Arithmetic and Geometric Sequences and Series Practice Questions — ACT — Page 2

Question 21Question →

The first term of an arithmetic sequence is $\frac{1}{2}$ , and the second term is $\frac{5}{6}$ . The first term of a geometric sequence is $2^2$ , and the common ratio is $2^3$ . Let $A$ be the third term of the arithmetic sequence, and let $G$ be the third term of the geometric sequence. If a third value, $V$ , is defined as $\frac{5}{6}$ less than twice $G$ , what is the value of $V + A$ ?

\frac{1541}{3}

512

\frac{1537}{3}

-510

\frac{24577}{3}

Show answer & explanation

Answer: $\frac{1537}{3}$

Answer

\frac{1537}{3}

The correct answer is

\frac{1537}{3}

because calculating the third term of the arithmetic sequence yields

A = \frac{7}{6}

and the third term of the geometric sequence yields

G = 256

. Translating the relationship for

V

yields

V = 2(256) - \frac{5}{6} = 512 - \frac{5}{6}

. Finding the sum of

V

and

A

yields

512 - \frac{5}{6} + \frac{7}{6} = 512\frac{1}{3} = \frac{1537}{3}

.

Step-by-Step Solution

1

Calculate the common difference

d

of the arithmetic sequence.

d = a_2 - a_1 = \frac{5}{6} - \frac{1}{2} = \frac{1}{3}

The common difference is the difference between any term and the preceding term in an arithmetic sequence.

2

Calculate the third term

A

of the arithmetic sequence.

A = a_1 + 2d = \frac{1}{2} + 2\left(\frac{1}{3}\right) = \frac{7}{6}

The

n

-th term of an arithmetic sequence is given by

a_n = a_1 + (n-1)d

.

3

Calculate the third term

G

of the geometric sequence.

G = g_1 \cdot r^2 = 2^2 \cdot (2^3)^2 = 2^2 \cdot 2^6 = 2^8 = 256

The

n

-th term of a geometric sequence is given by

g_n = g_1 \cdot r^{n-1}

.

4

Set up and solve for

V

using the algebraic relationship described.

V = 2G - \frac{5}{6} = 2(256) - \frac{5}{6} = 512 - \frac{5}{6}

The phrase '5/6 less than twice G' translates to

2G - \frac{5}{6}

.

5

Compute the sum of

V

and

A

.

V + A = \left(512 - \frac{5}{6}\right) + \frac{7}{6} = 512 + \frac{2}{6} = 512 + \frac{1}{3} = \frac{1537}{3}

Substitute the values of

V

and

A

and simplify the resulting fractional expression.

Key Concept

Arithmetic and Geometric Sequences and Series

Estimated Time:1m 30s

Question 22Question →

A geometric sequence consists of positive terms and has a first term of $12$ and a common ratio of $r$ . An arithmetic sequence has a first term of $5$ and a common difference of $d$ . If the $3\text{rd}$ term of the geometric sequence is $3$ and the $4\text{th}$ term of the arithmetic sequence is $7$ , what is the value of $r + d$ ?

\frac{3}{5}

\frac{11}{8}

\frac{7}{6}

-\frac{15}{2}

\frac{35}{48}

Show answer & explanation

Answer: $\frac{7}{6}$

Answer

\frac{7}{6}

The correct value is

\frac{7}{6}

. This is found by first calculating the common ratio of the geometric sequence, where

12r^2 = 3 \Rightarrow r^2 = \frac{1}{4} \Rightarrow r = \frac{1}{2}

, and the common difference of the arithmetic sequence, where

5 + 3d = 7 \Rightarrow 3d = 2 \Rightarrow d = \frac{2}{3}

. Adding these two fractions with a common denominator yields

\frac{3}{6} + \frac{4}{6} = \frac{7}{6}

.

Step-by-Step Solution

1

Find the common ratio

r

of the geometric sequence.

r = \frac{1}{2}

The formula for the

n\text{th}

term of a geometric sequence is

g_n = g_1 \cdot r^{n-1}

. For the

3\text{rd}

term,

g_3 = 12r^2 = 3

, which simplifies to

r^2 = \frac{1}{4}

. Since the sequence has positive terms, we take the positive square root to get

r = \frac{1}{2}

.

2

Find the common difference

d

of the arithmetic sequence.

d = \frac{2}{3}

The formula for the

n\text{th}

term of an arithmetic sequence is

a_n = a_1 + (n-1)d

. For the

4\text{th}

term,

a_4 = 5 + 3d = 7

, which simplifies to

3d = 2

, or

d = \frac{2}{3}

.

3

Calculate the sum of

r

and

d

.

r + d = \frac{7}{6}

Adding the two values with a common denominator of

6

gives

\frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6}

.

Key Concept

Arithmetic and Geometric Sequences and Series

Estimated Time:1m 30s

Question 23Question →

A sequence of numbers $t_1, t_2, t_3, \dots$ is defined by $t_1 = 3$ and $t_{n+1} = 3t_n - 2^n$ for all integers $n \geq 1$ . What is the value of the fourth term, $t_4$ ?

5

9

43

62

119

Show answer & explanation

Answer: 43

Answer

The value of the fourth term is 43.

