Vector illustration of an arithmetic sequence: number line with equal steps, rising linear graph, and three step bars
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Arithmetic Sequence Formula: A Complete Guide with Real-Life Examples

An arithmetic sequence is a list of numbers that increases or decreases by a constant amount, called the common difference. The nth term equals first term plus (n−1) times the difference: aₙ = a₁ + (n−1)d. The sum of n terms is Sₙ = n/2 × (first + last).

Table of Contents

  1. What Is an Arithmetic Sequence?

  2. Core Formulas and How They Connect

  3. A Step-by-Step Strategy That Always Works

  4. Real-Life Applications and Worked Examples

  5. Common Pitfalls, Extensions, and Teaching Tips

1) What Is an Arithmetic Sequence?

An arithmetic sequence is a sequence of numbers where each term is obtained by adding a fixed value to the previous one. That fixed value is the common difference, usually denoted by d. If the first term is a₁, then the second is a₂ = a₁ + d, the third is a₃ = a₁ + 2d, and so on. Because the change between terms is constant, arithmetic sequences model situations with steady, linear growth or decline.

Notation and meaning. We write the sequence as a₁, a₂, a₃, … with n indexing the term number. The subscript tells you the position of a term, not its value. The value of the nth term is aₙ, and it depends on the starting value (a₁) and the step size (d). The index n almost always starts at 1 in textbook problems, unless stated otherwise.

Visual intuition. If you plot the points (n,an)(n, aₙ) on a coordinate plane, you get a straight line. In fact, the arithmetic sequence is the discrete counterpart of the linear function y = mx + b: the slope m corresponds to d, and the intercept b relates to the starting value. This connection makes arithmetic sequences ideal for modeling budgeting with fixed weekly contributions, regular seating patterns, construction layouts with uniform spacing, and other predictable stepwise changes.

Quick checks to confirm it’s arithmetic.

  • The difference between consecutive terms is the same: ak+1ak=da_{k+1} – a_k = d for all relevant kk.

  • Graph points lie on a straight line when plotted against their term positions.

  • A missing term can be found by consistent addition or subtraction of d.

2) Core Formulas and How They Connect

The formulas below are the backbone of every arithmetic-sequence problem. While they look different, they describe the same structure from different angles.

Goal Formula Use When Short Example
Nth term (explicit) aₙ = a₁ + (n−1)d You know the first term and the common difference, or you can deduce them, and you want a specific term. If a₁ = 7, d = 3, then a₁₀ = 7 + 9×3 = 34.
Nth term (recursive) a₁ given; aₙ = aₙ₋₁ + d You prefer building terms step by step (useful for spreadsheets or code). Start at 7; keep adding 3 to get 10, 13, 16…
Sum of n terms Sₙ = n/2 × (a₁ + aₙ) You need the total of the first n terms and can find aₙ. First 10 terms of 7,10,13,… sum to 10/2×(7+34)=5×41=20510/2 × (7 + 34) = 5 × 41 = 205.
Sum without aₙ Sₙ = n/2 × [2a₁ + (n−1)d] You don’t know aₙ yet but you know a₁ and d. With a₁=7, d=3, n=10, S10=10/2×[14+27]=5×41=205S_{10} = 10/2 × [14 + 27] = 5 × 41 = 205.

Why these work. The sum formula averages the first and last terms and multiplies by the number of terms. The explicit nth-term formula comes from repeated addition: after n1n−1 steps of size d starting at a₁, you land at a₁ + (n−1)d.

Choosing the right formula.

  • If the position is the goal (e.g., “Find the 25th term”), think aₙ = a₁ + (n−1)d.

  • If the total is the goal (e.g., “How many tiles in the first 15 rows?”), think Sₙ, using whichever version fits the given information.

  • If terms are defined by process (e.g., “Start at 40, add 12 each month”), a recursive view is convenient for simulation or coding.

3) A Step-by-Step Strategy That Always Works

When faced with an arithmetic-sequence problem, slowing down to identify what you know prevents most errors. Use the following compact routine; it is short enough to apply mentally after a bit of practice.

