The nth term of a sequence is given by $an^2 + bn$ where a and b are integers. The 2nd term of the sequence is -2 The 4th term of the sequence is 12 (a) Find the 6th term of the sequence. Here are the first five terms of a different quadratic sequence. 0 2 6 12 20 (b) Find an expression, in terms of n, for the nth term of this sequence.

GCSE Edexcel Maths Right-Angled Triangles - Pythagoras & Trigonometry: The nth term of a sequence is gi

The nth term of a sequence is given by $an^2 + bn$ where a and b are integers - Edexcel - GCSE Maths - Question 17 - 2018 - Paper 3

Question 17

The nth term of a sequence is given by $an^2 + bn$ where a and b are integers. The 2nd term of the sequence is -2 The 4th term of the sequence is 12 (a) Find the 6... show full transcript

Worked Solution & Example Answer:The nth term of a sequence is given by $an^2 + bn$ where a and b are integers - Edexcel - GCSE Maths - Question 17 - 2018 - Paper 3

Step 1

Find the 6th term of the sequence.

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Answer

To find the 6th term, we start by establishing a pair of simultaneous equations using the known terms of the sequence.

From the 2nd term: $a(2^2) + b(2) = -2$ This simplifies to: $4a + 2b = -2$ (Equation 1)
From the 4th term: $a(4^2) + b(4) = 12$ This simplifies to: $16a + 4b = 12$ (Equation 2)

Now we can solve these two equations simultaneously.

First, let's manipulate Equation 1 to express b in terms of a:

ightarrow b = -1 - 2a$$ Substituting this expression for b into Equation 2: $$16a + 4(-1 - 2a) = 12$$ $$16a - 4 - 8a = 12$$ $$8a - 4 = 12$$ $$8a = 16 \ ightarrow a = 2$$ Now substituting back into the expression for b: $$b = -1 - 2(2) \ ightarrow b = -5$$ Now that we have both a and b, we can find the 6th term of the sequence: $$T_n = 2n^2 - 5n$$ Plugging n = 6: $$T_6 = 2(6^2) - 5(6) \ ightarrow T_6 = 72 - 30 = 42$$ Therefore, the 6th term of the sequence is 42.

Step 2

Find an expression, in terms of n, for the nth term of this sequence.

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Answer

The sequence given is 0, 2, 6, 12, 20. To find the nth term, we can observe the differences between the terms:

First differences: 2, 4, 6, 8
Second differences: 2, 2, 2

Since the second differences are constant, this indicates a quadratic sequence. Let the nth term be of the form:

$T_n = an^2 + bn + c$

Using the first three terms to construct equations:

For n = 1: $T_1 = a(1^2) + b(1) + c = 0 \rightarrow a + b + c = 0$
For n = 2: $T_2 = a(2^2) + b(2) + c = 2 \rightarrow 4a + 2b + c = 2$
For n = 3: $T_3 = a(3^2) + b(3) + c = 6 \rightarrow 9a + 3b + c = 6$

We now have three simultaneous equations:

$a + b + c = 0$ (Equation 1)
$4a + 2b + c = 2$ (Equation 2)
$9a + 3b + c = 6$ (Equation 3)

From Equation 1, express c: $c = -a - b$

Substituting c into Equations 2 and 3:

From Equation 2: $4a + 2b - a - b = 2 \rightarrow 3a + b = 2 \rightarrow b = 2 - 3a$ (Equation 4)
From Equation 3: $9a + 3b - a - b = 6 \rightarrow 8a + 2b = 6$ Substitute Equation 4 into this: $8a + 2(2 - 3a) = 6 \rightarrow 8a + 4 - 6a = 6 \rightarrow 2a = 2 \rightarrow a = 1$

Now substituting back to find b and c: $b = 2 - 3(1) = -1$ $c = -1 - 1 = -2$

Thus, the nth term is: $T_n = n^2 - n - 2$ This expression represents the nth term of the sequence.

The nth term of a sequence is given by $an^2 + bn$ where a and b are integers - Edexcel - GCSE Maths - Question 17 - 2018 - Paper 3

Worked Solution & Example Answer:The nth term of a sequence is given by $an^2 + bn$ where a and b are integers - Edexcel - GCSE Maths - Question 17 - 2018 - Paper 3

Find the 6th term of the sequence.

Find an expression, in terms of n, for the nth term of this sequence.

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