Sequences and series

Distinguish a sequence from a sum

A sequence is an ordered list of terms, often defined by a rule. A series is a sum of terms from a sequence. In 3, 7, 11, 15, the first term is 3 and each later term increases by 4. The position matters: the third term is 11. Not every pattern follows an arithmetic or geometric rule. A few displayed terms can fit several possible formulas, so use the rule stated by the question or justify the assumption before extending an apparent pattern.

Let Tₙ denote the nth term. In an arithmetic progression, or AP, the difference between consecutive terms is a constant d. Starting with first term a, the nth term is Tₙ = a + (n − 1)d. The factor n − 1 appears because reaching the nth term from the first requires n − 1 steps. For a = 5 and d = 3, the tenth term is 5 + 9 × 3 = 32. Using a + nd would incorrectly give the eleventh term.

Find unknown positions and AP parameters

If a sequence begins 7, 12, 17, then a = 7 and d = 5. To find which term is 62, solve 7 + (n − 1)5 = 62, giving n = 12. Check that n is a positive integer. If the result were 12.4, 62 would not occur exactly in that AP. A question about position differs from a question about term value: one asks for n, the other for Tₙ. State which quantity your final number represents.

Two known terms can determine a and d. If T₄ = 14 and T₉ = 34, then a + 3d = 14 and a + 8d = 34. Subtracting gives 5d = 20, so d = 4 and a = 2. Negative differences are allowed: 20, 17, 14, 11 is an AP with d = −3. The formula still works until any contextual restriction applies, such as a stock level that cannot become negative.

Derive and use the AP sum

Write Sₙ = a + (a + d) +... + l, where l is the nth term. Write it again in reverse order and add term by term. Each pair totals a + l, and there are n pairs, so. Therefore Sₙ. The sum is the number of terms multiplied by the average of the first and last terms. The derivation explains why the formula applies even when the terms decrease.

For the first 20 positive even integers, a = 2, d = 2 and l = 40. Their sum is 20(2 + 40)/2 = 420. In an application, a student saves ₦500 in the first week and increases the weekly amount by ₦100. The tenth week's deposit is ₦1,400, but the total deposited over ten weeks is. Confusing the last deposit with the accumulated sum is a common reading error.

Recognise a geometric progression

A geometric progression, or GP, has constant ratio r between consecutive terms, provided the relevant division is defined. Its nth term is Tₙ = arⁿ⁻¹. In 3, 6, 12, 24, a = 3 and r = 2, so T₈ = 3 × 2⁷ = 384. A negative ratio alternates signs: 2, −6, 18, −54 has r = −3. A ratio between 0 and 1 gives shrinking positive terms when a is positive. The word geometric refers to multiplicative growth, not a requirement for a drawn shape.

For r ≠ 1, the first n terms sum to Sₙ = a(1 − rⁿ)/(1 − r), equivalently a(rⁿ − 1)/(r − 1). To see why, subtract rSₙ from Sₙ: intermediate terms cancel, leaving a − arⁿ. For r = 1, every term equals a and Sₙ = na; the usual fraction would divide by zero. For a = 2, r = 3 and n = 4, the sum is 2(81 − 1)/2 = 80, agreeing with 2 + 6 + 18 + 54.

Know when an infinite sum exists

A GP has a finite infinite sum when |r| < 1. Then rⁿ approaches zero, so S∞ = a/(1 − r). For 8 + 4 + 2 +..., a = 8 and, giving S∞ = 16. The terms never become exactly zero, yet their partial sums approach a finite limit. If r = 2, the terms grow and the finite infinite-sum formula is invalid. A negative ratio such as −1/2 also converges, with alternating partial sums approaching the limit.

Compound growth is a GP. A population model starting at 2,000 and increasing 5% per year gives 2,000(1.05)ᵗ after t years. Be careful about indexing: if the initial value is called the first term, the fifth term corresponds to four growth intervals. If a question says “after five years”, there are five multiplications. Similar reasoning models reducing-balance depreciation with a multiplier below 1. These models assume a fixed rate over the specified period; the formula does not establish that real rates stay constant.

Self-check: find the twelfth term of 4, 7, 10...; the sum of its first twelve terms; and the infinite sum of 12 + 4 + 4/3 +.... The answers are 37, 246 and 18. In the last example, which meets the convergence condition. Before substituting, state a, d or r, and n. Afterwards, check the result against a few directly listed terms and decide whether the question asked for one term, a finite total or a limiting total.

Insert means and distinguish indexing

To insert three arithmetic means between 4 and 20, there will be five terms and four equal steps. Therefore d = (20 − 4)/4 = 4, and the inserted terms are 8, 12 and 16. Counting only the inserted terms as the number of gaps would give the wrong difference. For one positive geometric mean G between positive numbers 4 and 36, so G = 12. Without a positive-mean restriction, 4, −12, 36 also forms a GP, with ratio −3.

A recursively stated rule needs an initial term. If T₁ = 2 and Tₙ₊₁ = 3Tₙ, the terms are 2, 6, 18 and so on, agreeing with Tₙ = 2 × 3ⁿ⁻¹. A rule Tₙ₊₁ = Tₙ + 3 instead produces an AP. Writing the first few terms is a useful way to detect whether a recurrence uses addition or multiplication before selecting a sum formula.

For the alternating GP 6, −3, 1.5, ... , the sum to infinity is 6/(1 + 1/2) = 4. Successive partial sums oscillate around that limiting value.