Find year when threshold exceeded

9. In the first month after opening, a mobile phone shop sold 300 phones. A model for future sales assumes that the number of phones sold will increase by \(5 \%\) per month, so that \(300 \times 1.05\) will be sold in the second month, \(300 \times 1.05 ^ { 2 }\) in the third month, and so on. Using this model, calculate

1. the number of phones sold in the 24th month,
2. the total number of phones sold over the whole 24 months. This model predicts that, in the \(N\) th month, the number of phones sold in that month exceeds 3000 for the first time.
3. Find the value of \(N\).

9. The resident population of a city is 130000 at the end of Year 1 A model predicts that the resident population of the city will increase by \(2 \%\) each year, with the populations at the end of each year forming a geometric sequence.

1. Show that the predicted resident population at the end of Year 2 is 132600
2. Write down the value of the common ratio of the geometric sequence. The model predicts that Year \(N\) will be the first year which will end with the resident population of the city exceeding 260000
3. Show that $$N > \frac { \log _ { 10 } 2 } { \log _ { 10 } 1.02 } + 1$$
4. Find the value of \(N\).
   

9. A cyclist aims to travel a total of 1200 km over a number of days. He cycles 12 km on day 1
He increases the distance that he cycles each day by \(6 \%\) of the distance cycled on the previous day, until he reaches the total of 1200 km .

1. Show that on day 8 he cycles approximately 18 km . He reaches his total of 1200 km on day \(N\), where \(N\) is a positive integer.
2. Find the value of \(N\). The cyclist stops when he reaches 1200 km .
3. Find the distance that he cycles on day \(N\). Give your answer to the nearest km .

5. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. A colony of bees is being studied. The number of bees in the colony at the start of the study was 30000 Three years after the start of the study, the number of bees in the colony is 34000 A model predicts that the number of bees in the colony will increase by \(p \%\) each year, so that the number of bees in the colony at the end of each year of study forms a geometric sequence. Assuming the model,

1. find the value of \(p\), giving your answer to 2 decimal places. According to the model, at the end of \(N\) years of study the number of bees in the colony exceeds 75000
2. Find, showing all steps in your working, the smallest integer value of \(N\).

7. (i) A geometric sequence has first term 4 and common ratio 6 Given that the \(n ^ { \text {th } }\) term is greater than \(10 ^ { 100 }\), find the minimum possible value of \(n\).
(ii) A different geometric sequence has first term \(a\) and common ratio \(r\). Given that

• the second term of the sequence is - 6
• the sum to infinity of the series is 25
  1. show that

$$25 r ^ { 2 } - 25 r - 6 = 0$$

Write down the solutions of $$25 r ^ { 2 } - 25 r - 6 = 0$$ Hence,

state the value of \(r\), giving a reason for your answer,

find the sum of the first 4 terms of the series. \includegraphics[max width=\textwidth, alt={}, center]{124ee19f-8a49-42df-9f4b-5a1cc2139be9-23_70_37_2617_1914}

4. The first term of a geometric series is 5 and the common ratio is 1.2 For this series find, to 1 decimal place,

1. 1. the \(20 ^ { \text {th } }\) term,
   2. the sum of the first 20 terms. The sum of the first \(n\) terms of the series is greater than 3000
2. Calculate the smallest possible value of \(n\).

8. A trading company made a profit of \(\pounds 50000\) in 2006 (Year 1). A model for future trading predicts that profits will increase year by year in a geometric sequence with common ratio \(r , r > 1\). The model therefore predicts that in 2007 (Year 2) a profit of \(\pounds 50000 r\) will be made.

1. Write down an expression for the predicted profit in Year \(n\). The model predicts that in Year \(n\), the profit made will exceed \(\pounds 200000\).
2. Show that \(n > \frac { \log 4 } { \log r } + 1\). Using the model with \(r = 1.09\),
3. find the year in which the profit made will first exceed \(\pounds 200000\),
4. find the total of the profits that will be made by the company over the 10 years from 2006 to 2015 inclusive, giving your answer to the nearest \(\pounds 10000\).

11 There is a flowerhead at the end of each stem of an oleander plant. The next year, each flowerhead is replaced by three stems and flowerheads, as shown in Fig. 11. \begin{figure}[h] \end{figure}

1. How many flowerheads are there in year 5?
2. How many flowerheads are there in year \(n\) ?
3. As shown in Fig. 11, the total number of stems in year 2 is 4, (that is, 1 old one and 3 new ones). Similarly, the total number of stems in year 3 is 13 , (that is, \(1 + 3 + 9\) ). Show that the total number of stems in year \(n\) is given by \(\frac { 3 ^ { n } - 1 } { 2 }\).
4. Kitty's oleander has a total of 364 stems. Find
   (A) its age,
   (B) how many flowerheads it has.
5. Abdul's oleander has over 900 flowerheads. Show that its age, \(y\) years, satisfies the inequality \(y > \frac { \log _ { 10 } 900 } { \log _ { 10 } 3 } + 1\).
   Find the smallest integer value of \(y\) for which this is true.

