Understand · explore · practise

Finite geometric series

Derive and use the finite geometric series formula, including negative ratios, unknown term counts, sums between two indices and logarithmic thresholds.

Before you startGeometric sequences, logarithms and quadratics

01 / Adding a finite geometric list

The sum formula counts n terms, ending at ar^(n − 1).

For a geometric sequence with first term a and ratio r, the first n terms add to Sₙ. The last included term is ar^(n − 1), because the first term has index 1.

Sₙ = a + ar + ar² + … + ar^(n − 1)

For example, 3 + 6 + 12 + 24 + 48 + 96 has six terms. The final term is 3 × 2⁵; the sum is 189. A finite sum exists for every real ratio, including negative ratios and ratios above 1.

02 / Derive the formula

Multiplying by r lines up the terms that cancel.

Multiply every term of Sₙ by r. The new row is ar + ar² + … + arⁿ. Subtract that row from the original. All the middle terms cancel, leaving a − arⁿ.

(1 − r)Sₙ = a(1 − rⁿ)
Sₙ = a(1 − rⁿ)/(1 − r), for r ≠ 1

Multiplying numerator and denominator by −1 gives the equivalent formula a(rⁿ − 1)/(r − 1). Use whichever makes the arithmetic easier. Notice the exponent n in the sum formula, compared with n − 1 in the term formula.

Shift, subtract, cancelExplore

The first term stays at 3. Multiplying the whole sum by r shifts its terms one place. Subtract matching columns.

S₄ = 45.

There are 4 terms in each row; the extra end term is ar⁴.

Derivation step Common ratio Number of terms n 4

Watch the middle terms cancel

Pause, replay or seek freely. The notes explain the same idea and stay in view.

03 / Zero, one and negative ratios

Check the ratio before dividing.

If r = 1, the subtraction step only says 0 = 0. Every term is a, so use Sₙ = na instead. The displayed fraction would be 0/0 and is undefined.

If r = 0, the first term is a and all later terms are zero: Sₙ = a for every positive n. If a = 0, every term and every partial sum is zero for any ratio.

Sum the first 5 terms of 12, −6, 3, …Worked example

a = 12, r = −½, n = 5

Keep the negative ratio inside brackets.

S₅ = 12[1 − (−½)⁵]/[1 − (−½)]

The numerator contains 1 + 1/32 and the denominator is 3/2.

S₅ = 33/4

Check directly: 12 − 6 + 3 − 3/2 + 3/4 = 33/4.

For r = −1, pairs cancel: Sₙ = 0 for even n and Sₙ = a for odd n. An alternating sequence does not imply that its partial sums increase.

04 / When the last term is given

Find the number of terms before calculating the sum.

Add 4 + 12 + 36 + … + 972Worked example

4 × 3^(n − 1) = 972 ⇒ 3^(n − 1) = 243 = 3⁵

Use the term formula to locate the endpoint.

n = 6

There are six terms, not five.

S₆ = 4(3⁶ − 1)/(3 − 1) = 1456

The sum formula now uses exponent 6.

Always check that the proposed endpoint belongs to the sequence. If its candidate index is not a positive integer, the stated finite list is inconsistent.

05 / Find n or a from a total

Rearrange the finite sum before taking logarithms.

How many terms of 3, 6, 12,… sum to 765?Worked example

3(2ⁿ − 1) = 765

Substitute the known first term and ratio.

2ⁿ = 256 = 2⁸

Divide by 3 and then add 1.

n = 8

The answer is a positive integer; substitution verifies the total.

Given r = 3 and S₄ = 200, find aWorked example

200 = a(3⁴ − 1)/(3 − 1) = 40a

The ratio and number of terms fix the multiplier.

a = 5

The sum is 5 + 15 + 45 + 135 = 200.

If logarithms give a non-integer n in an exact-total question, rounding does not produce a solution. A nearest total is a different question.

06 / Find an unknown ratio

A sum condition can allow more than one sequence.

Given S₂ = 6 and S₃ = 14, find real a and rWorked example

a(1 + r) = 6 and ar² = S₃ − S₂ = 8

Subtracting sums isolates the third term.

a = 6/(1 + r), with r ≠ −1

The first equation rules out r = −1.

6r² = 8(1 + r) ⇒ (3r + 2)(r − 2) = 0

Keep both real candidates.

r = 2, a = 2; or r = −⅔, a = 18

Check: 2 + 4 = 6, then +8 = 14; or 18 − 12 = 6, then +8 = 14.

Starting with expanded finite sums can avoid dividing by 1 − r prematurely. Always check exceptional ratios in the original conditions.

07 / The least n above a target

A positive next term makes the partial sum increase.

First sum above 1000 when a = 5 and r = 1.2Worked example

Sₙ = 5(1.2ⁿ − 1)/0.2 = 25(1.2ⁿ − 1)

Write the sum, rather than just the nth term.

Sₙ > 1000 ⇒ 1.2ⁿ > 41

Divide and rearrange before logging.

n > ln 41/ln 1.2 ≈ 20.368

The least possible integer is 21.

S₂₀ ≈ 933.440 and S₂₁ ≈ 1125.128

Both adjacent values confirm the first crossing; all terms are positive.

A target need not be reachable. For a = 8 and r = ½, Sₙ = 16(1 − 2^(−n)) is always below 16 for finite n. Its value is exactly 15 at n = 4, and first exceeds 15 at n = 5.

08 / Sums between two indices

Count the terms in the selected part.

For uₙ = 3 × 2^(n − 1), add terms 4 through 9Worked example

First selected term = u₄ = 24

The selected block is itself geometric with ratio 2.

Number of selected terms = 9 − 4 + 1 = 6

Both endpoints are included.

Selected sum = 24(2⁶ − 1) = 1512

Restart the finite-sum formula at the selected first term.

Alternatively S₉ − S₃ = 1533 − 21 = 1512

With a = 9 and r = −1, find S₂₀ and S₂₁. Is there a least n for which Sₙ > 9?

Hint

Pair successive terms.

Worked solution

S₂₀ = 0 and S₂₁ = 9. Every partial sum is either 0 or 9, so none exceeds 9.

10 / Recap

A finite geometric sum needs no convergence condition.

For r ≠ 1, use Sₙ = a(1 − rⁿ)/(1 − r).
For r = 1, use Sₙ = na.
The sum exponent is n; the last-term exponent is n − 1.
Negative ratios require brackets and may give alternating sums.
Check every valid ratio and every integer term count.
For terms p to q, use S_q − S_p₋₁.

Review geometric sequences →

Section 1 of 10 · Adding a finite geometric list