Sequences and Series Practice Problems

A sequence is a list of numbers arranged in a specific order, following a particular rule. Each number in the sequence is called a term. Sequences can be finite, meaning they have a definite number of terms, or infinite, meaning they continue indefinitely.

Some of the common sequences or progressions are:

• Arithmetic Sequence
• Geometric Sequence
• Harmonic Sequence

What are Series?

A series is the sum of the terms in a sequence. In mathematics, we often encounter different types of series, each defined by specific rules for summing its terms. Here are some common types of series with examples:

• S_n ​= 1 + 1/2 ​ + 1/3​ + . . . + 1/n
• e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots

Formula for Sequences and Series

Some important formulas related to different series and sequences are:

Where,

• a₁​ or a: first term
• a_n​: n^th term
• d: common difference in arithmetic sequence
• r: common ratio in geometric sequence
• n: number of terms
• S_n​: sum of the first n terms

Read More about Sequences and Series.

Solved Problems on Sequences and Series

Problem 1: Find the 10^th term of the arithmetic sequence where the first term a₁​ is 5 and the common difference d is 3.

Solution:

Using the formula for the nth term of an arithmetic sequence:
a_n = a₁​ + (n - 1)d

For the 10th term (\(n = 10\)):

• a₁₀​ = 5 + (10-1) × 3
• ⇒ a₁₀​ = 5 + 9 × 3
• ⇒ a₁₀​ = 5 + 27
• ⇒ a₁₀​ = 32

Thus, the 10^th term is 32.

Problem 2: Find the 15^th term of the arithmetic sequence where the first term a₁​ is 7 and the common difference d is 4.

Solution:

Using the formula for the n^th term of an arithmetic sequence:
a_n​ = a₁​ + (n - 1)d

For the 15th term ( n = 15):

• a₁₅​ = 7 + (15 - 1) × 4
• ⇒ a₁₅​ = 7 + 14 × 4
• ⇒ a₁₅​ = 7 + 56
• ⇒ a₁₅​ = 63

Thus, the 15^th term is 63.

Problem 3: Find the sum of the first 12 terms of the arithmetic series where the first term a1​ is 3 and the common difference d is 5.

Solution:

Using the formula for the sum of the first n terms of an arithmetic series:

S_n​ = (n/2)(a₁​ + a_n​)

First, find the 12th term (a₁₂):

• a₁₂​ = 3 + (12-1) × 5
• ⇒ a₁₂​ = 3 + 11 × 5
• ⇒ a₁₂​ = 3 + 55 = 58

Now, find the sum S₁₂​:

• S₁₂ = (12/2)(3 + 58)
• ⇒ S₁₂ = 6 × 61 = 366

Thus, the sum of the first 12 terms is 366.

Problem 4: Find the 10th term of the arithmetic sequence: 3, 7, 11, 15, ...

Solution:

a₁ = 3 (first term)

d = 7 - 3 = 4 (common difference)

a₁₀ = a₁ + (n - 1)d = 3 + (10-1)4 = 3 + 36 = 39

Answer: The 10^th term is 39.

Problem 5: Find the sum of the first 20 terms of the arithmetic series where the first term a₁​ is 2 and the common difference d is 4.

Solution:

Using the formula for the sum of the first n terms of an arithmetic series:
S_n​ = (n/2)(a₁​ + a_n​)

First, find the 20^th term:

• a₂₀​ = 2 + (20-1) × 4
• ⇒ a₂₀ = 2 + 19 × 4
• ⇒ a₂₀ = 2 + 76
• ⇒ a₂₀ = 78

Now, find the sum S₂₀:

• S₂₀ = (20/2)(2 + 78)
• ⇒ S₂₀ = 10 × 80
• ⇒ S₂₀ = 800

Thus, the sum of the first 20 terms is 800.

Problem 6: Find the sum of the first 20 terms of the arithmetic series: 5 + 8 + 11 + 14 + ...

Solution:

• a₁ = 5 (first term)
• d = 8 - 5 = 3 (common difference)

a₂₀ = a₁ + (n-1)d = 5 + (20-1)3 = 5 + 57 = 62
S₂₀ = (n/2)(a₁ + a₂₀) = (20/2)(5 + 62) = 10 * 67 = 670

Answer: The sum of the first 20 terms is 670.

Problem 7: Find the 5th term of the geometric sequence where the first term​ is 3 and the common ratio r is 2.

Solution:

Using the formula for the nth term of a geometric sequence:
a_n = a r^n-1

For the 5^th term (\(n = 5):

• ⇒ a₅ = 3 × 2^5-1
• ⇒ a₅ = 3 × 2⁴
• ⇒ a₅ = 3 × 16
• ⇒ a₅ = 48

Thus, the 5^th term is 48.

Problem 8: Find the 8th term of the geometric sequence where the first term is 2 and the common ratio r is 3.

Solution:

Using the formula for the nth term of a geometric sequence:
a_n = a r^n-1

For the 8^th term:

a₈ = 2 × 3^{8 - 1}
⇒ a₈ = 2 × 3⁷ = 2 × 2187 = 4374

Thus, the 8^th term is 4374.

