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AI HL · 12.1

Sequences & Series

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12.1.10

Sequences and series: revision

12.1.10

Sequences and series: revision

Target: I can use a formula for \(S_n\) to find \(S_1\), \(S_2\), \(u_1\), \(u_2\) and the common difference. (+4 more)

By the end of this section

Key points

Arithmetic sequences

  • \(u_n=u_1+(n-1)d\). The common difference is \(d=u_2-u_1\). From two terms, \(d=\frac{u_m-u_k}{m-k}\), then find \(u_1\) from either term.
  • \(S_n=\frac n2\left(2u_1+(n-1)d\right)\), or \(S_n=\frac n2(u_1+u_n)\) when you know the last term.
  • Write down \(u_1\) and d before you start. Most errors come from using n instead of \(n-1\).

A series given by a formula for \(S_n\)

  • \(S_1\) is the first term, so \(u_1=S_1\).
  • \(S_2=u_1+u_2\), so \(u_2=S_2-S_1\). In general \(u_n=S_n-S_{n-1}\).
  • The common difference is \(d=u_2-u_1\). If you have time, find \(u_3=S_3-S_2\) to check the difference is the same.
  • After that, treat it like any arithmetic sequence: \(u_n\), \(S_n\), and the calculator for inequalities and equations.

Using the calculator for n

  • For "greatest n with \(u_n<k\)", solve \(u_1+(n-1)d=k\), or use a table. Then take the whole number below the answer, and check it in the formula.
  • For "\(S_n=\) a value", enter \(S_n\) as a function of n and solve. A quadratic gives two roots: reject a negative or non-whole n.
  • Two unknowns, \(u_1\) and d, need two equations. Write both, then solve them together on the calculator.

Geometric sequences and linking the two

  • \(u_n=u_1r^{n-1}\), with \(r=\frac{u_2}{u_1}\). \(S_n=\frac{u_1(1-r^n)}{1-r}\).
  • When two sequences are linked, write down exactly which terms of the first one become \(u_1\) and \(u_2\) of the second. Then find r.

Infinite geometric series

  • If \(|r|<1\), \(S_\infty=\frac{u_1}{1-r}\). If \(|r|\ge1\) there is no sum to infinity.
  • In context, check \(|r|<1\) first, then decide whether the first term stands alone or fits the pattern.

On the calculator

Greatest n with \(u_n\) below a value

  • Enter \(u_n\) as a function of n and look at a table, or solve \(u_n=k\).
  • n must be a whole number. Take the whole number below the solution when the inequality is "less than".
  • Check the two values of n either side in the formula.

Screenshot and video to come

Solving \(S_n\) = a value

  • Write \(S_n\) in terms of n, using the formula \(S_n=\frac n2(2u_1+(n-1)d)\).
  • Solve \(S_n=\) the value for n. If there are two answers, one is usually negative: reject it.

Screenshot and video to come

Solving two equations for \(u_1\) and d

  • Write each piece of information as an equation in \(u_1\) and d.
  • Use the simultaneous equation solver, or solve one equation for \(u_1\) and substitute.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A series given by \(S_n\)

The sum of the first n terms of an arithmetic sequence is \(S_n=3n^2+2n\). Find (a) \(S_1\) and \(S_2\), (b) \(u_1\) and \(u_2\), (c) the common difference d.

Working
  1. (a) Put in n = 1, 2

    \(S_1=3(1)^2+2(1)=5\), \(S_2=3(2)^2+2(2)=16\)

    Substitute into the formula.

  2. (b) First term

    \(u_1=S_1=5\)

    One term, so the sum is the term.

  3. (b) Second term

    \(u_2=S_2-S_1=16-5=11\)

    \(S_2\) is \(u_1+u_2\).

  4. (c) Difference

    \(d=u_2-u_1=11-5=6\)

    Check: \(u_3=S_3-S_2=33-16=17\), and \(17-11=6\).

Video walkthrough coming
Greatest n with \(u_n\) below a value

An arithmetic sequence has \(u_1=5\) and \(d=6\). (a) Find \(u_{10}\). (b) Find the greatest value of n for which \(u_n<200\).

Working
  1. (a) \(u_{10}\)

    \(u_{10}=5+9(6)=59\)

    \(u_n=u_1+(n-1)d\).

  2. (b) Inequality

    \(5+(n-1)(6)<200\)

    Put the formula for \(u_n\) in.

  3. Solve

    From the calculator: \(n<33.5\)

    Solve \(5+6(n-1)=200\) or use a table.

  4. Whole number

    \(n=33\)

    Check: \(u_{33}=197<200\) and \(u_{34}=203>200\).

Video walkthrough coming
Solving \(S_n\) = a value

For the sequence with \(S_n=3n^2+2n\), find the value of n for which \(S_n=1240\).

Working
  1. Equation

    \(3n^2+2n=1240\)

    Set the formula equal to the value.

  2. Solve

    From the calculator: \(n=20\) or \(n=-20.67\)

    A quadratic has two roots.

  3. Choose

    \(n=20\)

    A number of terms is a positive whole number. Check: \(S_{20}=3(400)+40=1240\).

Video walkthrough coming
d and \(u_1\) from two terms

An arithmetic sequence has \(u_4=18\) and \(u_{11}=-10\). Find the common difference and the first term.

Working
  1. Two equations

    \(u_1+3d=18\) and \(u_1+10d=-10\)

    \(u_n=u_1+(n-1)d\) for \(n=4\) and \(n=11\).

  2. Subtract

    \(7d=-28\Rightarrow d=-4\)

    The difference of the terms is 7 steps of d.

  3. First term

    \(u_1=18-3(-4)=30\)

    Put d back into either equation. Check: \(30+10(-4)=-10\).

Video walkthrough coming
A term and a sum

An arithmetic sequence has \(u_1=30\) and \(d=-4\). Find (a) \(u_{60}\), (b) \(S_{100}\).

Working
  1. (a) The term

    \(u_{60}=30+59(-4)=-206\)

    \(u_n=u_1+(n-1)d\).

  2. (b) The sum

    \(S_{100}=\frac{100}{2}\left(2(30)+99(-4)\right)\)

    \(S_n=\frac n2(2u_1+(n-1)d)\).

  3. Evaluate

    \(S_{100}=50(60-396)=-16\,800\)

    From the calculator.

Video walkthrough coming
Arithmetic to geometric

The second term of an arithmetic sequence is 8 and the common difference is 2. The first and fourth terms of this sequence are the first two terms of a geometric sequence. Find the sixth term of the geometric sequence.

Working
  1. The arithmetic terms

    \(u_1=8-2=6\), \(u_4=6+3(2)=12\)

    Find the two terms you need.

  2. Geometric ratio

    \(r=\frac{12}{6}=2\)

    Second term over first.

  3. Sixth term

    \(u_6=6(2)^5=192\)

    \(u_n=u_1r^{n-1}\).

Video walkthrough coming
A sum to infinity from an arithmetic sequence

An arithmetic sequence has \(u_1=12\) and \(d=-3\). Its first and third terms are the first two terms of an infinite geometric series. Find the sum to infinity of the geometric series.

Working
  1. The arithmetic terms

    \(u_1=12\), \(u_3=12+2(-3)=6\)

    The first and the third.

  2. Geometric ratio

    \(r=\frac{6}{12}=0.5\)

    \(|r|<1\), so the series has a sum.

  3. Sum to infinity

    \(S_\infty=\frac{12}{1-0.5}=24\)

    Use \(\frac{u_1}{1-r}\).

Video walkthrough coming