Step 1—Identify the pattern. Decide whether the situation grows by equal increments. If increases vary or multiply, it may not be arithmetic.

Step 2—Name the parameters. Record a₁ (starting value), d (common difference), and n (term position or count of terms). If one is missing, express it in terms of the others using the explicit formula.

Step 3—Pick the formula.

  • Target a term → aₙ formula.

  • Target a total → Sₙ formula.

  • Need iteration or coding → recursive form.

Step 4—Check units and indexing. Confirm whether the first relevant term is the 1st or the 0th in your context. For totals, verify that n truly counts the number of terms being added.

Mini-case (strategy in action). A mentor program offers stipends that form an arithmetic sequence: 120,135, 150, … with common difference15. A student wants to know the 10th stipend and the total of the first 10 stipends. Identify a₁=120, d=15, n=10. Then a₁₀ = 120 + 9×15 = 255. For the total, S₁₀ = 10/2 × (120 + 255) = 5 × 375 = 1,875. The routine cleanly delivers both answers.

4) Real-Life Applications and Worked Examples

Arithmetic sequences show up in planning, construction, schooling, and personal finance whenever changes are uniform. The examples here balance clarity and practicality so you can see the formulas at work without heavy abstraction.

Example 1: Saving with fixed contributions (no interest).
You deposit 50 every week into an envelope, starting with50 in week 1. The deposits form the sequence 50, 100, 150, … where each term represents your cumulative money added (not your bank balance with interest). After n weeks, the weekly addition forms an arithmetic sequence with a₁=50, d=50. If you want the total deposited after 12 weeks, that total is the sum of the sequence’s terms:
S12=12/2×[2×50+(121)×50]=6×[100+550]=6×650=3,900S_{12} = 12/2 × [2×50 + (12−1)×50] = 6 × [100 + 550] = 6 × 650 = 3{,}900.
Interpreting it: you have contributed 3,900 over 12 weeks. If you track only each week's deposit (50 each week), then the sequence of deposits is constant and the total is just 50 × 12. The example shows how clarifying what each term represents avoids misinterpretation. Example 2: Classroom seating with regular gaps.A teacher arranges chairs in rows. The first row has 18 seats, and each successive row has 2 fewer seats to maintain sight lines: 18, 16, 14, … This is an arithmetic sequence with a₁=18 and d=−2.  <ul data-start="6879" data-end="7237">  	<li data-start="6879" data-end="6952"> Find the 9th row's seats. <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>a</mi><mn>9</mn></msub><mo>=</mo><mn>18</mn><mo>+</mo><mo stretchy="false">(</mo><mn>9</mn><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mo>=</mo><mn>18</mn><mo>−</mo><mn>16</mn><mo>=</mo><mn>2</mn></mrow><annotation encoding="application/x-tex">a_9 = 18 + (9−1)(−2) = 18 − 16 = 2</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord"><span class="mord mathnormal">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mrel">=</span></span><span class="base"><span class="mord">18</span><span class="mbin">+</span></span><span class="base"><span class="mopen">(</span><span class="mord">9</span><span class="mbin">−</span></span><span class="base"><span class="mord">1</span><span class="mclose">)</span><span class="mopen">(</span><span class="mord">−</span><span class="mord">2</span><span class="mclose">)</span><span class="mrel">=</span></span><span class="base"><span class="mord">18</span><span class="mbin">−</span></span><span class="base"><span class="mord">16</span><span class="mrel">=</span></span><span class="base"><span class="mord">2</span></span></span></span>. </li>  	<li data-start="6953" data-end="7237"> Total seats in the first 9 rows. <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>S</mi><mn>9</mn></msub><mo>=</mo><mn>9</mn><mi mathvariant="normal">/</mi><mn>2</mn><mo>×</mo><mo stretchy="false">(</mo><mn>18</mn><mo>+</mo><mn>2</mn><mo stretchy="false">)</mo><mo>=</mo><mn>4.5</mn><mo>×</mo><mn>20</mn><mo>=</mo><mn>90</mn></mrow><annotation encoding="application/x-tex">S_9 = 9/2 × (18 + 2) = 4.5 × 20 = 90</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord"><span class="mord mathnormal">S</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">9</span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mrel">=</span></span><span class="base"><span