Linear and Geometric Sequences for GCSE, IGCSE & A Level

E.g.1: Find the nth term of the sequence: 3, 7, 11, 15, ...

The common difference is 4. 

Let N = an + b, where N is the nth term and a an b are constants to be found.

3 = a(1) + b = a + b

7 =  a(2) + b = 2a + b

Solving the two equations simultaneously, we get a = 4 and b = -1

So, the nth term, N = 4n - 1.

E.g.2 :  Given the nth term of a sequence is 5n + 2, find the 10th term. Check whether 301 is a term of this sequence.

Substitute n = 10: 5(10) + 2 = 52.

301 = 5n + 2

5n = 299

n = 59.8

Since n is not a whole number, 301 is not a term of the sequence.

E.g.3: Find the nth term of the sequence: 1, 4, 9, 16, ...

The sequence of differences is: 3, 5, 7, ... which is linear. The main sequence is, therefore, quadratic.

Let N = an² + bn + c, where n is the term. a, b and c are constants  to be found.

1 = a(1²) + b(1) + c = a + b + c → 1)

4 = a(2²) + b(2) + c = 4a + 2b + c →2)

9 = a(3²) + b(3) + c = 9a + 3b + c → 3)

2) - 1) → 3 = 3a + b → 4)

3) - 2) → 5 = 5a + b → 5)

5) - 4) → 2 = 2a → a = 1

Sub in 5) 

5 = 5 + b → b = 0

Sub in 1)

1 = 1 + 0 + c → c = 0

The nth term is,

N = n²

E.g.4: Find the nth term of the sequence: 2, 5, 10, 17, ...

The sequence of differences is: 3, 5, 7, ... which is linear. The main sequence is, therefore, quadratic.

Let N = an² + bn + c, where n is the term. a, b and c are constants  to be found.

2 = a(1²) + b(1) + c = a + b + c → 1)

5 = a(2²) + b(2) + c = 4a + 2b + c →2)

10 = a(3²) + b(3) + c = 9a + 3b + c → 3)

2) - 1) → 3 = 3a + b → 4)

3) - 2) → 5 = 5a + b → 5)

5) - 4) → 2 = 2a → a = 1

Sub in 5) 

5 = 5 + b → b = 0

Sub in 1)

2 = 1 + 0 + c → c = 1

The nth term is,

N = n² + n

E.g.5: Find the nth term of the sequence: 3, 7, 13, 21, ...

The sequence of differences is: 4, 6, 8, ... which is linear. The main sequence is, therefore, quadratic.

Let N = an² + bn + c, where n is the term. a, b and c are constants  to be found.

3 = a(1²) + b(1) + c = a + b + c → 1)

7 = a(2²) + b(2) + c = 4a + 2b + c →2)

13 = a(3²) + b(3) + c = 9a + 3b + c → 3)

2) - 1) → 4 = 3a + b → 4)

3) - 2) → 6 = 5a + b → 5)

5) - 4) → 2 = 2a → a = 1

Sub in 5) 

6 = 5 + b → b = 1

Sub in 1)

3 = 1 + 1 + c → c = 1

The nth term is,

N = n² + n + n

E.g.6: Given the nth term of a quadratic sequence is 2n² - 3n + 4, find the 5th term.

Substitute n = 5: 2(5)² - 3(5) + 4 = 39.

E.g.7: Determine if the sequence 2, 5, 10, 17, ... is linear, quadratic, or neither.

The sequence of differences is: 3, 5, 7, ... which is linear. Therefore, the sequence is quadratic.

E.g.8: Find the nth term of the sequence: 2, 3, 6, 11, ...

The sequence of differences is: 1, 3, 5, ... which is linear. Therefore, the main sequence is quadratic.

Let N = an² + bn + c, where n is the term. a, b and c are constants  to be found.

2 = a(1²) + b(1) + c = a + b + c → 1)

3 = a(2²) + b(2) + c = 4a + 2b + c →2)

6 = a(3²) + b(3) + c = 9a + 3b + c → 3)

2) - 1) → 1 = 3a + b → 4)

3) - 2) → 3 = 5a + b → 5)

5) - 4) → 2 = 2a → a = 1

Sub in 5) 

3 = 5 + b → b = -2

Sub in 1)

2 = 1 - 2 + c → c = 3

The nth term is,

N = n² -2n + 3

E.g.9: Given the nth term of a sequence is 3n² - 2n, find the term that is equal to 65.

3n² - 2n = 65

3n² - 2n - 65 = 0

To factorize the equation 3n² - 2n - 65 = 0, we need to find two numbers that multiply to -195 (3 * -65) and add up to -2.

These numbers are -15 and 13.

So, we can rewrite the equation as: 3n² - 15n + 13n - 65 = 0

Now, we can factor by grouping: 3n(n - 5) + 13(n - 5) = 0

We can see that (n - 5) is a common factor. So, we can factor it out: (n - 5)(3n + 13) = 0

Now, we can solve for n by setting each factor equal to zero: n - 5 = 0 or 3n + 13 = 0

Solving the first equation, we get: n = 5

Solving the second equation, we get: 3n = -13 n = -13/3

Therefore, the solutions to the equation 3n² - 2n - 65 = 0 are n = 5 and n = -13/3.

Since the term cannot be a fraction, n = 5.

It is the 5th term of the sequence.

E.g.10: Find the nth term of the sequence: 1, 5, 13, 25, ...

The sequence of differences is: 4, 8, 12, ... which is linear.

Let N = an² + bn + c, where n is the term. a, b and c are constants  to be found.

1 = a(1²) + b(1) + c = a + b + c → 1)

5 = a(2²) + b(2) + c = 4a + 2b + c →2)

13 = a(3²) + b(3) + c = 9a + 3b + c → 3)

2) - 1) → 4 = 3a + b → 4)

3) - 2) → 8 = 5a + b → 5)

5) - 4) → 4 = 2a → a = 2

Sub in 5) 

8 = 10 + b → b = -2

Sub in 1)

1 = 2 - 2 + c → c = 1

The nth term is,

N = 2n² -2n + 1