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 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 t... 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 of the sequence defined by $an^2 + bn$ , we first need to find the values of $a$ and $b$ . We have the following equations based on the given terms:

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

Now, we have the system of equations:

$4a + 2b = -2$
$16a + 4b = 12$

We can simplify the first equation by dividing it by 2: $2a + b = -1 ag{1}$

Now we can eliminate $b$ from the second equation by substituting in equation (1). From (1), we have: $b = -1 - 2a$

Substituting this into the second equation: $16a + 4(-1 - 2a) = 12$ Expanding this gives: $16a - 4 - 8a = 12$ Combining like terms: $8a - 4 = 12$ Adding 4 to both sides: $8a = 16$ Dividing by 8: $a = 2$

Now substituting $a = 2$ back into (1): $2(2) + b = -1$ Which simplifies to:

=> b = -5$$ So, we have $a = 2$ and $b = -5$. Now, we can find the 6th term: $$2(6^2) + (-5)(6) = 2(36) - 30 = 72 - 30 = 42$$ Thus, 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 a quadratic sequence with the first five terms being 0, 2, 6, 12, and 20. To find the nth term, we identify that the first differences of the terms are:

First differences: 2, 4, 6, 8

These differences show an arithmetic sequence with a common difference of 2. Therefore, the second differences (which are constant) confirm that this is indeed a quadratic sequence.

Thus, we can express the nth term as: $T_n = An^2 + Bn + C$

To find the coefficients $A$ , $B$ , and $C$ , we can substitute values from the sequence:

For $n=1$ , $T_1 = 0$ : $A(1^2) + B(1) + C = 0$ $A + B + C = 0 ag{1}$
For $n=2$ , $T_2 = 2$ : $A(2^2) + B(2) + C = 2$ $4A + 2B + C = 2 ag{2}$
For $n=3$ , $T_3 = 6$ : $A(3^2) + B(3) + C = 6$ $9A + 3B + C = 6 ag{3}$

Now we have a system of equations:

$A + B + C = 0$
$4A + 2B + C = 2$
$9A + 3B + C = 6$

Subtracting equation (1) from (2) and then from equation (3):

From (2) - (1): $3A + B = 2 ag{4}$
From (3) - (1): $8A + 2B = 6 ag{5}$

Now divide equation (5) by 2: $4A + B = 3 ag{6}$

Now we can subtract equation (4) from equation (6): $4A + B - (3A + B) = 3 - 2$ $A = 1$

Substituting $A = 1$ back into equation (4): $3(1) + B = 2$ $B = -1$

Now substituting $A$ and $B$ into equation (1): $1 - 1 + C = 0$ $C = 0$

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

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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