Sequences and series: exam review Exam review: second paper Worked solutions

An arithmetic sequence has \(u_{4}=17\) and \(u_{12}=65\). [5]

1. (a) Find the common difference. [2]

   • M1 for \(8d=65-17\)
   • A1 6

2. (b) Find the first term. [1]

   • A1 -1

3. (c) Find the sum of the first 20 terms. [2]

   • M1 for \(\frac{20}{2}\left(2(-1)+19(6)\right)\)
   • A1 1120

A bakery sells 850 loaves in its first week. In each later week it sells 6% more loaves than in the week before. [6]

1. (a) Find the number of loaves sold in week 8. Give your answer to the nearest whole number. [2]

   • M1 for \(850(1+0.06)^{7}\)
   • A1 1278

2. (b) Find the total number of loaves sold in the first 10 weeks. Give your answer to the nearest whole number. [2]

   • M1 for \(\frac{850(1.06^{10}-1)}{0.06}\)
   • A1 11204

3. (c) Find the first week in which more than 1500 loaves are sold. [2]

   • M1 for \(850(1.06)^{n-1}>1500\), by table or solver
   • A1 week 11

The first three terms of a geometric series are \(54+43.2+34.56+\dots\) [5]

1. (a) Write down the common ratio. [1]

   • A1 0.8

2. (b) Find the 8th term. [2]

   • M1 for \(54(0.8)^7\)
   • A1 11.3

3. (c) Find the sum to infinity. [2]

   • M1 for \(\frac{54}{1-0.8}\)
   • A1 270

Two questions on infinite geometric series. [5]

1. (a) Write \(0.\overline{72}\) as a fraction in lowest terms. [3]

   • M1 for \(u_1=\frac{72}{100},\ r=\frac{1}{100}\)
   • M1 for \(\frac{72/100}{99/100}=\frac{72}{99}\)
   • A1 \(\frac{8}{11}\)

2. (b) The series \((x+5)+(x+5)^2+(x+5)^3+\dots\) converges. Find the range of values of \(x\). [2]

   • M1 for \(-1<(x+5)<1\)
   • A1 \(-6<x<-4\)

The sum of the first \(n\) terms of an arithmetic sequence is \(S_n=\frac{3n^2+9n}{2}\). [13]

1. (a) Find \(S_1\) and \(S_2\). [2]

   • A1 \(S_1=6\)
   • A1 \(S_2=15\)

2. (b) Hence write down \(u_1\) and \(u_2\), and find the common difference. [2]

   • A1 \(u_1=6\) and \(u_2=15-6=9\)
   • A1 \(d=3\)

3. (c) Find \(u_{20}\). [2]

   • M1 for \(6+19(3)\)
   • A1 63

4. (d) Find the greatest value of \(n\) for which \(u_n<100\). [2]

   • M1 for \(6+(n-1)(3)<100\), by solver or table
   • A1 n = 32

5. (e) Find the value of \(n\) for which \(S_n=1050\). [2]

   • M1 for \(\frac{3n^2+9n}{2}=1050\), by solver
   • A1 n = 25 (reject the negative root)

6. (f) The first and third terms of this arithmetic sequence are the first two terms of a geometric sequence. Find the sixth term of the geometric sequence. [3]

   • M1 for \(u_3=12\)
   • M1 for \(r=\frac{12}{6}=2\)
   • A1 \(u_6=6(2)^5=192\)

A ball is dropped from a height of 12 m onto a hard floor. After each bounce it rises to 60% of the height of the bounce before. [12]

1. (a) Find the height the ball reaches after the 6th bounce. [2]

   • M1 for \(12(0.6)^{6}\)
   • A1 0.560

2. (b) Find the number of bounces after which the ball first rises to less than 0.2 m. [3]

   • M1 for \(12(0.6)^n<0.2\)
   • M1 for a table or solver (or logarithms)
   • A1 9

3. (c) Show that the total distance the ball has travelled when it hits the floor for the third time is 35.04 m. [3]

   • M1 for the first drop 12 m
   • M1 for up and down after bounce 1: \(2(7.2)=14.4\) and after bounce 2: \(2(4.32)=8.64\)
   • A1 for \(12+14.4+8.64=35.04\) (shown)

4. (d) Find the total distance the ball travels before it comes to rest. [4]

   • M1 for a geometric series with \(u_1=14.4\), \(r=0.6\)
   • M1 for \(|r|<1\) and \(S_\infty=\frac{14.4}{1-0.6}\)
   • M1 for adding the first drop 12
   • A1 48

A geometric series has second term 21 and sum to infinity 100. [12]

1. (a) Show that \(100r^2-100r+21=0\). [2]

   • M1 for \(u_1=\frac{21}{r}\) in \(\frac{u_1}{1-r}=100\)
   • A1 for \(21=100r(1-r)\), giving \(100r^2-100r+21=0\) (shown)

2. (b) Find the two possible values of the common ratio. [2]

   • M1 for solving the quadratic (solver)
   • A1 \(r=0.3\) and \(r=0.7\)

3. (c) Find the corresponding values of the first term. [2]

   • M1 for \(u_1=\frac{21}{r}\)
   • A1 \(70\) and \(30\)

4. (d) For the series with the larger common ratio, find the sum of the first 10 terms. [3]

   • M1 for \(S_{10}=\frac{30(1-0.7^{10})}{1-0.7}\)
   • A1 97.2

5. (e) For this series, find the least value of \(n\) for which \(S_n>99\). [3]

   • M1 for \(\frac{30(1-0.7^n)}{1-0.7}>99\)
   • M1 for a table or solver
   • A1 n = 13