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Number and Algebra - Arithmetic and Geometric sequences and series

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A sequence is an ordered list of numbers where each position is denoted by nn (n=1n=1 for the first term). Visually, if you plot the term number nn on the x-axis and the term value unu_n on the y-axis, the sequence appears as a series of discrete dots that are not connected by a continuous line.

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Arithmetic Sequences follow a linear pattern where each term is found by adding a common difference dd to the previous term (un+1=un+du_{n+1} = u_n + d). Graphically, these points fall perfectly on a straight line with a slope equal to dd. If d>0d > 0, the line slopes upward; if d<0d < 0, it slopes downward.

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Geometric Sequences follow an exponential pattern where each term is found by multiplying the previous term by a common ratio rr (un+1=un×ru_{n+1} = u_n \times r). Visually, if r>1r > 1, the points form a curve that grows increasingly steep (exponential growth). If 0<r<10 < r < 1, the points form a curve that levels out toward the x-axis (exponential decay).

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An Arithmetic Series represents the sum of the terms in an arithmetic sequence. A visual way to think of the sum SnS_n is the 'staircase' method: if you stack blocks representing the value of each term, the total area of the blocks is the sum. Reversing the sequence and adding it to itself creates a rectangle of height (u1+un)(u_1 + u_n) and length nn, which is why the formula is halved.

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A Geometric Series is the sum of terms in a geometric sequence. Unlike arithmetic series which grow infinitely, some geometric series can approach a finite total. This occurs when the sequence is 'converging' (i.e., ∣r∣<1|r| < 1). Visually, the additions become so small that the total sum approaches a horizontal limit or asymptote.

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Sigma Notation (∑\sum) is a shorthand way to write a long sum. The symbol ∑\sum is the Greek capital letter Sigma. The 'index' at the bottom (e.g., k=1k=1) tells you where to start, the number at the top (e.g., nn) tells you where to stop, and the expression to the right is the rule for each term.

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The Common Difference vs. Common Ratio: To find dd in an arithmetic sequence, subtract the first term from the second (u2−u1u_2 - u_1). To find rr in a geometric sequence, divide the second term by the first (u2u1\frac{u_2}{u_1}). Identifying these constants is the first step in solving any sequence problem.

📐Formulae

un=u1+(n−1)du_n = u_1 + (n - 1)d (Arithmetic nthn^{th} term)

Sn=n2(u1+un)S_n = \frac{n}{2}(u_1 + u_n) (Arithmetic sum using first and last terms)

Sn=n2[2u1+(n−1)d]S_n = \frac{n}{2}[2u_1 + (n - 1)d] (Arithmetic sum using u1u_1 and dd)

un=u1rn−1u_n = u_1 r^{n-1} (Geometric nthn^{th} term)

Sn=u1(rn−1)r−1S_n = \frac{u_1(r^n - 1)}{r - 1} (Geometric sum when r>1r > 1)

Sn=u1(1−rn)1−rS_n = \frac{u_1(1 - r^n)}{1 - r} (Geometric sum when r<1r < 1)

S∞=u11−rS_{\infty} = \frac{u_1}{1 - r} (Sum to infinity for a convergent geometric series, where ∣r∣<1|r| < 1)

💡Examples

Problem 1:

In an arithmetic sequence, the 3rd3^{rd} term is 1212 and the 10th10^{th} term is 4747. Find the first term u1u_1, the common difference dd, and the sum of the first 1010 terms S10S_{10}.

Solution:

Step 1: Set up two equations using un=u1+(n−1)du_n = u_1 + (n-1)d. u3=u1+2d=12u_3 = u_1 + 2d = 12 u10=u1+9d=47u_{10} = u_1 + 9d = 47 Step 2: Subtract the first equation from the second: (u1+9d)−(u1+2d)=47−12(u_1 + 9d) - (u_1 + 2d) = 47 - 12 7d=35  ⟹  d=57d = 35 \implies d = 5 Step 3: Substitute d=5d=5 back into u3u_3: u1+2(5)=12  ⟹  u1+10=12  ⟹  u1=2u_1 + 2(5) = 12 \implies u_1 + 10 = 12 \implies u_1 = 2 Step 4: Calculate S10S_{10} using Sn=n2(u1+un)S_n = \frac{n}{2}(u_1 + u_n): S10=102(2+47)=5×49=245S_{10} = \frac{10}{2}(2 + 47) = 5 \times 49 = 245.

Explanation:

This approach uses simultaneous equations to find the two unknown variables (u1u_1 and dd) that define the arithmetic sequence. Once these are known, any specific term or sum can be calculated using the standard formulas.

Problem 2:

A geometric sequence has a first term of 100100 and a common ratio of 0.80.8. Find the 5th5^{th} term and the sum to infinity.

Solution:

Step 1: Identify given values: u1=100u_1 = 100, r=0.8r = 0.8. Step 2: Calculate the 5th5^{th} term using un=u1rn−1u_n = u_1 r^{n-1}: u5=100×(0.8)5−1=100×(0.8)4u_5 = 100 \times (0.8)^{5-1} = 100 \times (0.8)^4 u5=100×0.4096=40.96u_5 = 100 \times 0.4096 = 40.96 Step 3: Determine if S∞S_{\infty} exists. Since ∣0.8∣<1|0.8| < 1, the series converges. Step 4: Use S∞=u11−rS_{\infty} = \frac{u_1}{1 - r}: S∞=1001−0.8=1000.2=500S_{\infty} = \frac{100}{1 - 0.8} = \frac{100}{0.2} = 500.

Explanation:

To solve geometric problems, first check if the common ratio allows for a sum to infinity. We use the power rule for the general term and division for the sum to infinity, which represents the limit the sum approaches as more terms are added.