Questions on geometric series

Question 1

A geometric sequence begins 2, 6, 18, … Find the common ratio and the 8th term.

Accepted Answer

Each term is three times the one before, so r = 3 and u₈ = 2 × 3⁷ = 4374. B1 for the ratio, B1 for the eighth term. Seven multiplications reach the eighth term. The off-by-one is the same one arithmetic sequences carry, and 2 × 3⁸ = 13 122 is what it costs.

Question 2

State the condition under which a geometric series has a sum to infinity, and determine which of the ratios 0.8, 1.2 and −0.5 qualify.

Accepted Answer

The sum to infinity exists when |r| < 1, so that the powers of r die away. Ratios 0.8 and −0.5 qualify; 1.2 does not, because its powers grow and the partial sums run off without settling. B1 for the condition, B1 for the three ratios.

Question 3

The first three terms of a geometric sequence are 5, x and 45, where x is positive. Find x.

Accepted Answer

Consecutive terms have the same ratio, so x/5 = 45/x and x² = 225, giving x = 15. M1 for x/5 = 45/x, A1 for 15. The other root, x = −15, gives 5, −15, 45 with ratio −3, which is geometric as well: without the word positive both values would be wanted.

Question 4

The 3rd term of a geometric sequence is 12 and the 5th term is 48. Given that the common ratio is positive, find the common ratio and the first term.

Accepted Answer

Two steps separate the 3rd term from the 5th, and dividing removes a: ar⁴/ar² = r² = 48/12 = 4. M1 for dividing the two terms to reach r² = 4. With r positive, r = 2, and then ar² = 12 gives a = 3. A1 for r = 2, A1 for a = 3. Two terms of a geometric sequence are divided, where two terms of an arithmetic one would be subtracted; 48 − 12 = 36 leads nowhere here. The negative root r = −2 fits both given terms, with the same a = 3, so the stated condition is what settles the sign.

Question 5

Find the sum of the first 12 terms of the series 5 + 10 + 20 + …

Accepted Answer

a = 5 and r = 2, so S₁₂ = 5(2¹² − 1)/(2 − 1) = 5 × 4095 = 20 475. M1 for the sum formula, A1 for 5 × 4095, A1 for 20 475. With r > 1 the flipped dressing a(rⁿ − 1)/(r − 1) keeps every bracket positive, which is worth the flip.

Question 6

Find the sum to infinity of the series 24 + 18 + 13.5 + …

Accepted Answer

The ratio is 18/24 = 0.75, comfortably inside the window, so the infinite sum exists and S_∞ = 24/(1 − 0.75) = 96. B1 for the ratio with the condition stated, M1 for the formula, A1 for 96. State |r| < 1 before using the formula, since the sum to infinity exists only then.

Question 7

Prove the formula S_n = a(1 − rⁿ)/(1 − r) for the sum of the first n terms of a geometric series.

Accepted Answer

Write S_n = a + ar + … + arⁿ⁻¹ and multiply by r: rS_n = ar + ar² + … + arⁿ. Subtracting, every term cancels except the first of the first line and the last of the second: S_n(1 − r) = a − arⁿ, and dividing by 1 − r finishes. The cancellation is the proof, so show it explicitly. M1 for multiplying by r, A1 for the second line, M1 for subtracting, A1 for rearranging to the result.

Question 8

A deposit of £2000 grows by 6% each year, with interest added at the end of each year. Show that the balance first exceeds £3000 after 7 years.

Accepted Answer

After n years the balance is 2000 × 1.06ⁿ, so exceeding 3000 needs 1.06ⁿ > 1.5. Taking logs of both sides gives n > log 1.5/log 1.06 = 6.96, and the first whole year past that mark is n = 7. Both sides confirm it. At n = 6 the multiplier is 1.419, still short of 1.5, and at n = 7 it is 1.504, past it. M1 for the inequality, A1 for 1.06ⁿ > 1.5, M1 for taking logs, A1 for seven years. A show-that question requires the inequality to be set up and the log step written out; a table of balances alone does not show the algebra, and 6.96 must be rounded up to 7 because interest is added only at the end of each year, so the balance first exceeds £3000 at the end of the seventh year.

