Arithmetic sequences and series Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Arithmetic sequences and series questions. See exactly how to solve problems on arithmetic-series, sequences.

arithmetic-seriessequences
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.

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
An arithmetic sequence has first term a=7a=7 and common difference d=2d=2. Find the 10{10}th term.

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=7+(n−1)(2)u_n=7+(n-1)(2)

    Substitute a and d ready for evaluation.

  3. Substitute n=10 into the nth-term formula

    u10=7+(10−1)(2)=25u_{10}=7+(10-1)(2)=25

    This gives the required term.

Answer
2525
Question 3
2 markseasy
In an arithmetic sequence the 1{1}th term is 55 and the 2{2}th term is 99. Find the common difference dd.

Worked solution

  1. Write the two terms using the nth-term formula

    u1=a+0d,u2=a+1du_{1}=a+0d,\quad u_{2}=a+1d

    Both terms share the same a and d.

  2. Subtract to eliminate a

    u2−u1=(1)du_{2}-u_{1}=(1)d

    The difference isolates a multiple of d.

  3. Divide the change in value by the change in position

    d=9−52−1=4d=\frac{9-5}{2-1}=4

    The common difference is the change per step.

Answer
44
Question 4
2 markseasy
An arithmetic sequence has first term a=2a=2 and common difference d=5d=5. Find the 8{8}th term.

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=2+(n−1)(5)u_n=2+(n-1)(5)

    Substitute a and d ready for evaluation.

  3. Substitute n=8 into the nth-term formula

    u8=2+(8−1)(5)=37u_{8}=2+(8-1)(5)=37

    This gives the required term.

Answer
3737
Question 5
2 markseasy
Find the sum of the first 1010 terms of the arithmetic series with first term a=1a=1 and common difference d=1d=1.

Worked solution

  1. Write down the sum formula

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

    The sum of an arithmetic series uses a, d and n.

  2. Insert the known values

    S10=102[2(1)+(9)(1)]S_{10}=\frac{10}{2}\left[2(1)+(9)(1)\right]

    Substitute a, d and n ready to evaluate.

  3. Substitute n=10 into the sum formula

    S10=102[2(1)+(9)(1)]=55S_{10}=\frac{10}{2}\left[2(1)+(9)(1)\right]=55

    This evaluates the required sum.

Answer
5555

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