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Each year, Andy pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 1

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Each year, Andy pays into a savings scheme. In year one he pays in £600. His payments increase by £120 each year so that he pays £720 in year two, £840 in year thr... show full transcript

Worked Solution & Example Answer:Each year, Andy pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 1

Step 1

Find out how much Andy pays into the savings scheme in year ten.

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Answer

To solve for Andy’s payment in year ten, we use the formula for the nth term of an arithmetic sequence:

an=a+(n1)da_n = a + (n-1)d

Where:

  • aa is the first term (600600)
  • dd is the common difference (120120)
  • nn is the term number (1010)

Plugging in the values:

a10=600+(101)×120a_{10} = 600 + (10-1) \times 120

Calculating the term:

a10=600+9×120=600+1080=1680a_{10} = 600 + 9 \times 120 = 600 + 1080 = 1680

Thus, Andy pays £1680 into the savings scheme in year ten.

Step 2

Find the value of N.

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Answer

Let d=80d = 80 for Kim.

For Kim’s payments:

  • In the first year, she pays £130.
  • Yearly increase is £80.

To find the total amount Kim has paid by year N, the sum formula for the first N terms is applied:

SKim=N2×[2×130+(N1)×80]S_{Kim} = \frac{N}{2} \times [2 \times 130 + (N-1) \times 80]

Now, for Andy:

  • Total payments for Andy by year N:

SAndy=N2×[2×600+(N1)×120]S_{Andy} = \frac{N}{2} \times [2 \times 600 + (N-1) \times 120]

Setting up the equation based on the provided information:

At year N, Andy has paid twice as much as Kim:

SAndy=2×SKimS_{Andy} = 2 \times S_{Kim}

This leads to:

N2×[2×600+(N1)×120]=2×N2×[2×130+(N1)×80]\frac{N}{2} \times [2 \times 600 + (N-1) \times 120] = 2 \times \frac{N}{2} \times [2 \times 130 + (N-1) \times 80]

This simplifies down to:

20N=360N20N = 360N

Solving for NN gives:

N=18N = 18

Therefore, the value of N is 18.

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