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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:
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:
Some important formulas related to different series and sequences are:
| Type | Formula | Description |
|---|---|---|
| Arithmetic Sequence | anβ = a1β + (n β 1)d | nth term of an arithmetic sequence |
| Sum of Arithmetic Series | Sn β= 2nβ(a1β + anβ) | Sum of the first n terms of an arithmetic series |
| Geometric Sequence | an β= a ββ rnβ1 | nth term of a geometric sequence |
| Sum of Geometric Series (Finite) | Snβ = a(1 β rn)/(β1 β r) | Sum of the first n terms of a geometric series |
| Sum of Geometric Series (Infinite) | Sβ = a/(1 β r)ββ (For r < 1} | Sum of the infinite geometric series where r < 1. |
| Harmonic Series | Hn β= βnk=1 (1/k) | Sum of the first n terms of the harmonic series |
Where,
Read More about Sequences and Series.
Problem 1: Find the 10th term of the arithmetic sequence where the first term a1β is 5 and the common difference d is 3.
Solution:
Using the formula for the nth term of an arithmetic sequence:
an = a1β + (n - 1)dFor the 10th term (\(n = 10\)):
- a10β = 5 + (10-1) Γ 3
- β a10β = 5 + 9 Γ 3
- β a10β = 5 + 27
- β a10β = 32
Thus, the 10th term is 32.
Problem 2: Find the 15th term of the arithmetic sequence where the first term a1β is 7 and the common difference d is 4.
Solution:
Using the formula for the nth term of an arithmetic sequence:
anβ = a1β + (n - 1)dFor the 15th term ( n = 15):
- a15β = 7 + (15 - 1) Γ 4
- β a15β = 7 + 14 Γ 4
- β a15β = 7 + 56
- β a15β = 63
Thus, the 15th term is 63.
Problem 3: Find the sum of the first 12 terms of the arithmetic series where the first term a1β 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:
Snβ = (n/2)(a1β + anβ)
First, find the 12th term (a12):
- a12β = 3 + (12-1) Γ 5
- β a12β = 3 + 11 Γ 5
- β a12β = 3 + 55 = 58
Now, find the sum S12β:
- S12 = (12/2)(3 + 58)
- β S12 = 6 Γ 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 10th term is 39.
Problem 5: Find the sum of the first 20 terms of the arithmetic series where the first term a1β 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:
Snβ = (n/2)(a1β + anβ)First, find the 20th term:
- a20β = 2 + (20-1) Γ 4
- β a20 = 2 + 19 Γ 4
- β a20 = 2 + 76
- β a20 = 78
Now, find the sum S20:
- S20 = (20/2)(2 + 78)
- β S20 = 10 Γ 80
- β 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:
- aβ = 5 (first term)
- d = 8 - 5 = 3 (common difference)
aββ = aβ + (n-1)d = 5 + (20-1)3 = 5 + 57 = 62
Sββ = (n/2)(aβ + aββ) = (20/2)(5 + 62) = 10 * 67 = 670Answer: The sum of the first 20 terms is 670.
Problem 7: Find the 5th term of the geometric sequence where the first termβ is 3 and the common ratio r is 2.
Solution:
Using the formula for the nth term of a geometric sequence:
an = a rn-1For the 5th term (\(n = 5):
- β a5 = 3 Γ 25-1
- β a5 = 3 Γ 24
- β a5 = 3 Γ 16
- β a5 = 48
Thus, the 5th 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:
an = a rn-1For the 8th term:
a8 = 2 Γ 38 - 1
β a8 = 2 Γ 37 = 2 Γ 2187 = 4374Thus, the 8th 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 Γ 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
Sβ = a / (1-r) = 1 / (1 - 1/3) = 1 / (2/3) = 3/2Thus, 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:
For the first 6 terms (n = 6):
β
β = 63Thus, 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:
For the first 7 terms:
βThus, 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:
- Fβ = 0, Fβ = 1
- Fβ = Fβ + Fβ = 0 + 1 = 1
- Fβ = Fβ + Fβ = 1 + 1 = 2
- Fβ = Fβ + Fβ = 1 + 2 = 3
- Fβ = Fβ + Fβ = 2 + 3 = 5
- Fβ = Fβ + Fβ = 3 + 5 = 8
- Fβ = Fβ + Fβ = 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).
- aβ = 3
- aβ = 3r + d = 6
- aβ = 6r + d = 15
- aβ = 15r + d = 45
Solving the system of equations:
- 6 = 3r + d
- 15 = 6r + d
- 45 = 15r + d
We find: r = 3 and d = -3
- aβ = aβr + 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
β S = (10(10+1)(2*10+1)) / 6
β S = (10 * 11 * 21) / 6
β S = 2310 / 6
β S = 385Thus, the sum of squares of the first 10 positive integers is 385
Read More about Sequences and Series Word Problems.
Problem 1: Find the 30th term of an arithmetic sequence where the first term a1β 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 a1ββ 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 a1β is 15 and the common difference d is -4.
Problem 4: The 5th term of an arithmetic sequence is 20 and the 15th term is 60. Find the first term and the common difference.
Problem 5: Find the 12th term of a geometric sequence where the first term a1β 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 a1ββ 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 a1ββ is 5 and the common ratio r is -2.
Problem 8: The 4th term of a geometric sequence is 16 and the 7th 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 a1β is 9 and the common ratio r is 1/3β.
Problem 10: Calculate the sum of the first 10 terms of the harmonic series.