7. A student completes a mathematics course and begins to work through past exam papers. He completes the first paper in 2 hours and the second in 1 hour 54 minutes. Assuming that the times he takes to complete successive papers form a geometric sequence,

1. find, to the nearest minute, how long he will take to complete the fifth paper,
2. show that the total time he takes to complete the first eight papers is approximately 13 hours 28 minutes,
3. find the least number of papers he must work through if he is to complete a paper in less than one hour.

8

1. The first term of a geometric progression is 50 and the common ratio is 0.8 . Use logarithms to find the smallest value of \(k\) such that the value of the \(k\) th term is less than 0.15 .
2. In a different geometric progression, the second term is - 3 and the sum to infinity is 4 . Show that there is only one possible value of the common ratio and hence find the first term. \section*{Question 9 begins on page 4.}

4 An artist is creating a design for a large painting. The design includes a set of steps of varying heights. In the painting the lowest step has height 20 cm and the height of each other step is \(5 \%\) less than the height of the step immediately below it. In the painting the total height of the steps is 205 cm , correct to the nearest centimetre. Determine the number of steps in the design.

The first term of a geometric series is 120. The sum to infinity of the series is 480.

1. Show that the common ratio, \(r\), is \(\frac{3}{4}\). [3]
2. Find, to 2 decimal places, the difference between the 5th and 6th terms. [2]
3. Calculate the sum of the first 7 terms. [2]

The sum of the first \(n\) terms of the series is greater than 300.

1. Calculate the smallest possible value of \(n\). [4]

A trading company made a profit of £50 000 in 2006 (Year 1). A model for future trading predicts that profits will increase year by year in a geometric sequence with common ratio \(r, r > 1\). The model therefore predicts that in 2007 (Year 2) a profit of £50 000r will be made.

1. Write down an expression for the predicted profit in Year \(n\). [1]

The model predicts that in Year \(n\), the profit made will exceed £200 000.

1. Show that \(n > \frac{\log 4}{\log r} + 1\). [3]

Using the model with \(r = 1.09\),

1. find the year in which the profit made will first exceed £200 000, [2]
2. find the total of the profits that will be made by the company over the 10 years from 2006 to 2015 inclusive, giving your answer to the nearest £10 000. [3]

The amounts of oil pumped from an oil well in each of the years 2001 to 2004 formed a geometric progression with common ratio 0.9. The amount pumped in 2001 was 100 000 barrels.

1. Calculate the amount pumped in 2004. [2]

It is assumed that the amounts of oil pumped in future years will continue to follow the same geometric progression. Production from the well will stop at the end of the first year in which the amount pumped is less than 5000 barrels.

1. Calculate in which year the amount pumped will fall below 5000 barrels. [4]
2. Calculate the total amount of oil pumped from the well from the year 2001 up to and including the final year of production. [3]

\includegraphics{figure_12} A branching plant has stems, nodes, leaves and buds. • There are 7 leaves at each node. • From each node, 2 new stems grow. • At the end of each final stem, there is a bud. Fig. 12 shows one such plant with 3 stages of nodes. It has 15 stems, 7 nodes, 49 leaves and 8 buds.

1. One of these plants has 10 stages of nodes.
   1. How many buds does it have? [2]
   2. How many stems does it have? [2]
2. 1. Show that the number of leaves on one of these plants with \(n\) stages of nodes is $$7(2^n - 1).$$ [2]
   2. One of these plants has \(n\) stages of nodes and more than 200000 leaves. Show that \(n\) satisfies the inequality \(n > \frac{\log_{10} 200007 - \log_{10} 7}{\log_{10} 2}\). Hence find the least possible value of \(n\). [4]

Jill has 3 daughters and no sons. They are generation 1 of Jill's descendants. Each of her daughters has 3 daughters and no sons. Jill's 9 granddaughters are generation 2 of her descendants. Each of her granddaughters has 3 daughters and no sons; they are descendant generation 3. Jill decides to investigate what would happen if this pattern continues, with each descendant having 3 daughters and no sons.

1. How many of Jill's descendants would there be in generation 8? [2]
2. How many of Jill's descendants would there be altogether in the first 15 generations? [3]
3. After \(n\) generations, Jill would have over a million descendants altogether. Show that \(n\) satisfies the inequality $$n > \frac{\log_{10}2000003}{\log_{10}3} - 1.$$ Hence find the least possible value of \(n\). [4]
4. How many fewer descendants would Jill have altogether in 15 generations if instead of having 3 daughters, she and each subsequent descendant has 2 daughters? [3]

The first term of a geometric series is 120. The sum to infinity of the series is 480.

1. Show that the common ratio, \(r\), is \(\frac{3}{4}\). [3]
2. Find, to 2 decimal places, the difference between the 5th and 6th term. [2]
3. Calculate the sum of the first 7 terms. [2]

The sum of the first \(n\) terms of the series is greater than 300.

1. Calculate the smallest possible value of \(n\). [4]

A geometric progression with common ratio \(r\) consists of positive terms. The sum of the first four terms is five times the sum of the first two terms.

1. Find an equation in \(r\) and deduce that \(r = 2\). [3]
2. Given that the fifth term is 192, find the value of the first term. [1]
3. Find the smallest value of \(n\) such that the sum of the first \(n\) terms of the progression exceeds \(10^{64}\). [3]