Sequences and series: exam review Exam review Worked solutions

An arithmetic sequence has \(u_{3}=19\) and \(u_{10}=47\). [5]

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

   • M1 for \(7d=47-19\)
   • A1 4

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

   • A1 11

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

   • M1 for \(\frac{20}{2}\left(2(11)+19(4)\right)\)
   • A1 980

A library lends 1200 books in its first month. In each later month it lends 4% more books than in the month before. [6]

1. (a) Find the number of books lent in month 9. Give your answer to the nearest whole number. [2]

   • M1 for \(1200(1+0.04)^{8}\)
   • A1 1642

2. (b) Find the total number of books lent in the first 12 months. Give your answer to the nearest whole number. [2]

   • M1 for \(\frac{1200(1.04^{12}-1)}{0.04}\)
   • A1 18031

3. (c) Find the first month in which more than 2000 books are lent. [2]

   • M1 for \(1200(1.04)^{n-1}>2000\), by table or solver
   • A1 month 15

The first three terms of a geometric series are \(40+30+22.5+\dots\) [5]

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

   • A1 0.75

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

   • M1 for \(40(0.75)^7\)
   • A1 5.34

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

   • M1 for \(\frac{40}{1-0.75}\)
   • A1 160

Two questions on infinite geometric series. [5]

1. (a) Write \(0.\overline{36}\) as a fraction in lowest terms. [3]

   • M1 for \(u_1=\frac{36}{100},\ r=\frac{1}{100}\)
   • M1 for \(\frac{36/100}{99/100}=\frac{36}{99}\)
   • A1 \(\frac{4}{11}\)

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

   • M1 for \(-1<(x-3)<1\)
   • A1 \(2<x<4\)

The sum of the first \(n\) terms of an arithmetic sequence is \(S_n=3n^2+5n\). [13]

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

   • A1 \(S_1=8\)
   • A1 \(S_2=22\)

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

   • A1 \(u_1=8\) and \(u_2=22-8=14\)
   • A1 \(d=6\)

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

   • M1 for \(8+19(6)\)
   • A1 122

4. (d) Find the greatest value of \(n\) for which \(u_n<150\). [2]

   • M1 for \(8+(n-1)(6)<150\), by solver or table
   • A1 n = 24

5. (e) Find the value of \(n\) for which \(S_n=1300\). [2]

   • M1 for \(3n^2+5n=1300\), by solver
   • A1 n = 20 (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 \(u_3=20\)
   • M1 for \(r=\frac{20}{8}=2.5\)
   • A1 \(u_6=8(2.5)^5=781.25\)

A ball is dropped from a height of 10 m onto a hard floor. After each bounce it rises to 75% of the height of the bounce before. [12]

1. (a) Find the height the ball reaches after the 5th bounce. [2]

   • M1 for \(10(0.75)^{5}\)
   • A1 2.373

2. (b) Find the number of bounces after which the ball first rises to less than 0.5 m. [3]

   • M1 for \(10(0.75)^n<0.5\)
   • M1 for a table or solver (or logarithms)
   • A1 11

3. (c) Show that the total distance the ball has travelled when it hits the floor for the third time is 36.25 m. [3]

   • M1 for the first drop 10 m
   • M1 for up and down after bounce 1: \(2(7.5)=15\) and after bounce 2: \(2(5.625)=11.25\)
   • A1 for \(10+15+11.25=36.25\) (shown)

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

   • M1 for a geometric series with \(u_1=15\), \(r=0.75\)
   • M1 for \(|r|<1\) and \(S_\infty=\frac{15}{1-0.75}\)
   • M1 for adding the first drop 10
   • A1 70

A geometric series has second term 12 and sum to infinity 75. [12]

1. (a) Show that \(25r^2-25r+4=0\). [2]

   • M1 for \(u_1=\frac{12}{r}\) in \(\frac{u_1}{1-r}=75\)
   • A1 for \(12=75r(1-r)\), giving \(25r^2-25r+4=0\) (shown)

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

   • M1 for solving the quadratic (solver)
   • A1 \(r=0.2\) and \(r=0.8\)

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

   • M1 for \(u_1=\frac{12}{r}\)
   • A1 \(60\) and \(15\)

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

   • M1 for \(S_{10}=\frac{15(1-0.8^{10})}{1-0.8}\)
   • A1 66.9

5. (e) For this series, find the least value of \(n\) for which \(S_n>74\). [3]

   • M1 for \(\frac{15(1-0.8^n)}{1-0.8}>74\)
   • M1 for a table or solver
   • A1 n = 20