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Geometric series questions
Multiply by the same ratio at each step and you have a geometric sequence, growing rapidly if the ratio is greater than one and shrinking towards zero if it is between −1 and 1. Multiplying the sum by r and subtracting gives the sum formula, an infinite series has a finite total when |r| < 1, and logarithms answer the questions about how many terms are needed.
7 original questions · 21 marks · the geometric series notes · Sequences and series
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
A geometric sequence begins 2, 6, 18, … Find the common ratio and the 8th term.
Worked 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.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.
Worked 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.Find the sum of the first 12 terms of the series 5 + 10 + 20 + …
Worked 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.Find the sum to infinity of the series 24 + 18 + 13.5 + …
Worked 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. Endless additions still reach a finite total, because each one is a fixed fraction of the last and the leftovers shrink geometrically. State |r| < 1 before using the formula; that check is a mark in its own right.Prove the formula Sn = a(1 − rn)/(1 − r) for the sum of the first n terms of a geometric series.
Worked answer
Write Sn = a + ar + … + arn−1 and multiply by r: rSn = ar + ar2 + … + arn. Subtracting, every term cancels except the first of the first line and the last of the second: Sn(1 − r) = a − arn, and dividing by 1 − r finishes. The cancellation is the proof; showing it explicitly is what the marks are for. M1 for multiplying by r, A1 for the second line, M1 for subtracting, A1 for rearranging to the result.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.
Worked 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 wants the inequality set up and the log step written out; a table of balances alone is worth about half the marks, and rounding 6.96 down to 6 throws away the rest.Use a geometric series to write the recurring decimal 0.727272… as a fraction in lowest terms.
Worked 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. Every recurring decimal is an infinite geometric series in disguise, and that is the reason every one of them is a fraction. Writing 0.72 and 0.01 rather than 72 and 100 keeps the series honest.
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