TMUA: Arithmetic and geometric series

Arithmetic and geometric series.

Practise arithmetic and geometric series →Timed TMUA paper

What the specification covers

Key ideas

Common mistakes

Exam tip

Write down \(S_n\) for each given \(n\) as a linear equation in \(a\) and \(d\) before doing any arithmetic.

Worked example

Worked example

The sum of the first 10 terms of an arithmetic sequence is 145 and the sum of the first 20 terms is 590. What is the 30th term?

  1. \(85\)
  2. \(87\)
  3. \(88\)
  4. \(90\)
  5. \(91\)

Answer: C: \(88\)

\(S_{10}=5(2a+9d)=145\Rightarrow 2a+9d=29\). \(S_{20}=10(2a+19d)=590\Rightarrow 2a+19d=59\). Subtracting, \(10d=30\), \(d=3\), \(a=1\). The 30th term is \(a+29d=88\).

Practice questions

Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.

Question 1

A convergent geometric series has sum to infinity equal to 4 times its first term. The sum of its first two terms is 7. What is its sum to infinity?

  1. \(\frac{64}{7}\)
  2. \(16\)
  3. \(12\)
  4. \(28\)
  5. \(\frac{28}{3}\)
Show the answer and solution

Answer: B: \(16\)

\(\dfrac{a}{1-r}=4a\) with \(a\ne0\) gives \(r=\tfrac34\). Then \(a(1+\tfrac34)=7\), so \(a=4\) and the sum to infinity is \(4a=16\).

Question 2

What is the sum of all the positive integers less than 200 that are not multiples of 3?

  1. \(6633\)
  2. \(13267\)
  3. \(13333\)
  4. \(13266\)
  5. \(19900\)
Show the answer and solution

Answer: B: \(13267\)

\(1+2+\dots+199=\tfrac{199\times200}{2}=19900\). The multiples of 3 below 200 are \(3,6,\dots,198\): 66 terms with sum \(3\times\tfrac{66\times67}{2}=6633\). The answer is \(19900-6633=13267\).

Question 3

The sum of the first 8 terms of an arithmetic sequence is \(56\) and the sum of the first 16 terms is \(240\). What is term number 50?

  1. \(98\)
  2. \(99\)
  3. \(100\)
  4. \(96\)
  5. \(49\)
Show the answer and solution

Answer: A: \(98\)

\(S_{8}=\tfrac{8}{2}(2a+7d)=56\) and \(S_{16}=\tfrac{16}{2}(2a+15d)=240\). Solving, \(d=2\) and \(a=0\). Term 50 is \(a+49d=98\).

Question 4

The sum of the first 10 terms of an arithmetic sequence is \(85\) and the sum of the first 30 terms is \(1155\). What is term number 40?

  1. \(115\)
  2. \(109\)
  3. \(113\)
  4. \(112\)
  5. \(118\)
Show the answer and solution

Answer: D: \(112\)

\(S_{10}=\tfrac{10}{2}(2a+9d)=85\) and \(S_{30}=\tfrac{30}{2}(2a+29d)=1155\). Solving, \(d=3\) and \(a=-5\). Term 40 is \(a+39d=112\).

Question 5

The sum of the first 8 terms of an arithmetic sequence is \(96\) and the sum of the first 16 terms is \(320\). What is term number 40?

  1. \(84\)
  2. \(81\)
  3. \(85\)
  4. \(83\)
  5. \(44\)
Show the answer and solution

Answer: D: \(83\)

\(S_{8}=\tfrac{8}{2}(2a+7d)=96\) and \(S_{16}=\tfrac{16}{2}(2a+15d)=320\). Solving, \(d=2\) and \(a=5\). Term 40 is \(a+39d=83\).

Question 6

The sum of the first 10 terms of an arithmetic sequence is \(185\) and the sum of the first 30 terms is \(1455\). What is term number 25?

  1. \(74\)
  2. \(41\)
  3. \(78\)
  4. \(80\)
  5. \(77\)
Show the answer and solution

Answer: E: \(77\)

\(S_{10}=\tfrac{10}{2}(2a+9d)=185\) and \(S_{30}=\tfrac{30}{2}(2a+29d)=1455\). Solving, \(d=3\) and \(a=5\). Term 25 is \(a+24d=77\).

Question 7

The sum of the first 5 terms of an arithmetic sequence is \(25\) and the sum of the first 15 terms is \(225\). What is term number 25?

  1. \(51\)
  2. \(25\)
  3. \(50\)
  4. \(47\)
  5. \(49\)
Show the answer and solution

Answer: E: \(49\)

\(S_{5}=\tfrac{5}{2}(2a+4d)=25\) and \(S_{15}=\tfrac{15}{2}(2a+14d)=225\). Solving, \(d=2\) and \(a=1\). Term 25 is \(a+24d=49\).

Question 8

The sum of the first 5 terms of an arithmetic sequence is \(15\) and the sum of the first 15 terms is \(195\). What is term number 25?

  1. \(48\)
  2. \(23\)
  3. \(47\)
  4. \(45\)
  5. \(49\)
Show the answer and solution

Answer: C: \(47\)

\(S_{5}=\tfrac{5}{2}(2a+4d)=15\) and \(S_{15}=\tfrac{15}{2}(2a+14d)=195\). Solving, \(d=2\) and \(a=-1\). Term 25 is \(a+24d=47\).

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