To find

t_4

, we use the recursive formula

t_{n+1} = 3t_n - 2^n

. Substituting

n=1

gives

t_2 = 3(3) - 2 = 7

. Substituting

n=2

gives

t_3 = 3(7) - 4 = 17

. Substituting

n=3

gives

t_4 = 3(17) - 8 = 43

.

Step-by-Step Solution

1

Calculate the second term,

t_2

, by substituting

n = 1

and

t_1 = 3

into the formula

t_{n+1} = 3t_n - 2^n

.

t_2 = 3t_1 - 2^1 = 3(3) - 2 = 7

To progress to the fourth term, we must first find each preceding term in the sequence.

2

Calculate the third term,

t_3

, by substituting

n = 2

and

t_2 = 7

into the formula.

t_3 = 3t_2 - 2^2 = 3(7) - 4 = 17

Using the value of the second term allows us to find the third term.

3

Calculate the fourth term,

t_4

, by substituting

n = 3

and

t_3 = 17

into the formula.

t_4 = 3t_3 - 2^3 = 3(17) - 8 = 43

This completes the recursive process to find the target term.

Key Concept

Evaluating terms of a sequence defined by a recursive formula.

Estimated Time:1m 30s

Question 24Question →

A geometric sequence of positive terms has a first term of $9$ and a third term of $4$ . What is the sum of the first $4$ terms of this sequence?

\frac{1}{3}

\frac{27}{4}

19

21

\frac{65}{3}

Show answer & explanation

Answer: $\frac{65}{3}$

Answer

The correct answer is

\frac{65}{3}

.

To find the sum of the first 4 terms of a geometric sequence with

a_1 = 9

and

a_3 = 4

, we first determine the common ratio

r

. Since

a_3 = a_1 \cdot r^2

, we have

4 = 9 \cdot r^2

, which gives

r^2 = \frac{4}{9}

. Because the terms are positive,

r = \frac{2}{3}

. The first 4 terms are

9

,

6

,

4

, and

\frac{8}{3}

. Summing these terms gives

9 + 6 + 4 + \frac{8}{3} = 19 + \frac{8}{3} = \frac{65}{3}

.

Step-by-Step Solution

1

Find the common ratio

r

of the geometric sequence.

r = \frac{2}{3}

Since the sequence is geometric, the third term is related to the first term by

a_3 = a_1 \cdot r^2

. Substituting the given values yields

4 = 9 \cdot r^2

, which simplifies to

r^2 = \frac{4}{9}

. Because all terms in the sequence are positive,

r

must be positive, so

r = \sqrt{\frac{4}{9}} = \frac{2}{3}

.

2

Calculate the first 4 terms of the sequence.

a_1 = 9

,

a_2 = 6

,

a_3 = 4

,

a_4 = \frac{8}{3}

Multiply each term by the common ratio

r = \frac{2}{3}

to find the subsequent term:

a_1 = 9

,

a_2 = 9 \cdot \frac{2}{3} = 6

,

a_3 = 6 \cdot \frac{2}{3} = 4

, and

a_4 = 4 \cdot \frac{2}{3} = \frac{8}{3}

.

3

Sum the first 4 terms of the sequence.

Sum =

\frac{65}{3}

Add the four terms:

9 + 6 + 4 + \frac{8}{3} = 19 + \frac{8}{3} = \frac{57}{3} + \frac{8}{3} = \frac{65}{3}

.

Key Concept

Calculating the sum of the first

n

terms of a geometric sequence given its first and third terms.

Question 25Question →

A geometric sequence has a first term of $\frac{1}{2}$ and a common ratio of $\frac{1}{4}$ . What is the sum of the first 3 terms of this sequence?

\frac{1}{14}

\frac{13}{16}

\frac{21}{32}

\frac{1}{32}

\frac{9}{128}

Show answer & explanation

Answer: $\frac{21}{32}$

Answer

The sum of the first 3 terms of this sequence is

\frac{21}{32}

.

The sum of the first three terms of a geometric sequence is calculated by finding each individual term and then adding them together. The first term is

\frac{1}{2}

. The second term is obtained by multiplying the first term by the common ratio:

\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

. The third term is obtained by multiplying the second term by the common ratio:

\frac{1}{8} \times \frac{1}{4} = \frac{1}{32}

. To add these terms, we find a common denominator of 32:

\frac{16}{32} + \frac{4}{32} + \frac{1}{32} = \frac{21}{32}

.

Step-by-Step Solution

1

Identify the first three terms of the geometric sequence using the formula

a_n = a_1 \cdot r^{n-1}

.

The first term

a_1

is given as

\frac{1}{2}

. The second term is

a_2 = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}

. The third term is

a_3 = \frac{1}{8} \cdot \frac{1}{4} = \frac{1}{32}

.

Before calculating the sum, each individual term to be summed must be determined.

2

Find a common denominator to add the three fractional terms.

The least common multiple of the denominators 2, 8, and 32 is 32. Express the terms with this common denominator:

a_1 = \frac{16}{32}

,

a_2 = \frac{4}{32}

, and

a_3 = \frac{1}{32}

.

Adding fractions requires a common denominator.

3

Sum the adjusted fractions.

\frac{16}{32} + \frac{4}{32} + \frac{1}{32} = \frac{21}{32}

.

This yields the total sum of the first three terms.

Key Concept

Calculating the sum of a finite geometric series by finding and summing individual terms.

Arithmetic and Geometric Sequences and Series

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