left, students must take the original amount, divide by 3 or multiply by $$\frac{1}{3}$$, and then subtract that portion from the original amount. An easier way to do this is to just multiply by the amount remaining. If $$\frac{1}{3}$$ is eliminated, $$\frac{2}{3}$$ remains.

Question 11.
Find an explicit form f(n) for each of the following arithmetic sequences (assume a is some real number and x is some real number).
a. – 34, – 22, – 10, 2, …
f(n) = – 34 + 12(n – 1) = 12n – 46, where n ≥ 1

b. $$\frac{1}{5}$$, $$\frac{1}{10}$$, 0, – $$\frac{1}{10}$$, …
f(n) = $$\frac{1}{5}$$ – $$\frac{1}{10}$$(n – 1) = $$\frac{3}{10}$$ – $$\frac{1}{10}$$n, where n ≥ 1

c. x + 4, x + 8, x + 12, x + 16, …
f(n) = x + 4 + 4(n – 1) = x + 4n, where n ≥ 1

d. a, 2a + 1, 3a + 2, 4a + 3, …
f(n) = a + (a + 1)(n – 1) = a + an – a + n – 1 = an + n – 1, where n ≥ 1

Question 12.
Consider the arithmetic sequence 13, 24, 35, ….
a. Find an explicit form for the sequence in terms of n.
f(n) = 13 + 11(n – 1) = 11n + 2, where n ≥ 1

b. Find the 40th term.
f(40) = 442

c. If the nth term is 299, find the value of n.
299 = 11n + 2 → n = 27

Question 13.
If – 2, a, b, c, 14 forms an arithmetic sequence, find the values of a, b, and c.
14 = – 2 + (5 – 1)d
16 = 4d
d = 4

a = – 2 + 4 = 2
b = 2 + 4 = 6
c = 6 + 4 = 10

Question 14.
3 + x, 9 + 3x, 13 + 4x, … is an arithmetic sequence for some real number x.
a. Find the value of x.
The difference between term 1 and term 2 can be expressed as (9 + 3x) – (3 + x) = 6 + 2x.
The difference between term 2 and term 3 can be expressed as (13 + 4x) – (9 + 3x) = 4 + x.
Since the sequence is known to be arithmetic, the difference between term 1 and term 2 must be equal to the difference between term 2 and term 3. Thus, 6 + 2x = 4 + x, and x = – 2; therefore, the sequence is
1, 3, 5, ….

b. Find the 10th term of the sequence.
f(n) = 1 + 2(n – 1) = 2n – 1, where n ≥ 1
f(10) = 19

Question 15.
Find an explicit form f(n) of the arithmetic sequence where the 2nd term is 25 and the sum of the 3rd term and 4th term is 86.
a,25,b,c
25 = a + (2 – 1)d
25 = a + d

b = 25 + d
b = a + 2d

c = 25 + 2d
c = a + 3d
b + c = (a + 2d) + (a + 3d) = 2a + 5d = 86
a + d = 25
Solving this system: d = 12, a = 13, so f(n) = 13 + 12(n – 1), where n ≥ 1
b = 13 + 2(12) = 37
c = 13 + 3(12) = 49
OR
b + c = (25 + d) + (25 + 2d) = 50 + 3d = 86 → d = 12
25 = a + 12 → a = 13
b = 25 + 12 → b = 37
c = 25 + 2(12) → c = 49
So, f(n) = 13 + 12(n – 1).

Question 16.
Challenge: In the right triangle figure below, the lengths of the sides a cm, b cm, and c cm of the right triangle form a finite arithmetic sequence. If the perimeter of the triangle is 18 cm, find the values of a, b, and c.

a + b + c = 18 b = a + d c = a + 2d
a + (a + d) + (a + 2d) = 18
3a + 3d = 18
a + d = 6 = b
Now, do not forget that it is a right triangle, so the Pythagorean theorem must apply: a2 + b2 = c2.
Since we know that b = 6, the perimeter equation becomes a + c = 12 once b is substituted. So, substituting
c = 12 – a and b into the Pythagorean theorem equation gives us a2 + 36 = (12 – a)2, which gives us the following answer: a = $$\frac{9}{2}$$, b = 6, c = $$\frac{15}{2}$$.

Question 17.
Find the common ratio and an explicit form in each of the following geometric sequences.
a. 4, 12, 36, 108, …
r = 3 f(n) = 4(3)(n – 1), where n ≥ 1

b. 162, 108, 72, 48, …
r = $$\frac{108}{162}$$ = $$\frac{2}{3}$$ f(n) = 162($$\frac{2}{3}$$)(n – 1), where n ≥ 1

c. $$\frac{4}{3}$$, $$\frac{2}{3}$$, $$\frac{1}{3}$$, $$\frac{1}{6}$$, …
r = $$\frac{1}{2}$$ f(n) = ($$\frac{4}{3}$$) ($$\frac{1}{2}$$)(n – 1) = ($$\frac{4}{3}$$) (2)(1 – n), where n ≥ 1

d. xz, x2 z3, x3 z5, x4 z7, …
r = xz2 f(n) = xz(xz2 )(n – 1), where n ≥ 1

Question 18.
The first term in a geometric sequence is 54, and the 5th term is 2/3. Find an explicit form for the geometric sequence.
$$\frac{2}{3}$$ = 54(r)4
$$\frac{1}{81}$$ = r4
r = $$\frac{1}{3}$$ or – $$\frac{1}{3}$$
f(n) = 54($$\frac{1}{3}$$)(n – 1)

Question 19.
If 2, a, b, – 54 forms a geometric sequence, find the values of a and b.
a = 2r
b = 2(r)2
– 54 = 2(r)3
– 27 = r3
– 3 = r, so a = – 6 and b = 18

Question 20.
Find the explicit form f(n) of a geometric sequence if f(3) – f(1) = 48 and $$\frac{f(3)}{f(1)}$$ = 9.
f(3) = f(1) (r)2
$$\frac{f(3)}{f(1)}$$ = r2 = 9
r = 3 or – 3 f(1) r2 – f(1) = 48
f(1)(r2 – 1) = 48
f(1)(8) = 48
f(1) = 6
f(n) = 6(3)(n – 1), where n ≥ 1 or f(n) = 6( – 3)(n – 1), where n ≥ 1

### Eureka Math Algebra 1 Module 3 Lesson 3 Exit Ticket Answer Key

Question 1.
Write the first three terms in the following sequences. Identify them as arithmetic or geometric.
a. A(n + 1) = A(n) – 5 for n ≥ 1 and A(1) = 9
9, 4, – 1 Arithmetic

b. A(n + 1) = $$\frac{1}{2}$$ A(n) for n ≥ 1 and A(1) = 4
4, 2, 1 Geometric

c. A(n + 1) = A(n)÷10 for n ≥ 1 and A(1) = 10
10, 1, $$\frac{1}{10}$$ or 10, 1, 0.1 Geometric

Question 2.
Identify each sequence as arithmetic or geometric. Explain your answer, and write an explicit formula for the sequence.
a. 14, 11, 8, 5, …
Arithmetic
– 3 pattern
17 – 3n, where n starts at 1

b. 2, 10, 50, 250, …
c. – $$\frac{1}{2}$$, – $$\frac{3}{2}$$, – $$\frac{5}{2}$$, – $$\frac{7}{2}$$, …
$$\frac{1}{2}$$ – n, where n starts at 1