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🎯⭐ INTERACTIVE LESSON

Arithmetic and Geometric Series

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Arithmetic and Geometric Series - Complete Interactive Lesson

Part 1: From Sequences to Series & Sigma Notation

➕ Arithmetic & Geometric Series

Part 1 of 5 — From Sequences to Series & Sigma Notation


Topics in This Part

Section
Sequence vs. Series
Sigma (Summation) Notation
Expanding and Evaluating a Sum

🔑 Key Concept: A sequence is an ordered list of numbers. A series is what you get when you add the terms of a sequence. This lesson is about adding them up — quickly, with formulas, instead of one term at a time.

Sequence vs. Series

Start with the sequence 3, 7, 11, 15, 193,\ 7,\ 11,\ 15,\ 19.

  • As a sequence, we just list the terms.
  • As a series, we add them: 3+7+11+15+19=553 + 7 + 11 + 15 + 19 = 55.
Looks likeResult
Sequence3, 7, 11, 15, 193,\ 7,\ 11,\ 15,\ 19a list
Series3+7+11+15+193 + 7 + 11 + 15 + 19a single number (5555)

A partial sum is the sum of the first nn terms, written SnS_n:

S1=3,S2=3+7=10,S3=3+7+11=21S_1 = 3,\quad S_2 = 3+7 = 10,\quad S_3 = 3+7+11 = 21

💡 The whole point of this lesson is to find SnS_n without adding every term by hand. Adding 100100 numbers is slow; a formula is instant.

Concept Check 🎯

Sigma (Summation) Notation

Mathematicians abbreviate a sum with the Greek capital letter sigma, Σ\Sigma:

∑k=1nak=a1+a2+a3+⋯+an\sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n

Read it as "the sum of aka_k as kk goes from 11 to nn."

  • kk is the index (a counter).
  • The number below Σ\Sigma is the starting value of kk.
  • The number above Σ\Sigma is the ending value of kk.
  • aka_k is the rule — plug in each kk and add the results.

Example

∑k=14(2k+1)=3⏟k=1+5⏟k=2+7⏟k=3+9⏟k=4=24\sum_{k=1}^{4} (2k + 1) = \underbrace{3}_{k=1} + \underbrace{5}_{k=2} + \underbrace{7}_{k=3} + \underbrace{9}_{k=4} = 24

⚠️ Watch the index range. ∑k=14\sum_{k=1}^{4} has four terms (k=1,2,3,4k = 1, 2, 3, 4). A sum like ∑k=04\sum_{k=0}^{4} would have five terms. Count carefully: (top) − (bottom) + 1.

Decode the Notation 🔽

Consider the sum ∑k=15(3k−2)\displaystyle\sum_{k=1}^{5} (3k - 2).

Evaluate the Sum 🧮

Write out the terms and add them.

1) ∑k=13(4k)= ?\displaystyle\sum_{k=1}^{3} (4k) = \,? 2) ∑k=14(k2)= ?\displaystyle\sum_{k=1}^{4} (k^2) = \,? 3) ∑k=24(k+5)= ?\displaystyle\sum_{k=2}^{4} (k + 5) = \,?

Part 2: The Arithmetic Series Formula

➕ Arithmetic & Geometric Series

Part 2 of 5 — The Arithmetic Series Formula


🔑 The Idea: An arithmetic series adds terms that change by a constant common difference dd. There's a beautiful shortcut — pair the first and last terms — that sums them all at once.

Gauss's Pairing Trick

Legend says young Gauss added 1+2+⋯+1001 + 2 + \cdots + 100 in seconds. The trick: pair the ends.

S=1+2+3+⋯+98+99+100S=100+99+98+⋯+3+2+1\begin{aligned} S &= 1 + 2 + 3 + \cdots + 98 + 99 + 100 \\ S &= 100 + 99 + 98 + \cdots + 3 + 2 + 1 \end{aligned}

Add the two lines term by term — every pair sums to 101101, and there are 100100 pairs:

2S=100×101=10100  ⇒  S=101002=50502S = 100 \times 101 = 10100 \;\Rightarrow\; S = \frac{10100}{2} = 5050

This works for any arithmetic series, giving the formula:

Sn=n2 (a1+an)S_n = \frac{n}{2}\,(a_1 + a_n)

💡 In words: the sum equals the number of terms times the average of the first and last term. (a1a_1 is the first term, ana_n is the last term, nn is how many terms.)

Concept Check 🎯

Worked Examples

Example 1 — first and last term known

Find 4+7+10+⋯+314 + 7 + 10 + \cdots + 31 (common difference d=3d = 3).

First find nn. Since an=a1+(n−1)da_n = a_1 + (n-1)d: 31=4+(n−1)(3)  ⇒  27=3(n−1)  ⇒  n−1=9  ⇒  n=1031 = 4 + (n-1)(3) \;\Rightarrow\; 27 = 3(n-1) \;\Rightarrow\; n - 1 = 9 \;\Rightarrow\; n = 10

Now apply the sum formula: S10=102(4+31)=5×35=175S_{10} = \frac{10}{2}(4 + 31) = 5 \times 35 = 175

Example 2 — last term unknown

Sum the first 2020 terms of 5,8,11,14,…5, 8, 11, 14, \dots (a1=5a_1 = 5, d=3d = 3).

