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Here are the first four terms of a sequence - OCR - GCSE Maths - Question 12 - 2019 - Paper 4

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Here are the first four terms of a sequence. -1 4 9 14 Write an expression for the $n$th term of this sequence. (b) The $n$th term of another sequence is give... show full transcript

Worked Solution & Example Answer:Here are the first four terms of a sequence - OCR - GCSE Maths - Question 12 - 2019 - Paper 4

Step 1

Write an expression for the $n$th term of this sequence.

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Answer

To find the expression for the nnth term of the sequence, we first observe the pattern in the given terms: -1, 4, 9, 14.

Calculate the differences between successive terms:

  • 4 - (-1) = 5
  • 9 - 4 = 5
  • 14 - 9 = 5

The first differences are constant, indicating this is a linear sequence. We can express the nnth term in the form:

tn=an+bt_n = an + b

Using the known terms, we can set up the following equations:

  1. For n=1n=1: t1=a(1)+b=1t_1 = a(1) + b = -1.
  2. For n=2n=2: t2=a(2)+b=4t_2 = a(2) + b = 4.

Substituting from the first equation into the second, we find:

Step 2

Find the value of a and the value of b.

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Answer

We are given the expression for the nnth term as ( an^2 + bn ). To find the values of aa and bb, we can use the information provided:

  1. The third term is equal to 9: a(32)+b(3)=99a+3b=93a+b=3 (Equation 1)a(3^2) + b(3) = 9 \Rightarrow 9a + 3b = 9 \Rightarrow 3a + b = 3 \text{ (Equation 1)}

  2. The sixth term is equal to 126: a(62)+b(6)=12636a+6b=1266a+b=21 (Equation 2)a(6^2) + b(6) = 126 \Rightarrow 36a + 6b = 126 \Rightarrow 6a + b = 21 \text{ (Equation 2)}

Now, we can solve these equations simultaneously: Subtract Equation 1 from Equation 2: (6a+b)(3a+b)=2133a=18a=6.(6a + b) - (3a + b) = 21 - 3 \Rightarrow 3a = 18 \Rightarrow a = 6. Now substituting a=6a = 6 back into Equation 1: 3(6)+b=318+b=3b=15.3(6) + b = 3 \Rightarrow 18 + b = 3 \Rightarrow b = -15.

Thus, the final values are:

  • a=6a = 6
  • b=15b = -15.

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