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 three and so on, so that his payments form an arithmetic sequence. (a) Find out how much Andy pays into the savings scheme in year ten. Kim starts paying money into a different savings scheme at the same time as Andy. In year one she pays in £130. Her payments increase each year so that she pays £210 in year two, £290 in year three and so on, so that her payments form a different arithmetic sequence. At the end of year N, Andy has paid, in total, twice as much money into his savings scheme as Kim has paid, in total, into her savings scheme. (b) Find the value of N.

A-Level Edexcel Maths Pure Sequences & Series: Each year, Andy pays into a savi

Each year, Andy pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 1

Question 6

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 three... show full transcript

Worked Solution & Example Answer:Each year, Andy pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 1

Step 1

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

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Answer

To determine the amount Andy pays in year ten, we use the formula for the nth term of an arithmetic sequence:

$a_n = a + (n-1)d$

Where:

$a$ is the first term, which is £600,
$d$ is the common difference, which is £120,
$n$ is the term number, which is 10.

Plugging in the values:

$a_{10} = 600 + (10-1) \times 120$

Calculating this:

$a_{10} = 600 + 9 \times 120$

$a_{10} = 600 + 1080$

$a_{10} = 1680$

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

Step 2

Find the value of N.

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Answer

First, we need to establish the total contributions for both Andy and Kim.

For Andy, the total contribution after N years is given by: $S_A = \frac{N}{2} \times (2a + (N-1)d)$ Substituting the values: $S_A = \frac{N}{2} \times (2 \times 600 + (N-1) \times 120)$

For Kim, her total contribution after N years is: $S_K = \frac{N}{2} \times (2 \times 130 + (N-1) \times 80)$ Substituting the values: $S_K = \frac{N}{2} \times (2 \times 130 + (N-1) \times 80)$

According to the problem, Andy's total is twice Kim's total: $S_A = 2S_K$

This leads to: $\frac{N}{2} \times (2 \times 600 + (N-1) \times 120) = 2 \left( \frac{N}{2} \times (2 \times 130 + (N-1) \times 80) \right)$

We can simplify this to find N. Solving this equation gives:

After simplifying and solving, we find that:
- $N = 18$ .

To summarize, the value of N is 18.

Each year, Andy pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 1

Worked Solution & Example Answer:Each year, Andy pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 1

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

Find the value of N.

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