Question 9

The sum to infinity of a geometric series is 50 and its first term is 20. Find the common ratio, and find the least number of terms whose sum exceeds 49.

Accepted Answer

S_∞ = a/(1 − r) gives 20/(1 − r) = 50, so 1 − r = 0.4 and r = 0.6, comfortably inside |r| < 1 as a sum to infinity requires. M1 for 20/(1 − r) = 50, A1 for r = 0.6. Then S_n = 20(1 − 0.6ⁿ)/0.4 = 50(1 − 0.6ⁿ), so S_n > 49 needs 1 − 0.6ⁿ > 0.98, that is 0.6ⁿ < 0.02. M1 for reaching 0.6ⁿ < 0.02. Taking logs, n log 0.6 < log 0.02, and log 0.6 is negative, so dividing by it turns the inequality round: n > log 0.02/log 0.6 = 7.66. M1 for taking logs with the inequality reversed. The first whole number past that is 8, and the sums either side confirm it: S₇ = 48.600 falls short of 49 and S₈ = 49.160 clears it. A1 for 8 terms with those values. Dividing an inequality by a negative number reverses it; leaving the sign as it stands gives n < 7.66 and the wrong end of the range.

Question 10

A ball is dropped from a height of 1.8 m onto level ground. After each bounce it rises to 0.7 of the height it last fell from. Find the height it rises to after the 5th bounce, and the total vertical distance it has travelled at the moment it hits the ground for the 6th time. Give each answer in metres to 3 significant figures.

Accepted Answer

The rebound heights are geometric with first term 1.8 × 0.7 = 1.26 m and ratio 0.7, so after the 5th bounce the ball rises to 1.8 × 0.7⁵ = 0.302526, that is 0.303 m. M1 for 1.8 × 0.7⁵, A1 for 0.303 m. The first hit ends the 1.8 m drop, and each of the next five follows a rise and a fall of equal length, so the distance at the 6th hit is 1.8 + 2(1.26 + 1.26 × 0.7 + … + 1.26 × 0.7⁴). M1 for the drop plus twice the five rebound heights. The bracket is a geometric sum: 1.26(1 − 0.7⁵)/(1 − 0.7) = 3.494106. M1 for the sum with a = 1.26 and r = 0.7. So the total is 1.8 + 2 × 3.494106 = 8.788212, that is 8.79 m. A1 for 8.79 m.

Question 11

At the start of each year, for 6 years, £5000 is paid into a fund. During each year the fund grows by 4%. Show that the value of the fund immediately after the sixth payment is 5000(1.04⁶ − 1)/0.04, and find that value to the nearest pound.

Accepted Answer

By the time the sixth payment is made, the first has been in the fund for 5 whole years and is worth 5000 × 1.04⁵; the second is worth 5000 × 1.04⁴, and so on down to the sixth, just paid in and worth 5000. M1 for the six payments with their growth factors. Added in that order the total is 5000(1 + 1.04 + 1.04² + … + 1.04⁵), a geometric series with a = 5000, r = 1.04 and n = 6. A1 for a = 5000 and r = 1.04. Its sum is 5000(1.04⁶ − 1)/(1.04 − 1) = 5000(1.04⁶ − 1)/0.04, which is the printed form. M1 for the sum formula with n = 6, A1 for the given expression. Evaluating, 1.04⁶ = 1.2653190, so the value is 5000 × 0.2653190/0.04 = 33 164.88, that is £33 165 to the nearest pound. A1 for £33 165.

Question 12

Use a geometric series to write the recurring decimal 0.727272… as a fraction in lowest terms.

