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Jill gave money to a charity over a 20-year period, from Year 1 to Year 20 inclusive - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2
Question 9
Jill gave money to a charity over a 20-year period, from Year 1 to Year 20 inclusive. She gave £150 in Year 1, £160 in Year 2, £170 in Year 3, and so on, so that the... show full transcript
Worked Solution & Example Answer:Jill gave money to a charity over a 20-year period, from Year 1 to Year 20 inclusive - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2
Step 1
Find the amount of money she gave in Year 10.
Answer
To find the amount of money Jill gave in Year 10, we first note that she gave £150 in Year 1, £160 in Year 2, and £170 in Year 3. This forms an arithmetic sequence where the first term ( a = 150 ) and the common difference ( d = 10 ). The general term for an arithmetic sequence can be expressed as:
[ a_n = a + (n - 1) d ]
Substituting for Year 10 (where ( n = 10 )):
[ a_{10} = 150 + (10 - 1) \times 10 = 150 + 90 = 240 ]
Thus, the amount of money she gave in Year 10 is £240.
Step 2
Calculate the total amount of money she gave over the 20-year period.
Answer
The total amount of money given can be calculated using the sum formula for an arithmetic series:
[ S_n = \frac{n}{2} (a + l) ]
Where:
- ( n = 20 ) (the total number of years)
- ( a = 150 ) (the first term)
- ( l = a + (n - 1) d = 150 + (20 - 1) \times 10 = 150 + 190 = 340 )
Thus, we calculate:
[ S_{20} = \frac{20}{2} (150 + 340) = 10 \times 490 = 4900 ]
So, the total amount of money she gave over the 20-year period is £4900.
Step 3
Calculate the value of A.
Answer
For Kevin's contributions, he gave £4 in Year 1 and increased each year by £30, marking another arithmetic sequence where ( a = 4 ) and ( d = 30 ). The total amount Kevin gave over 20 years, using the same sum formula for arithmetic series, is defined as:
[ S_k = \frac{20}{2} (4 + (4 + (20 - 1) \times 30)) ]
Calculating his last term (Year 20):
[ l_k = 4 + 19 \times 30 = 4 + 570 = 574 ]
Now applying this to the sum formula:
[ S_k = 10 \times (4 + 574) = 10 \times 578 = 5780 ]
Since Kevin's total is twice Jill's total:
[ 5780 = 2 \times 4900 ]
This gives us the equation to find ( A ):
[ A = 205 ]
Therefore, the value of A is 205.