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International Baccalaureate · DP 2: Applications and Interpretation HL

Revision: Number, Algebra and Finance

⌚ ~50 min ▢ 4 blocks ✓ 18 exercises ➣ Prerequisites : Sequences and Series, Loans and Annuities
I Toolbox: Arithmetic Series
Recappreviously studied — click to reread
Proposition — Sum of an Arithmetic Sequence
The sum of the first terms of an arithmetic sequence is given by the formula:$$S_n = \text{Number of terms} \times \frac{(\text{First term} + \text{Last term})}{2}$$
  • If a sequence starts at \(u_0\) and ends at \(u_n\), there are \(\boldsymbol{n+1}\) terms. The formula is:$$S_n = \frac{n+1}{2}(u_0 + u_n)$$
  • If a sequence starts at \(u_1\) and ends at \(u_n\), there are \(\boldsymbol{n}\) terms. The formula is:$$S_n = \frac{n}{2}(u_1 + u_n)$$
Example
Compute the sum \(1 + 2 + 3 + \cdots + 100\).

The terms \(1, 2, 3, \ldots, 100\) form an arithmetic sequence starting at \(u_1 = 1\) and ending at \(u_{100} = 100\), so there are \(100\) terms: $$\begin{aligned}[t] 1 + 2 + \cdots + 100 &= \frac{100}{2}\,(1 + 100) && \text{(} S_n = \tfrac{n}{2}(u_1 + u_n)\text{)} \\ &= 50 \times 101 && \text{(simplify } \tfrac{100}{2}\text{)} \\ &= 5\,050. \end{aligned}$$

Proposition — Sum from the First Term and the Common Difference
Let \((u_n)\) be an arithmetic sequence with common difference \(d\). The sum of its first terms can also be written with the first term and the common difference only, without computing the last term:
  • If the sequence starts at \(u_1\), the sum of its first \(\boldsymbol{n}\) terms is:$$S_n = \frac{n}{2}\big(2u_1 + (n-1)d\big)$$
  • If the sequence starts at \(u_0\), the sum of its first \(\boldsymbol{n+1}\) terms is:$$S_n = \frac{n+1}{2}\big(2u_0 + nd\big)$$
Example
An arithmetic sequence has first term \(u_1 = 3\) and common difference \(d = 4\).
  1. Compute the sum \(S_{12}\) of its first \(12\) terms.
  2. How many terms, starting from \(u_1\), must be added for the sum to equal \(210\)?

  1. The last term is not needed. With \(u_1 = 3\), \(d = 4\) and \(n = 12\): $$\begin{aligned}[t] S_{12} &= \frac{12}{2}\big(2u_1 + 11d\big) && \text{(} S_n = \tfrac{n}{2}(2u_1 + (n-1)d)\text{)} \\ &= 6 \times (2 \times 3 + 11 \times 4) \\ &= 6 \times 50 \\ &= 300. \end{aligned}$$
  2. Here the number of terms \(n\) is the unknown. First, \(S_n\) in terms of \(n\): $$\begin{aligned}[t] S_n &= \frac{n}{2}\big(2 \times 3 + (n-1) \times 4\big) && \text{(} S_n = \tfrac{n}{2}(2u_1 + (n-1)d)\text{)} \\ &= \frac{n}{2}\,(4n + 2) && \text{(expand)} \\ &= 2n^2 + n. \end{aligned}$$ So we look for the number of terms \(n\) such that \(2n^2 + n = 210\). The terms \(3, 7, 11, \ldots\) are positive, so each added term makes the sum larger: \(S_n\) increases with \(n\). A table of values gives
    \(n\) \(8\) \(9\) \(10\) \(11\)
    \(S_n = 2n^2 + n\) \(136\) \(171\) \(210\) \(253\)
    \(S_{10} = 210\), and as \(S_n\) increases, no other number of terms gives \(210\): \(n = \boldsymbol{10}\), so \(10\) terms must be added.
    Check: \(S_{10} = \frac{10}{2}(2 \times 3 + 9 \times 4) = 5 \times 42 = 210\).

Skills to practice
  • Calculating an Arithmetic Series
II Toolbox: Upper and Lower Bounds
Recappreviously studied — click to reread
Definition — Lower and Upper Bounds
For a rounded value:
  • The Lower Bound (LB) is the smallest possible value.
  • The Upper Bound (UB) is the value that the possible values never reach: every possible value is strictly below it, so we write “\(<\)” for the upper bound. The upper bound itself is not a possible value, but it is the number used in calculations.
Example
The length of a book is given as \(L = 12\) cm, rounded to the nearest cm (\(u=1\)).
  1. Calculate the lower bound and the upper bound of the length.
  2. Write the inequality representing the possible values for the true length.

  1. Bounds:
    • Lower bound: \(12 - 0.5 = 11.5\) cm.
    • Upper bound: \(12 + 0.5 = 12.5\) cm.
  2. Inequality: Therefore: \(11.5 \le L < 12.5\).

Method — Bounds of a Calculation
When performing calculations with rounded values (area, perimeter, speed, \(\dots\)), errors propagate.
Let \(A\) and \(B\) be two positive quantities with bounds \(A\in[A_{\min},A_{\max})\) and \(B\in[B_{\min},B_{\max})\) (with \(B_{\min}>0\) if dividing by \(B\)).
Operation Upper bound Lower bound
\(A + B\) \(A_{\max} + B_{\max}\) \(A_{\min} + B_{\min}\)
\(A - B\) \(A_{\max} - B_{\min}\) \(A_{\min} - B_{\max}\)
\(A \times B\) \(A_{\max} \times B_{\max}\) \(A_{\min} \times B_{\min}\)
\(A \div B\) \(A_{\max} \div B_{\min}\) \(A_{\min} \div B_{\max}\)
For a measurement, the lower bound \(A_{\min}\) is a possible value, and the upper bound \(A_{\max}\) is never reached. For a sum or a product, the lower bound is reached and the upper bound is not. For a difference or a quotient, each bound combines a bound that is reached with one that is not, so neither bound is reached: for example \(A_{\min} - B_{\max} < A - B < A_{\max} - B_{\min}\). The two quantities are measured independently: any possible value of \(A\) can go with any possible value of \(B\).
Example
A rectangle has length \(l = 10\) cm and width \(w = 5\) cm, both measured to the nearest cm. Calculate the upper and lower bounds of the area.

  • Step 1: Determine the bounds.
    Since the measurements are to the nearest 1 cm, the accuracy is \(\pm 0.5\) cm.
    • Length \(l\): Lower bound \(= 9.5\), Upper bound \(= 10.5\).
    • Width \(w\): Lower bound \(= 4.5\), Upper bound \(= 5.5\).
  • Step 2: Calculate the areas.
    • Upper bound of the area (using the upper bounds of the sides): $$ A_{\max} = 10.5 \times 5.5 = 57.75 \text{ cm}^2 $$
    • Lower bound of the area (using the lower bounds of the sides): $$ A_{\min} = 9.5 \times 4.5 = 42.75 \text{ cm}^2 $$

Skills to practice
  • Finding Upper and Lower Bounds
III Exam-Style Questions
Skills to practice
  • Working with Technology: Number, Algebra and Finance
  • Working with Technology: Further Number and Algebra
  • Quiz