Find the 2020th term first: a20=5+(20−1)(3)=5+57=62a_{20} = 5 + (20-1)(3) = 5 + 57 = 62. S20=202(5+62)=10×67=670S_{20} = \frac{20}{2}(5 + 62) = 10 \times 67 = 670

⚠️ You need the last term ana_n to use n2(a1+an)\frac{n}{2}(a_1 + a_n). If it's not given, compute it with an=a1+(n−1)da_n = a_1 + (n-1)d first.

Build the Sum 🔽

You're summing the first 1212 terms of 2, 6, 10, 14,…2,\ 6,\ 10,\ 14,\dots (a1=2a_1 = 2, d=4d = 4).

Sum the Series 🧮

1) 3+6+9+⋯+303 + 6 + 9 + \cdots + 30 (that's 1010 terms). S= ?S = \,? 2) First 2525 terms of 1,3,5,7,…1, 3, 5, 7, \dots (odd numbers). S= ?S = \,? 3) ∑k=18(2k+1)\displaystyle\sum_{k=1}^{8}(2k+1), which is 3+5+⋯+173+5+\cdots+17. S= ?S = \,?

Part 3: Finite Geometric Series

➕ Arithmetic & Geometric Series

Part 3 of 5 — Finite Geometric Series


🔑 The Idea: A geometric series multiplies by a constant common ratio rr each step (2+6+18+54+⋯2 + 6 + 18 + 54 + \cdots). Its sum also has a clean closed form.

The Finite Geometric Sum Formula

For a geometric series with first term a1a_1 and common ratio r≠1r \ne 1, the sum of the first nn terms is:

Sn=a1⋅1−r n1−rS_n = a_1 \cdot \frac{1 - r^{\,n}}{1 - r}

Where it comes from (the shift-and-subtract trick):

Sn=a1+a1r+a1r2+⋯+a1rn−1r Sn=  a1r+a1r2+⋯+a1rn−1+a1r n\begin{aligned} S_n &= a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} \\ r\,S_n &= \qquad\; a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} + a_1 r^{\,n} \end{aligned}

Subtract the second line from the first — almost everything cancels:

Sn−r Sn=a1−a1r n  ⇒  Sn(1−r)=a1(1−r n)S_n - r\,S_n = a_1 - a_1 r^{\,n} \;\Rightarrow\; S_n(1 - r) = a_1(1 - r^{\,n})

Divide by (1−r)(1 - r) to get the formula.

💡 Identify a1a_1 (first term) and rr (divide any term by the one before it). To find rr: r=a2a1r = \dfrac{a_2}{a_1}.

Concept Check 🎯

Worked Examples

Example 1 — 2+6+18+54+1622 + 6 + 18 + 54 + 162

Here a1=2a_1 = 2, r=3r = 3, and there are n=5n = 5 terms:

S5=2⋅1−351−3=2⋅1−243−2=2⋅−242−2=2⋅121=242S_5 = 2 \cdot \frac{1 - 3^5}{1 - 3} = 2 \cdot \frac{1 - 243}{-2} = 2 \cdot \frac{-242}{-2} = 2 \cdot 121 = 242

✅ Check: 2+6+18+54+162=2422 + 6 + 18 + 54 + 162 = 242 ✓

Example 2 — a fractional ratio

Sum the first 44 terms of 8+4+2+1+⋯8 + 4 + 2 + 1 + \cdots (a1=8a_1 = 8, r=12r = \frac{1}{2}):

S4=8⋅1−(12)41−12=8⋅1−11612=8⋅151612=8⋅158=15S_4 = 8 \cdot \frac{1 - \left(\frac{1}{2}\right)^4}{1 - \frac{1}{2}} = 8 \cdot \frac{1 - \frac{1}{16}}{\frac{1}{2}} = 8 \cdot \frac{\frac{15}{16}}{\frac{1}{2}} = 8 \cdot \frac{15}{8} = 15

✅ Check: 8+4+2+1=158 + 4 + 2 + 1 = 15 ✓

Set Up the Geometric Sum 🔽

You're summing the first 55 terms of 3+6+12+24+⋯3 + 6 + 12 + 24 + \cdots

Compute the Sum 🧮

1) 1+2+4+8+16+321 + 2 + 4 + 8 + 16 + 32 (a1=1a_1=1, r=2r=2, n=6n=6). S= ?S = \,? 2) ∑k=135⋅3 k−1\displaystyle\sum_{k=1}^{3} 5\cdot 3^{\,k-1}, i.e. 5+15+455 + 15 + 45. S= ?S = \,? 3) First 33 terms of 9+3+1+⋯9 + 3 + 1 + \cdots (r=13r = \frac{1}{3}): 9+3+19 + 3 + 1. S= ?S = \,?

