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Arithmetic Sequences practice questions — GCSE Maths

12 free multiple-choice problems on arithmetic sequences, ordered to match GCSE Maths difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

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Problem #0131 International
Beginnersequences
In the arithmetic progression , and . Determine .

Problems & worked solutions

Problem #0131 International

Problem 1 Arithmetic Sequences

In the arithmetic progression (an)n1, a2=7 and a5=16. Determine a1.

Show answer & worked solution
  1. A. 3
  2. B. 4✓ correct
  3. C. 5
  4. D. 7

r=1673=3, so a1=a2r=73=4.

Problem #0132 International

Problem 2 Arithmetic Sequences

The numbers 5,8,11,14, form an arithmetic progression. The common difference r equals:

Show answer & worked solution
  1. A. 2
  2. B. 3✓ correct
  3. C. 4
  4. D. 5

r=a2a1=85=3.

Problem #0134 International

Problem 3 Arithmetic Sequences

In an arithmetic progression with a1=3 and r=4, the term an=47. Determine n.

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  1. A. 10
  2. B. 11
  3. C. 12✓ correct
  4. D. 13

47=3+(n1)4(n1)4=44n1=11n=12.

Problem #0133 International

Problem 4 Arithmetic Sequences

Compute the sum of the first 20 terms of the arithmetic progression (an)n1 with a1=2 and r=3.

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  1. A. 560
  2. B. 590
  3. C. 610✓ correct
  4. D. 620

S20=20(22+193)2=20612=1061=610.

Problem #0139 International

Problem 5 Arithmetic Sequences

In an arithmetic progression, a3=11 and a7=27. Determine a5.

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  1. A. 15
  2. B. 17
  3. C. 19✓ correct
  4. D. 22

a5=a3+a72=11+272=19.

Problem #0138 International

Problem 6 Arithmetic Sequences

A student deposits $50 in January and increases the deposit by $10 every following month. How much will be deposited in total over 12 months?

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  1. A. $960
  2. B. $1,080
  3. C. $1,260✓ correct
  4. D. $1,320

S12=12(250+1110)2=6210=1260.

Problem #0135 International

Problem 7 Arithmetic Sequences

The numbers x2, 5, x+4 are in arithmetic progression (in this order). Determine x.

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  1. A. 1
  2. B. 2
  3. C. 4✓ correct
  4. D. 7

5=(x2)+(x+4)210=2x+2x=4.

Problem #0137 International

Problem 8 Arithmetic Sequences

For the arithmetic progression with a1=1 and r=2, compute a5+a6+a7+a8+a9.

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  1. A. 35
  2. B. 45
  3. C. 65✓ correct
  4. D. 75

an=2n1, so a5++a9=9+11+13+15+17=65. Equivalently, this is the sum of 5 consecutive odd numbers starting at 9.

Problem #0140 International

Problem 9 Arithmetic Sequences

For an arithmetic progression with a1=5 and r=3, find the smallest n such that Sn500.

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  1. A. 16
  2. B. 17
  3. C. 18✓ correct
  4. D. 19

Sn=n(3n+7)2. Compute: S17=17582=493<500 and S18=18612=549500. Hence the smallest n is 18.

Problem #0136 International

Problem 10 Arithmetic Sequences

Three numbers in arithmetic progression have sum 15 and the sum of their squares is 83. The largest of them is:

Show answer & worked solution
  1. A. 5
  2. B. 6
  3. C. 7✓ correct
  4. D. 8

Set the terms as 5r,5,5+r. Then (5r)2+25+(5+r)2=50+2r2+25=83, so 2r2=8r=2. The largest term is 5+2=7.

2 more Arithmetic Sequences questions in the app

Also covered in Arithmetic Sequences practice across every exam.

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