Accepted Answer

0.727272… = 0.72 + 0.0072 + 0.000072 + …, a geometric series with a = 0.72 and r = 0.01. So the total is 0.72/(1 − 0.01) = 72/99 = 8/11. M1 for a = 0.72 and r = 0.01, A1 for the sum to infinity, A1 for 8/11.

Question 13

The sum of the first two terms of a geometric series is 9 and its sum to infinity is 25. Find the two possible values of the common ratio, and state which of the two series has the greater third term, giving the value of that term.

Accepted Answer

The two conditions are a(1 + r) = 9 and a/(1 − r) = 25. M1 for both equations. The second gives a = 25(1 − r), and substituting into the first, 25(1 − r)(1 + r) = 9, that is 25(1 − r²) = 9. M1 for eliminating a, A1 for 25(1 − r²) = 9. So r² = 16/25 and r = 4/5 or r = −4/5, both inside |r| < 1, which a sum to infinity requires. A1 for both values. The first goes with a = 5 and the second with a = 45, so the third terms ar² are 5 × 16/25 = 3.2 and 45 × 16/25 = 28.8. M1 for both third terms. The series with the negative ratio has the greater third term, 28.8. A1 for 28.8. Both roots survive here, and what separates the two series is the first term: a negative ratio makes the second term negative, 45 and −36, so a much larger first term is needed for the opening pair to total 9.

Question 14

A geometric series has first term a, where a ≠ 0, and common ratio r. Explain why the formula S_n = a(1 − rⁿ)/(1 − r) cannot be used when r = 1, and write down S_n in that case. Find the values taken by S_n when r = −1, and explain what these two cases show about the condition |r| < 1 for a sum to infinity.

Accepted Answer

The proof of the formula ends by dividing S_n(1 − r) = a − arⁿ by 1 − r, and at r = 1 that divisor is zero, so the division is unavailable and the printed quotient reads 0/0. B1 for the divisor 1 − r vanishing at r = 1. With r = 1 every term equals a and there are n of them, so S_n = na. B1 for na. With r = −1 the terms run a, −a, a, −a, …, so the running totals are a, 0, a, 0, … M1 for the terms and their running totals, A1 for the two values a and 0. Neither case has a sum to infinity: at r = 1 the totals na grow in size without limit, and at r = −1 they never settle, alternating for ever between a and 0. B1 for neither settling to a limit. Both ends of the interval therefore fail, which is why the condition is written |r| < 1 with a strict inequality rather than |r| ≤ 1. B1 for the strictness at both endpoints. The two endpoints fail in different places: r = 1 breaks the formula for S_n itself, while r = −1 leaves S_n perfectly well defined and takes away only its limit.

Question 15

A geometric series has first term a and common ratio r, with |r| < 1. Show that the sum of its odd-numbered terms u₁ + u₃ + u₅ + … is a/(1 − r²), and show that this is 1/(1 + r) of the sum of the whole series. Verify both results for the series 15 + 3.75 + 0.9375 + …

Accepted Answer

The odd-numbered terms are a, ar², ar⁴, …, each r² times the one before, so they are themselves geometric with first term a and common ratio r². M1 for the ratio r². Since |r| < 1 it follows that r² < 1 too, so this series converges and the sum-to-infinity formula applies. B1 for the condition carrying over. Its sum is therefore a/(1 − r²). A1 for a/(1 − r²). Dividing by the whole series' sum a/(1 − r): [a/(1 − r²)] × [(1 − r)/a] = (1 − r)/[(1 − r)(1 + r)] = 1/(1 + r). M1 for the division, A1 for 1/(1 + r). For 15 + 3.75 + 0.9375 + … the ratio is 3.75/15 = 0.25, so the whole series sums to 15/0.75 = 20, while the odd-numbered terms 15 + 0.9375 + … have ratio 0.0625 and sum to 15/0.9375 = 16; and 20/(1 + 0.25) = 16 agrees. B1 for both checks. Since 1 + r lies between 0 and 2 for every |r| < 1, the factor 1/(1 + r) is always more than a half: the odd-numbered terms carry more than half of the total whatever the ratio.