Show quadratic equation in n

3 An arithmetic progression has first term 4 and common difference \(d\). The sum of the first \(n\) terms of the progression is 5863.

1. Show that \(( n - 1 ) d = \frac { 11726 } { n } - 8\).
2. Given that the \(n\)th term is 139 , find the values of \(n\) and \(d\), giving the value of \(d\) as a fraction.

9. The first term of an arithmetic series is \(a\) and the common difference is \(d\). The 18th term of the series is 25 and the 21st term of the series is \(32 \frac { 1 } { 2 }\).

1. Use this information to write down two equations for \(a\) and \(d\).
2. Show that \(a = - 17.5\) and find the value of \(d\). The sum of the first \(n\) terms of the series is 2750 .
3. Show that \(n\) is given by $$n ^ { 2 } - 15 n = 55 \times 40 .$$
4. Hence find the value of \(n\).

5. Ben is saving for the deposit for a house over a period of 60 months. Ben saves \(\pounds 100\) in the first month and in each subsequent month, he saves \(\pounds 5\) more than the previous month, so that he saves \(\pounds 105\) in the second month, \(\pounds 110\) in the third month, and so on, forming an arithmetic sequence.

1. Find the amount Ben saves in the 40th month.
2. Find the total amount Ben saves over the 60 -month period. Lina is also saving for a deposit for a house.
   Lina saves \(\pounds 600\) in the first month and in each subsequent month, she saves \(\pounds 10\) less than the previous month, so that she saves \(\pounds 590\) in the second month, \(\pounds 580\) in the third month, and so on, forming an arithmetic sequence. Given that, after \(n\) months, Lina will have saved exactly \(\pounds 18200\) for her deposit,
3. form an equation in \(n\) and show that it can be written as $$n ^ { 2 } - 121 n + 3640 = 0$$
4. Solve the equation in part (c).
5. State, with a reason, which of the solutions to the equation in part (c) is not a sensible value for \(n\).

5. A company that owned a silver mine

• extracted 480 tonnes of silver from the mine in year 1
• extracted 465 tonnes of silver from the mine in year 2
• extracted 450 tonnes of silver from the mine in year 3
  and so on, forming an arithmetic sequence.
  1. Find the mass of silver extracted in year 14

After a total of 7770 tonnes of silver was extracted, the company stopped mining. Given that this occurred at the end of year \(N\),

show that $$N ^ { 2 } - 65 N + 1036 = 0$$

Hence, state the value of \(N\).

13. A construction company had a 30 -year programme to build new houses in Newtown. They began in the year 1991 (Year 1) and finished in 2020 (Year 30).
The company built 120 houses in year 1, 140 in year 2, 160 houses in year 3 and so on, so that the number of houses they built form an arithmetic sequence.
A total of 8400 new houses were built in year \(n\).
a. Show that $$n ^ { 2 } + 11 n - 840 = 0$$ b. Solve the equation $$n ^ { 2 } + 11 n - 840 = 0$$ and hence find in which year 8400 new houses were built.

3 The first term of an arithmetic series is 1 . The common difference of the series is 6 .

1. Find the tenth term of the series.
2. The sum of the first \(n\) terms of the series is 7400 .
   1. Show that \(3 n ^ { 2 } - 2 n - 7400 = 0\).
   2. Find the value of \(n\).

9 An arithmetic sequence has first term \(a\) and common difference \(d\). The sum of the first 36 terms of the sequence is equal to the square of the sum of the first 6 terms. 9

1. Show that \(4 a + 70 d = 4 a ^ { 2 } + 20 a d + 25 d ^ { 2 }\) 9
2. Given that the sixth term of the sequence is 25 , find the smallest possible value of \(a\).