Part 4: Infinite Geometric Series

➕ Arithmetic & Geometric Series

Part 4 of 5 — Infinite Geometric Series


🔑 The Big Surprise: You can add infinitely many numbers and still get a finite answer — but only for a geometric series when the terms shrink (∣r∣<1|r| < 1).

When Does an Infinite Sum Have a Value?

Take 12+14+18+116+⋯\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \cdots. The partial sums creep toward 11:

S1=0.5,S2=0.75,S3=0.875,S4=0.9375, ⋯→1S_1 = 0.5,\quad S_2 = 0.75,\quad S_3 = 0.875,\quad S_4 = 0.9375,\ \dots \to 1

They converge to a fixed number. This happens exactly when the common ratio satisfies ∣r∣<1|r| < 1 — each new term is a fraction of the one before, so the leftover shrinks to nothing.

Condition on rrBehaviorInfinite sum?
$r< 1$
$r= 1$
$r> 1$

In the formula Sn=a11−r n1−rS_n = a_1\dfrac{1 - r^{\,n}}{1 - r}, when ∣r∣<1|r| < 1 the term r n→0r^{\,n} \to 0 as n→∞n \to \infty. That leaves:

S∞=a11−r(valid only when ∣r∣<1)S_\infty = \frac{a_1}{1 - r}\qquad (\text{valid only when } |r| < 1)

⚠️ If ∣r∣≥1|r| \ge 1 the series diverges — there is no finite sum, and you must not plug into a11−r\frac{a_1}{1-r}.

Concept Check 🎯

Worked Examples

Example 1 — 12+14+18+⋯\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots

a1=12a_1 = \frac{1}{2}, r=12r = \frac{1}{2}, and ∣r∣<1|r| < 1 so it converges:

S∞=121−12=1212=1S_\infty = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1

Example 2 — a negative ratio

Sum 8−4+2−1+⋯8 - 4 + 2 - 1 + \cdots Here a1=8a_1 = 8 and r=−12r = -\frac{1}{2} (since −48=−12\frac{-4}{8} = -\frac{1}{2}), and ∣r∣=12<1|r| = \frac{1}{2} < 1:

S∞=81−(−12)=832=8⋅23=163S_\infty = \frac{8}{1 - \left(-\frac{1}{2}\right)} = \frac{8}{\frac{3}{2}} = 8 \cdot \frac{2}{3} = \frac{16}{3}

Example 3 — repeating decimal as a series

0.3‾=0.3+0.03+0.003+⋯0.\overline{3} = 0.3 + 0.03 + 0.003 + \cdots is geometric with a1=0.3a_1 = 0.3, r=0.1r = 0.1:

S∞=0.31−0.1=0.30.9=13S_\infty = \frac{0.3}{1 - 0.1} = \frac{0.3}{0.9} = \frac{1}{3}

💡 That's why 0.3‾=130.\overline{3} = \frac{1}{3} — an infinite geometric series in disguise!

Converge or Diverge? 🔽

For each series, decide whether the infinite sum exists.

Find the Infinite Sum 🧮

Use S∞=a11−rS_\infty = \dfrac{a_1}{1 - r} (fractions are fine, e.g. 16/3).

1) a1=4a_1 = 4, r=12r = \frac{1}{2}: S∞= ?S_\infty = \,? 2) a1=10a_1 = 10, r=15r = \frac{1}{5}: S∞= ?S_\infty = \,? 3) 9+3+1+13+⋯9 + 3 + 1 + \frac{1}{3} + \cdots (a1=9a_1 = 9, r=13r = \frac{1}{3}): S∞= ?S_\infty = \,?

Part 5: Mixed Practice & Mastery Check

➕ Arithmetic & Geometric Series

Part 5 of 5 — Mixed Practice & Mastery Check


You can now (1) read sigma notation, (2) sum an arithmetic series, (3) sum a finite geometric series, and (4) sum an infinite geometric series when it converges. Time to choose the right tool for each problem.

Quick Reference

SituationFormula
Arithmetic, first & last term knownSn=n2(a1+an)S_n = \dfrac{n}{2}(a_1 + a_n)
Find the last terman=a1+(n−1)da_n = a_1 + (n-1)d
Finite geometric (r≠1r \ne 1)Sn=a11−r n1−rS_n = a_1\dfrac{1 - r^{\,n}}{1 - r}
Infinite geometric ($r

Which type is it?

🔑 Arithmetic = constant difference (you add the same amount). Geometric = constant ratio (you multiply by the same amount). Check: is a2−a1a_2 - a_1 constant, or is a2/a1a_2 / a_1 constant?

⚠️ An infinite sum only has a value for a geometric series with ∣r∣<1|r| < 1. An infinite arithmetic series always diverges.

Pick the Right Tool 🔽

Classify each series and decide what to use.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.