Hard A-Level Arithmetic sequences and series Questions

Challenging, exam-style A-Level Arithmetic sequences and series questions with worked solutions. Stretch yourself on the hardest arithmetic-series, sequences problems.

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A-Level34 questionsStep-by-step solutions
Question 1
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
Question 2
8 markschallenging
Sequence A has first term 55 and common difference 44. Sequence B has first term 4040 and common difference 11. Which sequence has the larger tenth term?
Show worked solution

Worked solution

  1. Identify the first term

    a=5a=5

    The first term is where the sequence begins.

  2. Identify the common difference

    d=4d=4

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

  3. List the first few terms of the sequence

    5, 9, 13, 17, …5,\ 9,\ 13,\ 17,\ \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=5+(n−1)(4)u_n=5+(n-1)(4)

    Replace a and d with their known values.

  6. Simplify the nth-term expression

    un=4n+1u_n=4 n + 1

    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(5)+(n−1)(4)]S_n=\frac{n}{2}\left[2(5)+(n-1)(4)\right]

    Replace a and d with their known values.

  9. Simplify the sum expression

    Sn=2n2+3nS_n=2 n^{2} + 3 n

    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=9−5=4u_2-u_1=9-5=4

    A constant difference between consecutive terms confirms it is arithmetic.

  12. Evaluate the fifth term

    u5=5+4(4)=21u_5=5+4(4)=21

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

  13. Evaluate the tenth term

    u10=5+9(4)=41u_{10}=5+9(4)=41

    Substitute n=10 into the nth-term formula.

  14. Evaluate the sum of the first ten terms

    S10=230S_{10}=230

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

  15. Compare the two tenth terms

    41 vs 49 ⇒ Sequence B41\ \text{vs}\ 49\ \Rightarrow\ \text{Sequence B}

    The larger tenth term identifies the required sequence.

Answer
Sequence B
Question 3
8 markschallenging
An arithmetic sequence begins −8, 7, 22, …-8,\ 7,\ 22,\ \dots. State the common difference.
Show worked solution

Worked solution

  1. Identify the first term

    a=−8a=-8

    The first term is where the sequence begins.

  2. Identify the common difference

    d=15d=15

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

  3. List the first few terms of the sequence

    −8, 7, 22, 37, …-8,\ 7,\ 22,\ 37,\ \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=−8+(n−1)(15)u_n=-8+(n-1)(15)

    Replace a and d with their known values.

  6. Simplify the nth-term expression

    un=15n−23u_n=15 n - 23

    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(−8)+(n−1)(15)]S_n=\frac{n}{2}\left[2(-8)+(n-1)(15)\right]

    Replace a and d with their known values.

  9. Simplify the sum expression

    Sn=15n22−31n2S_n=\frac{15 n^{2}}{2} - \frac{31 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=7−−8=15u_2-u_1=7--8=15

    A constant difference between consecutive terms confirms it is arithmetic.

  12. Evaluate the fifth term

    u5=−8+4(15)=52u_5=-8+4(15)=52

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

  13. Evaluate the tenth term

    u10=−8+9(15)=127u_{10}=-8+9(15)=127

    Substitute n=10 into the nth-term formula.

  14. Evaluate the sum of the first ten terms

    S10=595S_{10}=595

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

  15. Subtract consecutive terms

    d=7−−8=15d=7--8=15

    The common difference is the gap between successive terms.

Answer
1515
Question 4
8 markschallenging
A sequence begins 6, 19, 32, 45, …6,\ 19,\ 32,\ 45,\ \dots. Is the sequence arithmetic, and if so what is its common difference?
Show worked solution

Worked solution

  1. Identify the first term

    a=6a=6

    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

    6, 19, 32, 45, …6,\ 19,\ 32,\ 45,\ \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=6+(n−1)(13)u_n=6+(n-1)(13)

    Replace a and d with their known values.

  6. Simplify the nth-term expression

    un=13n−7u_n=13 n - 7

    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(6)+(n−1)(13)]S_n=\frac{n}{2}\left[2(6)+(n-1)(13)\right]

    Replace a and d with their known values.

  9. Simplify the sum expression

    Sn=13n22−n2S_n=\frac{13 n^{2}}{2} - \frac{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=19−6=13u_2-u_1=19-6=13

    A constant difference between consecutive terms confirms it is arithmetic.

  12. Evaluate the fifth term

    u5=6+4(13)=58u_5=6+4(13)=58

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

  13. Evaluate the tenth term

    u10=6+9(13)=123u_{10}=6+9(13)=123

    Substitute n=10 into the nth-term formula.

  14. Evaluate the sum of the first ten terms

    S10=645S_{10}=645

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

  15. Check the differences between consecutive terms

    19−6=13,32−19=1319-6=13,\quad 32-19=13

    A constant difference confirms the sequence is arithmetic.

Answer
Yes; the common difference is 13
Question 5
8 markschallenging
Which of the following gives the sum of the first nn terms of an arithmetic series with first term aa and common difference dd?
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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=7d=7

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

  3. List the first few terms of the sequence

    3, 10, 17, 24, …3,\ 10,\ 17,\ 24,\ \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)(7)u_n=3+(n-1)(7)

    Replace a and d with their known values.

  6. Simplify the nth-term expression

    un=7n−4u_n=7 n - 4

    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)(7)]S_n=\frac{n}{2}\left[2(3)+(n-1)(7)\right]

    Replace a and d with their known values.

  9. Simplify the sum expression

    Sn=7n22−n2S_n=\frac{7 n^{2}}{2} - \frac{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=10−3=7u_2-u_1=10-3=7

    A constant difference between consecutive terms confirms it is arithmetic.

  12. Evaluate the fifth term

    u5=3+4(7)=31u_5=3+4(7)=31

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

  13. Evaluate the tenth term

    u10=3+9(7)=66u_{10}=3+9(7)=66

    Substitute n=10 into the nth-term formula.

  14. Evaluate the sum of the first ten terms

    S10=345S_{10}=345

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

  15. Recall the standard sum formula

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

    This is the derived formula for an arithmetic series sum.

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

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