A-Level Arithmetic sequences and series Practice Questions

Free A-Level Arithmetic sequences and series practice questions with full step-by-step worked solutions. Covers arithmetic-series, sequences. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
An arithmetic sequence has first term a=3a=3 and common difference d=4d=4. Find the 5{5}th term.
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Worked solution

  1. Write down the nth-term formula

    un=a+(n−1)du_n=a+(n-1)d

    The term formula uses the first term and common difference.

  2. Insert the known values

    un=3+(n−1)(4)u_n=3+(n-1)(4)

    Substitute a and d ready for evaluation.

  3. Substitute n=5 into the nth-term formula

    u5=3+(5−1)(4)=19u_{5}=3+(5-1)(4)=19

    This gives the required term.

Answer
1919
Question 2
2 markseasy
Describe the behaviour of the arithmetic sequence with first term a=1a=1 and common difference d=5d=5.
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Worked solution

  1. List the first few terms

    1, 6, 11, 16, …1,\ 6,\ 11,\ 16,\ \dots

    Watch whether the terms grow or shrink.

  2. Note the sign of the common difference

    d=5d=5

    The sign of d controls the trend.

  3. Read the behaviour from the common difference

    d=5 ⇒ increasingd=5\ \Rightarrow\ \text{increasing}

    A positive difference rises, a negative difference falls.

Answer
The sequence is increasing
Question 3
3 marksintermediate
An arithmetic sequence begins 100, 91, 82, …100,\ 91,\ 82,\ \dots. State the common difference.
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Worked solution

  1. Identify the first term

    a=100a=100

    The first term is where the sequence begins.

  2. Identify the common difference

    d=−9d=-9

    Each term is obtained by adding the common difference to the previous one.

  3. List the first few terms of the sequence

    100, 91, 82, 73, …100,\ 91,\ 82,\ 73,\ \dots

    Adding d repeatedly generates successive terms.

  4. Recall the formula for the nth term

    un=a+(n−1)du_n=a+(n-1)d

    Any term can be found from the first term and the common difference.

  5. Substitute a and d into the nth-term formula

    un=100+(n−1)(−9)u_n=100+(n-1)(-9)

    Replace a and d with their known values.

  6. Subtract consecutive terms

    d=91−100=−9d=91-100=-9

    The common difference is the gap between successive terms.

Answer
−9-9
Question 4
5 markshard
An arithmetic sequence begins 3, 14, 25, …3,\ 14,\ 25,\ \dots. State the common difference.
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Worked solution

  1. Identify the first term

    a=3a=3

    The first term is where the sequence begins.

  2. Identify the common difference

    d=11d=11

    Each term is obtained by adding the common difference to the previous one.

  3. List the first few terms of the sequence

    3, 14, 25, 36, …3,\ 14,\ 25,\ 36,\ \dots

    Adding d repeatedly generates successive terms.

  4. Recall the formula for the nth term

    un=a+(n−1)du_n=a+(n-1)d

    Any term can be found from the first term and the common difference.

  5. Substitute a and d into the nth-term formula

    un=3+(n−1)(11)u_n=3+(n-1)(11)

    Replace a and d with their known values.

  6. Simplify the nth-term expression

    un=11n−8u_n=11 n - 8

    Expanding gives a linear expression in n.

  7. Recall the formula for the sum of the first n terms

    Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}\left[2a+(n-1)d\right]

    This standard result sums an arithmetic series.

  8. Substitute a and d into the sum formula

    Sn=n2[2(3)+(n−1)(11)]S_n=\frac{n}{2}\left[2(3)+(n-1)(11)\right]

    Replace a and d with their known values.

  9. Simplify the sum expression

    Sn=11n22−5n2S_n=\frac{11 n^{2}}{2} - \frac{5 n}{2}

    Expanding gives a quadratic expression in n.

  10. Subtract consecutive terms

    d=14−3=11d=14-3=11

    The common difference is the gap between successive terms.

Answer
1111
Question 5
8 markschallenging
Describe the behaviour of the arithmetic sequence with first term a=1000a=1000 and common difference d=−13d=-13.
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Worked solution

  1. Identify the first term

    a=1000a=1000

    The first term is where the sequence begins.

  2. Identify the common difference

    d=−13d=-13

    Each term is obtained by adding the common difference to the previous one.

  3. List the first few terms of the sequence

    1000, 987, 974, 961, …1000,\ 987,\ 974,\ 961,\ \dots

    Adding d repeatedly generates successive terms.

  4. Recall the formula for the nth term

    un=a+(n−1)du_n=a+(n-1)d

    Any term can be found from the first term and the common difference.

  5. Substitute a and d into the nth-term formula

    un=1000+(n−1)(−13)u_n=1000+(n-1)(-13)

    Replace a and d with their known values.

  6. Simplify the nth-term expression

    un=1013−13nu_n=1013 - 13 n

    Expanding gives a linear expression in n.

  7. Recall the formula for the sum of the first n terms

    Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}\left[2a+(n-1)d\right]

    This standard result sums an arithmetic series.

  8. Substitute a and d into the sum formula

    Sn=n2[2(1000)+(n−1)(−13)]S_n=\frac{n}{2}\left[2(1000)+(n-1)(-13)\right]

    Replace a and d with their known values.

  9. Simplify the sum expression

    Sn=−13n22+2013n2S_n=- \frac{13 n^{2}}{2} + \frac{2013 n}{2}

    Expanding gives a quadratic expression in n.

  10. Recall the alternative sum formula

    Sn=n2(a+l)S_n=\frac{n}{2}\left(a+l\right)

    Here l is the last term; this form is useful when the last term is known.

  11. Verify the sequence is arithmetic

    u2−u1=987−1000=−13u_2-u_1=987-1000=-13

    A constant difference between consecutive terms confirms it is arithmetic.

  12. Evaluate the fifth term

    u5=1000+4(−13)=948u_5=1000+4(-13)=948

    Substitute n=5 into the nth-term formula as a check.

  13. Evaluate the tenth term

    u10=1000+9(−13)=883u_{10}=1000+9(-13)=883

    Substitute n=10 into the nth-term formula.

  14. Evaluate the sum of the first ten terms

    S10=9415S_{10}=9415

    Substitute n=10 into the sum formula as a check.

  15. Read the behaviour from the common difference

    d=−13 ⇒ decreasingd=-13\ \Rightarrow\ \text{decreasing}

    A positive difference rises, a negative difference falls.

Answer
The sequence is decreasing

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