A-Level Geometric sequences and series Practice Questions

Free A-Level Geometric sequences and series practice questions with full step-by-step worked solutions. Covers geometric-sequence, nth-term, common-ratio, geometric-series. Practise exam-style problems and check your method.

geometric-sequencenth-termcommon-ratiogeometric-seriessum-of-n-termssum-to-infinity
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A geometric sequence has first term 22 and common ratio 33. Find the 55th term u5u_{5}.
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Worked solution

  1. Identify the first term

    a=2a=2

    The first term of the sequence.

  2. Identify the common ratio

    r=3r=3

    The constant multiplier between terms.

  3. State the required term

    u5=162u_{5}=162

    This is the value of the requested term.

Answer
162162
Question 2
2 markseasy
A geometric series has first term 1010 and common ratio 25\frac{2}{5}. Which of the following is its sum to infinity?
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Worked solution

  1. Identify the first term

    a=10a=10

    The first term of the series.

  2. Identify the common ratio

    r=25r=\frac{2}{5}

    The constant multiplier between terms.

  3. Select the correct sum to infinity

    ⇒S∞=503\Rightarrow S_\infty=\frac{50}{3}

    Identify the correct value.

Answer
503\frac{50}{3}
Question 3
3 marksintermediate
A geometric series has first term 1515 and common ratio 13\frac{1}{3}. Which of the following is its sum to infinity?
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Worked solution

  1. Identify the first term

    a=15a=15

    The first term of the series.

  2. Identify the common ratio

    r=13r=\frac{1}{3}

    The constant multiplier between terms.

  3. Check the series converges

    ∣13∣<1\left|\frac{1}{3}\right|<1

    A sum to infinity exists only when |r|<1.

  4. Recall the sum to infinity formula

    S∞=a1−rS_\infty=\frac{a}{1-r}

    This formula applies for a convergent geometric series.

  5. Substitute the values

    S∞=151−(13)S_\infty=\frac{15}{1-\left(\frac{1}{3}\right)}

    Put a and r into the formula.

  6. Select the correct sum to infinity

    ⇒S∞=452\Rightarrow S_\infty=\frac{45}{2}

    Identify the correct value.

Answer
452\frac{45}{2}
Question 4
5 markshard
Series A has first term 44 and common ratio 12\frac{1}{2}; series B has first term 66 and common ratio 13\frac{1}{3}. Which series has the greater sum to infinity?
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Worked solution

  1. Consider series A

    a=4, r=12a=4,\ r=\frac{1}{2}

    The first series.

  2. Check series A converges

    ∣12∣<1\left|\frac{1}{2}\right|<1

    A sum to infinity exists for A.

  3. Sum to infinity of A

    SA=a1−rS_A=\frac{a}{1-r}

    Use the sum-to-infinity formula.

  4. Compute the denominator for A

    1−(12)=121-\left(\frac{1}{2}\right)=\frac{1}{2}

    One minus the ratio of A.

  5. Compute S_A

    SA=8S_A=8

    The sum to infinity of series A.

  6. Consider series B

    a=6, r=13a=6,\ r=\frac{1}{3}

    The second series.

  7. Check series B converges

    ∣13∣<1\left|\frac{1}{3}\right|<1

    A sum to infinity exists for B.

  8. Sum to infinity of B

    SB=a1−rS_B=\frac{a}{1-r}

    Use the sum-to-infinity formula.

  9. Compute the denominator for B

    1−(13)=231-\left(\frac{1}{3}\right)=\frac{2}{3}

    One minus the ratio of B.

  10. Select the correct comparison

    ⇒series B has the greater sum\Rightarrow \text{series B has the greater sum}

    State which series has the larger sum to infinity.

Answer
Series B has the greater sum to infinity
Question 5
8 markschallenging
Series A has first term 33 and common ratio 23\frac{2}{3}; series B has first term 55 and common ratio 12\frac{1}{2}. Which series has the greater sum to infinity?
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Worked solution

  1. Consider series A

    a=3, r=23a=3,\ r=\frac{2}{3}

    The first series.

  2. Check series A converges

    ∣23∣<1\left|\frac{2}{3}\right|<1

    A sum to infinity exists for A.

  3. Sum to infinity of A

    SA=a1−rS_A=\frac{a}{1-r}

    Use the sum-to-infinity formula.

  4. Compute the denominator for A

    1−(23)=131-\left(\frac{2}{3}\right)=\frac{1}{3}

    One minus the ratio of A.

  5. Compute S_A

    SA=9S_A=9

    The sum to infinity of series A.

  6. Consider series B

    a=5, r=12a=5,\ r=\frac{1}{2}

    The second series.

  7. Check series B converges

    ∣12∣<1\left|\frac{1}{2}\right|<1

    A sum to infinity exists for B.

  8. Sum to infinity of B

    SB=a1−rS_B=\frac{a}{1-r}

    Use the sum-to-infinity formula.

  9. Compute the denominator for B

    1−(12)=121-\left(\frac{1}{2}\right)=\frac{1}{2}

    One minus the ratio of B.

  10. Compute S_B

    SB=10S_B=10

    The sum to infinity of series B.

  11. Compare the two sums

    SA=9, SB=10S_A=9,\ S_B=10

    Place the two values side by side.

  12. Determine which is larger

    9<109 < 10

    Compare the numerical values.

  13. Interpret the comparison

    series B is larger\text{series B is larger}

    Translate the inequality into words.

  14. Restate the comparison

    SA<SBS_A < S_B

    The relationship between the sums.

  15. Select the correct comparison

    ⇒series B has the greater sum\Rightarrow \text{series B has the greater sum}

    State which series has the larger sum to infinity.

Answer
Series B has the greater sum to infinity

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