Problem 9: Find the 6th term of the geometric sequence: 2, 6, 18, 54, ...

Solution:

• a₁ = 2 (first term)
• r = 6/2 = 3 (common ratio)

a₆ = a₁ × r^(n-1) = 2 * 3^5 = 2 * 243 = 486

Thus, the 6th term is 486.

Problem 10: Find the sum of the infinite geometric series: 1 + 1/3 + 1/9 + 1/27 + ...

Solution:

• a = 1 (first term)
• r = 1/3 (common ratio)

As |r| < 1, so the series converges
S_∞ = a / (1-r) = 1 / (1 - 1/3) = 1 / (2/3) = 3/2

Thus, the sum of the infinite series is 3/2.

Problem 11: Find the sum of the first 6 terms of the geometric series where the first term is 1 and the common ratio r is 2.

Solution:

Using the formula for the sum of the first n terms of a geometric series:
S_n = a_1 \frac{1 - r^n}{1 - r}

For the first 6 terms (n = 6):
S_6 = 1 \cdot \frac{1 - 2^6}{1 - 2}
⇒ S_6 = \frac{1 - 64}{1 - 2}
⇒ S_6 = \frac{-63}{-1} = 63

Thus, the sum of the first 6 terms is 63.

Problem 12: Calculate the sum of the first 7 terms of the harmonic series.

Solution:

The harmonic series is given by:
H_n = \sum_{k=1}^n \frac{1}{k}

For the first 7 terms:
H_7 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}
⇒ H_7 = 1 + 0.5 + 0.333 + 0.25 + 0.2 + 0.167 + 0.143 \approx 2.593

Thus, the sum of the first 7 terms is approximately 2.593.

Problem 13: Find the 8th term of the Fibonacci sequence: 0, 1, 1, 2, 3, 5, ...

Solution:

• F₁ = 0, F₂ = 1
• F₃ = F₁ + F₂ = 0 + 1 = 1
• F₄ = F₂ + F₃ = 1 + 1 = 2
• F₅ = F₃ + F₄ = 1 + 2 = 3
• F₆ = F₄ + F₅ = 2 + 3 = 5
• F₇ = F₅ + F₆ = 3 + 5 = 8
• F₈ = F₆ + F₇ = 5 + 8 = 13

Thus, the 8th term of the Fibonacci sequence is 13.

Problem 14: Find the 5th term of the sequence: 3, 6, 15, 45, ...

Solution:

This is an arithmetic-geometric sequence where each term is multiplied by a constant (r) and then added to a constant (d).

• a₁ = 3
• a₂ = 3r + d = 6
• a₃ = 6r + d = 15
• a₄ = 15r + d = 45

Solving the system of equations:

• 6 = 3r + d
• 15 = 6r + d
• 45 = 15r + d

We find: r = 3 and d = -3

• a₅ = a₄r + d = 45 * 3 + (-3) = 132

Thus, the 5th term is 132.

Problem 15: Find the sum of the squares of the first 10 positive integers.

Solution:

Use the formula: S = (n(n+1)(2n+1)) / 6
⇒ S = (10(10+1)(2*10+1)) / 6
⇒ S = (10 * 11 * 21) / 6
⇒ S = 2310 / 6
⇒ S = 385

Thus, the sum of squares of the first 10 positive integers is 385

Read More about Sequences and Series Word Problems.

Practice Problems on Sequences and Series

Problem 1: Find the 30^th term of an arithmetic sequence where the first term a₁​ is 8 and the common difference ddd is 5.

Problem 2: Determine the sum of the first 40 terms of an arithmetic series where the first term a₁​​ is 10 and the common difference d is 3.

Problem 3: Find the number of terms in the arithmetic sequence 3, 7, 11, ..., if the last term is 123.

Problem 3: Find the sum of the first 25 terms of an arithmetic series where the first term a₁​ is 15 and the common difference d is -4.

Problem 4: The 5^th term of an arithmetic sequence is 20 and the 15^th term is 60. Find the first term and the common difference.

Problem 5: Find the 12^th term of a geometric sequence where the first term a₁​ is 6 and the common ratio r is 2.

Problem 6: Calculate the sum of the first 10 terms of a geometric series where the first term a₁​​ is 3 and the common ratio r is 0.5.

Problem 7: Find the sum of the first 7 terms of a geometric series where the first term a₁​​ is 5 and the common ratio r is -2.

Problem 8: The 4^th term of a geometric sequence is 16 and the 7^th term is 128. Find the first term and the common ratio.

Problem 9: Determine the sum of an infinite geometric series where the first term a₁​ is 9 and the common ratio r is 1/3​.

Problem 10: Calculate the sum of the first 10 terms of the harmonic series.

Answer Key

1. 153
2. 2740
3. 31
4. -825
5. First term a₁ = 4, Common difference d = 4
6. 12288
7. 5.994
8. 215
9. First term a₁ = 2, Common ratio r = 2
10. 13.5