Hard A-Level Geometric sequences and series Questions

Challenging, exam-style A-Level Geometric sequences and series questions with worked solutions. Stretch yourself on the hardest geometric-series, sum-of-n-terms, sum-to-infinity, geometric-sequence problems.

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A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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
Question 2
8 markschallenging
Series A has first term 1010 and common ratio 12\frac{1}{2}; series B has first term 88 and common ratio 14\frac{1}{4}. Which series has the greater sum to infinity?
Show worked solution

Worked solution

  1. Consider series A

    a=10, r=12a=10,\ 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=20S_A=20

    The sum to infinity of series A.

  6. Consider series B

    a=8, r=14a=8,\ r=\frac{1}{4}

    The second series.

  7. Check series B converges

    ∣14∣<1\left|\frac{1}{4}\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−(14)=341-\left(\frac{1}{4}\right)=\frac{3}{4}

    One minus the ratio of B.

  10. Compute S_B

    SB=323S_B=\frac{32}{3}

    The sum to infinity of series B.

  11. Compare the two sums

    SA=20, SB=323S_A=20,\ S_B=\frac{32}{3}

    Place the two values side by side.

  12. Determine which is larger

    20>32320 > \frac{32}{3}

    Compare the numerical values.

  13. Interpret the comparison

    series A is larger\text{series A 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 A has the greater sum\Rightarrow \text{series A has the greater sum}

    State which series has the larger sum to infinity.

Answer
Series A has the greater sum to infinity
Question 3
8 markschallenging
Consider the geometric series with first term 33 and common ratio −74- \frac{7}{4}. Does the series converge?
Show worked solution

Worked solution

  1. Identify the first term

    a=3a=3

    The first term of the series.

  2. Identify the common ratio

    r=−74r=- \frac{7}{4}

    The constant multiplier.

  3. Write the first few terms

    3, −214, 14716, −102964, …3,\ - \frac{21}{4},\ \frac{147}{16},\ - \frac{1029}{64},\ \ldots

    See how the terms behave.

  4. Recall the condition for convergence

    converges  ⟺  ∣r∣<1\text{converges} \iff \left|r\right|<1

    A geometric series converges exactly when |r|<1.

  5. Compute the size of the ratio

    ∣r∣=74\left|r\right|=\frac{7}{4}

    Find the magnitude of the common ratio.

  6. Compare with 1

    ∣−74∣≥1\left|- \frac{7}{4}\right| \ge 1

    Decide whether the magnitude is below 1.

  7. Decide convergence

    diverges\text{diverges}

    Apply the convergence test.

  8. Examine the size of a later term

    u5=7203256u_5=\frac{7203}{256}

    The terms are not shrinking.

  9. Terms do not tend to zero

    ∣un∣→∞\left|u_n\right|\to \infty

    Because |r|>=1 the terms grow without bound.

  10. A sum to infinity does not exist

    S∞ undefinedS_\infty \text{ undefined}

    There is no finite limit.

  11. Partial sums grow without bound

    Sn→±∞S_n\to \pm\infty

    The running total does not settle.

  12. Compute a further term

    u6=−504211024u_6=- \frac{50421}{1024}

    The terms keep getting larger in size.

  13. Conclude from the test

    ∣r∣≥1⇒diverges\left|r\right|\ge 1 \Rightarrow \text{diverges}

    The condition fails.

  14. Restate the outcome

    the series diverges\text{the series diverges}

    Final decision.

  15. Select the correct statement

    ⇒it diverges\Rightarrow \text{it diverges}

    State whether the series converges.

Answer
No; it diverges because ∣r∣≥1|r|\ge 1
Question 4
8 markschallenging
Consider the geometric series with first term 66 and common ratio 45\frac{4}{5}. Does the series converge?
Show worked solution

Worked solution

  1. Identify the first term

    a=6a=6

    The first term of the series.

  2. Identify the common ratio

    r=45r=\frac{4}{5}

    The constant multiplier.

  3. Write the first few terms

    6, 245, 9625, 384125, …6,\ \frac{24}{5},\ \frac{96}{25},\ \frac{384}{125},\ \ldots

    See how the terms behave.

  4. Recall the condition for convergence

    converges  ⟺  ∣r∣<1\text{converges} \iff \left|r\right|<1

    A geometric series converges exactly when |r|<1.

  5. Compute the size of the ratio

    ∣r∣=45\left|r\right|=\frac{4}{5}

    Find the magnitude of the common ratio.

  6. Compare with 1

    ∣45∣<1\left|\frac{4}{5}\right| < 1

    Decide whether the magnitude is below 1.

  7. Decide convergence

    converges\text{converges}

    Apply the convergence test.

  8. Recall the sum to infinity

    S∞=a1−rS_\infty=\frac{a}{1-r}

    A finite limit exists.

  9. Compute the denominator

    1−(45)=151-\left(\frac{4}{5}\right)=\frac{1}{5}

    One minus the common ratio.

  10. Compute the sum to infinity

    S∞=30S_\infty=30

    The series adds up to a finite value.

  11. Note the terms tend to zero

    un→0u_n\to 0

    Because |r|<1 the terms shrink to zero.

  12. Partial sums approach the limit

    Sn→30S_n\to 30

    The running total settles down.

  13. Conclude from the test

    ∣r∣<1⇒converges\left|r\right|<1 \Rightarrow \text{converges}

    The condition is satisfied.

  14. Restate the outcome

    the series converges\text{the series converges}

    Final decision.

  15. Select the correct statement

    ⇒it converges\Rightarrow \text{it converges}

    State whether the series converges.

Answer
Yes; it converges because ∣r∣<1|r|<1
Question 5
8 markschallenging
A geometric series has first term 77 and common ratio −25- \frac{2}{5}. Which of the following is its sum to infinity?
Show worked solution

Worked solution

  1. Identify the first term

    a=7a=7

    The first term of the series.

  2. Identify the common ratio

    r=−25r=- \frac{2}{5}

    The constant multiplier between terms.

  3. Check the series converges

    ∣−25∣<1\left|- \frac{2}{5}\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∞=71−(−25)S_\infty=\frac{7}{1-\left(- \frac{2}{5}\right)}

    Put a and r into the formula.

  6. Simplify the denominator

    1−(−25)=751-\left(- \frac{2}{5}\right)=\frac{7}{5}

    Work out one minus the common ratio.

  7. Compute the sum to infinity

    S∞=775=5S_\infty=\frac{7}{\frac{7}{5}}=5

    Divide the first term by one minus r.

  8. Write the sum as a single fraction

    S∞=775S_\infty=\frac{7}{\frac{7}{5}}

    The exact value as a fraction.

  9. Write out the first few terms

    7, −145, 2825, −56125, …7,\ - \frac{14}{5},\ \frac{28}{25},\ - \frac{56}{125},\ \ldots

    The terms get smaller because |r|<1.

  10. Compute the first partial sum

    S1=7S_1=7

    The sum of the first term only.

  11. Compute the second partial sum

    S2=215S_2=\frac{21}{5}

    Add the second term.

  12. Compute the third partial sum

    S3=13325S_3=\frac{133}{25}

    Add the third term.

  13. Note the partial sums approach the limit

    Sn→5S_n\to 5

    As n increases the partial sums tend to the sum to infinity.

  14. Confirm the exact value

    S∞=5 (exact)S_\infty=5\ (\text{exact})

    The exact sum to infinity.

  15. Select the correct sum to infinity

    ⇒S∞=5\Rightarrow S_\infty=5

    Identify the correct value.

Answer
55

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