Exercise practice: Sequences and Series

4.1 How the exercises are laid out

Exercise 8.1 builds terms and sums from nth-term and recursive definitions. Exercise 8.2 is the geometric progression set: general terms, finding which term equals a value, sums of finite progressions, two-condition problems, products of corresponding terms, inserted means and AM-GM questions. The miscellaneous exercise adds growth and interest word problems, a bacteria culture, depreciation and proof questions.

This page works representative questions from each block with full reasoning.

4.2 Exercise 8.1

Question 1 asks for the first five terms of the sequence with nth term n(n plus 2): the terms are 3, 8, 15, 24 and 35. Question 4 defines a1 equal to 3 and a(n) equal to 3a(n minus 1) plus 2; the terms are 3, 11, 35, 107 and 323, which can be checked one by one.

Question 7 asks for the 17th and 24th terms of the sequence whose nth term is 4n minus 3. The 17th is 65 and the 24th is 93. Question 9 asks for the ninth term of the sequence with nth term (minus 1) to the (n minus 1) times n cubed, which is positive for odd n, so the ninth term is 729.

A term question from a sum is also common. If the sum of the first n terms is 3n squared plus 5n, the nth term is the difference of consecutive sums, which gives 6n plus 2. This form appears in the previous year questions as well.

4.3 Exercise 8.2: general terms and sums

Question 1 asks for the 20th and nth terms of the progression 5 by 2, 5 by 4, 5 by 8 and so on. The first term is 5 by 2 and the ratio is 1 by 2, so the 20th term is 5 by 2 times 1 by 2 to the nineteenth, that is 5 by 2 to the twentieth.

Question 2 gives the eighth term as 192 and the ratio as 2, and asks for the 12th term. The first term is 192 divided by 2 to the seventh, that is 3 by 2, and the 12th term is 3 by 2 times 2 to the eleventh, that is 3072.

Question 12 gives three terms in geometric progression with sum 39 by 10 and product 1. Writing the terms as a by r, a and ar makes the product a cubed equal to 1, so a is 1, and the sum gives r plus 1 plus 1 by r equal to 39 by 10. Multiplying by r gives a quadratic with roots 5 by 2 and 2 by 5, and the corresponding progressions are 2 by 5, 1, 5 by 2 and the reverse.

In Question 14 the first three terms add to 16 and the following three add to 128. The sum of the next three is the sum to six terms minus the sum to three, which equals r cubed times the sum of the first three. So r cubed is 8, r is 2, and the first term follows from the three-term sum.

4.4 Exercise 8.2: later questions

Question 18 asks for the sum of 8, 88, 888 and so on to n terms. Write each term as 8 by 9 times (10 to the k minus 1) and sum the geometric part; the answer is 8 by 81 times (10 to the n plus 1 minus 10 minus 9n). The method is the same for 3, 33, 333 and 5, 55, 555.

Question 26 asks for two numbers inserted between 3 and 81 so that the four numbers form a geometric progression. The ratio r satisfies 3 times r cubed equal to 81, so r cubed is 27 and r is 3, giving the inserted numbers 9 and 27.

Question 29 uses the arithmetic mean A and the geometric mean G between two positive numbers. The numbers are the roots of x squared minus 2Ax plus G squared equal to 0, so they are A plus or minus the square root of (A plus G)(A minus G). This is the standard proof that the AM is at least the GM, because the roots are real only when A is at least G.

Question 32 uses an arithmetic mean of 8 and geometric mean of 5 for the roots of a quadratic. The sum of the roots is 16 and their product is 25, so the quadratic is x squared minus 16x plus 25 equal to 0.

4.5 Miscellaneous exercise

Question 1 uses a function with f(x plus y) equal to f(x) times f(y) and f(1) equal to 3. This makes f(n) equal to 3 to the n, and the sum from 1 to n equal to 120 leads to a geometric sum that gives n equal to 4.

Question 2 gives a geometric sum of 315 with first term 5 and ratio 2. The sum formula gives 5 times (2 to the n minus 1) equal to 315, so 2 to the n is 64 and n is 6; the last term is 5 times 2 to the fifth, that is 160.

Question 4 asks for three numbers in geometric progression whose sum is 56 and which become an arithmetic progression after subtracting 1, 7 and 21. Writing the numbers as a by r, a and ar and imposing the arithmetic condition gives a quadratic that yields the numbers 8, 16 and 32, or the reverse depending on the order of subtraction.

Question 12 asks for the 20th term of the series 2 times 4, 4 times 6, 6 times 8 and so on. The nth term is (2n) times (2n plus 2), so the 20th term is 40 times 42, that is 1680.

The growth and interest questions use the same formulas. A tractor paid in instalments, a machine depreciating by 20 percent each year and a simple interest deposit are all handled by writing the year-by-year amounts and summing the resulting arithmetic or geometric progression.