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How do you describe and sum arithmetic and geometric sequences, and when does an infinite series converge?

Arithmetic and geometric sequences and series, sigma notation, sum formulae, recurrence relations, increasing, decreasing and periodic sequences, and the sum to infinity of a convergent geometric series.

A focused answer to the OCR A-Level Mathematics A sequences and series content, covering arithmetic and geometric sequences, the sum formulae, sigma notation, recurrence relations, increasing, decreasing and periodic sequences, and the sum to infinity of a convergent geometric series.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
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What this dot point is asking

OCR wants you to work with arithmetic and geometric sequences and series: find the nnth term and the sum of nn terms, use sigma notation, generate sequences from recurrence relations, recognise increasing, decreasing and periodic sequences, identify when a geometric series converges, and find its sum to infinity.

The answer

Arithmetic sequences

An arithmetic sequence has a constant common difference dd. The nnth term and the sum are:

Geometric sequences

A geometric sequence has a constant common ratio rr. The nnth term and the sum are:

The sum to infinity exists only when ∣r∣<1|r| < 1; otherwise the terms do not shrink to zero and the series diverges. Stating this condition is often a mark in itself.

Sigma notation

βˆ‘r=1nur\displaystyle\sum_{r=1}^{n} u_r means "add the terms uru_r from r=1r = 1 to r=nr = n". For example βˆ‘r=14(2r+1)=3+5+7+9=24\sum_{r=1}^{4}(2r + 1) = 3 + 5 + 7 + 9 = 24.

Recurrence relations and sequence behaviour

A recurrence relation defines each term from previous ones, such as un+1=2unβˆ’3u_{n+1} = 2u_n - 3 with u1=5u_1 = 5. A sequence is increasing if un+1>unu_{n+1} > u_n for all nn, decreasing if un+1<unu_{n+1} < u_n, and periodic if the terms repeat in a fixed cycle, like un+1=1unu_{n+1} = \tfrac{1}{u_n} alternating between two values.

Examples in context

A geometric sum

Generating from a recurrence

Proving a sum using sigma results

A common Paper 1 task combines sigma notation with the standard sum formulae. The key facts are βˆ‘r=1n1=n\sum_{r=1}^{n} 1 = n and βˆ‘r=1nc=cn\sum_{r=1}^{n} c = cn for a constant cc. Combined with the arithmetic sum, these let you evaluate sums written in sigma form without listing every term.

Modelling with sequences

Sequences model real situations such as repayments, salaries that rise by a fixed percentage, or the bounce heights of a ball. A fixed yearly increase gives an arithmetic model; a fixed percentage change gives a geometric model. Identifying which model applies, then choosing the nnth-term or sum formula, is a frequent overarching-theme OT3 application. For a geometric decay such as a ball losing a fixed fraction of its height on each bounce, the total distance travelled is found with the sum to infinity when the ratio satisfies ∣r∣<1|r| < 1.

Try this

Q1. Find the 20th term of the arithmetic sequence 5,8,11,…5, 8, 11, \dots. [2 marks]

  • Cue. u20=5+19Γ—3=62u_{20} = 5 + 19 \times 3 = 62.

Q2. A geometric series has a=12a = 12, r=13r = \tfrac{1}{3}. Find its sum to infinity. [2 marks]

  • Cue. S∞=121βˆ’1/3=18S_\infty = \dfrac{12}{1 - 1/3} = 18.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20196 marksAn arithmetic series has first term 77 and common difference 44. Find the number of terms needed for the sum to first exceed 10001000.
Show worked answer β†’

The sum of nn terms is Sn=n2(2a+(nβˆ’1)d)S_n = \dfrac{n}{2}\big(2a + (n - 1)d\big) with a=7a = 7, d=4d = 4 (M1).

Sn=n2(14+4(nβˆ’1))=n2(4n+10)=n(2n+5)=2n2+5nS_n = \dfrac{n}{2}\big(14 + 4(n - 1)\big) = \dfrac{n}{2}(4n + 10) = n(2n + 5) = 2n^2 + 5n (A1).

Require 2n2+5n>10002n^2 + 5n > 1000, i.e. 2n2+5nβˆ’1000>02n^2 + 5n - 1000 > 0 (M1).

Solve 2n2+5nβˆ’1000=02n^2 + 5n - 1000 = 0: n=βˆ’5+25+80004=βˆ’5+80254β‰ˆ21.16n = \dfrac{-5 + \sqrt{25 + 8000}}{4} = \dfrac{-5 + \sqrt{8025}}{4} \approx 21.16 (M1, A1).

Since nn is a positive integer, the smallest value is n=22n = 22 (A1).

Markers reward the sum formula, the simplified quadratic, solving it, and rounding up to the next integer.

OCR 20215 marksA geometric series has first term aa and common ratio rr. The sum to infinity is 8080 and the second term is 1515. Find aa and rr.
Show worked answer β†’

Sum to infinity: a1βˆ’r=80\dfrac{a}{1 - r} = 80 (M1). Second term: ar=15ar = 15 (M1).

From the first, a=80(1βˆ’r)a = 80(1 - r). Substitute into ar=15ar = 15: 80(1βˆ’r)r=1580(1 - r)r = 15, so 80rβˆ’80r2=1580r - 80r^2 = 15 (M1).

Rearrange: 80r2βˆ’80r+15=080r^2 - 80r + 15 = 0, i.e. 16r2βˆ’16r+3=016r^2 - 16r + 3 = 0, giving (4rβˆ’1)(4rβˆ’3)=0(4r - 1)(4r - 3) = 0, so r=14r = \tfrac{1}{4} or r=34r = \tfrac{3}{4} (A1).

Then a=80(1βˆ’r)a = 80(1 - r) gives a=60a = 60 (for r=14r = \tfrac14) or a=20a = 20 (for r=34r = \tfrac34) (A1).

Markers reward both equations, eliminating aa, solving the quadratic in rr, and the matched pairs. Both ∣r∣<1|r| < 1, so both solutions are valid.

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