class="mord">9/2</span><span class="mbin">×</span></span><span class="base"><span class="mopen">(</span><span class="mord">18</span><span class="mbin">+</span></span><span class="base"><span class="mord">2</span><span class="mclose">)</span><span class="mrel">=</span></span><span class="base"><span class="mord">4.5</span><span class="mbin">×</span></span><span class="base"><span class="mord">20</span><span class="mrel">=</span></span><span class="base"><span class="mord">90</span></span></span></span>.The negative d simply means the sequence decreases by the same amount each step. A quick sense-check tells us the last row looks too small for comfort—useful insight when planning real layouts. </li> </ul> Example 3: Tiling a staircase.A staircase has 12 steps. The first step needs 6 tiles, and each higher step needs 1 more tile due to increased width: 6, 7, 8, …  <ul data-start="7424" data-end="7736">  	<li data-start="7424" data-end="7493"> How many tiles for the 12th step? <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>a</mi><mn>12</mn></msub><mo>=</mo><mn>6</mn><mo>+</mo><mn>11</mn><mo>×</mo><mn>1</mn><mo>=</mo><mn>17</mn></mrow><annotation encoding="application/x-tex">a_{12} = 6 + 11×1 = 17</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord"><span class="mord mathnormal">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">12</span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mrel">=</span></span><span class="base"><span class="mord">6</span><span class="mbin">+</span></span><span class="base"><span class="mord">11</span><span class="mbin">×</span></span><span class="base"><span class="mord">1</span><span class="mrel">=</span></span><span class="base"><span class="mord">17</span></span></span></span>. </li>  	<li data-start="7494" data-end="7736"> Total tiles for all steps. <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>S</mi><mn>12</mn></msub><mo>=</mo><mn>12</mn><mi mathvariant="normal">/</mi><mn>2</mn><mo>×</mo><mo stretchy="false">(</mo><mn>6</mn><mo>+</mo><mn>17</mn><mo stretchy="false">)</mo><mo>=</mo><mn>6</mn><mo>×</mo><mn>23</mn><mo>=</mo><mn>138</mn></mrow><annotation encoding="application/x-tex">S_{12} = 12/2 × (6 + 17) = 6 × 23 = 138</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord"><span class="mord mathnormal">S</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">12</span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mrel">=</span></span><span class="base"><span class="mord">12/2</span><span class="mbin">×</span></span><span class="base"><span class="mopen">(</span><span class="mord">6</span><span class="mbin">+</span></span><span class="base"><span class="mord">17</span><span class="mclose">)</span><span class="mrel">=</span></span><span class="base"><span class="mord">6</span><span class="mbin">×</span></span><span class="base"><span class="mord">23</span><span class="mrel">=</span></span><span class="base"><span class="mord">138</span></span></span></span>.A foreman can use this to order materials without overbuying, demonstrating how arithmetic sequences turn a qualitative build pattern into a precise quantity. </li> </ul> Example 4: Salary schedule with steady increments.A small firm pays a new hire42,000 in year 1 and increases salary by 1,800 each year for five years. That is a₁=42,000, d=1,800, n=5 for the first five post-hire years.  <ul data-start="7983" data-end="8289">  	<li data-start="7983" data-end="8052"> Year 5 salary. <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>a</mi><mn>5</mn></msub><mo>=</mo><mn>42,000</mn><mo>+</mo><mo stretchy="false">(</mo><mn>5</mn><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>×</mo><mn>1,800</mn><mo>=</mo><mn>49,200</mn></mrow><annotation encoding="application/x-tex">a_5 = 42{,}000 + (5−1)×1{,}800 = 49{,}200</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord"><span class="mord mathnormal">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mrel">=</span></span><span class="base"><span class="mord">42</span><span class="mord"><span class="mpunct">,</span></span><span class="mord">000</span><span class="mbin">+</span></span><span class="base"><span class="mopen">(</span><span class="mord">5</span><span class="mbin">−</span></span><span class="base"><span class="mord">1</span><span class="mclose">)</span><span class="mbin">×</span></span><span class="base"><span class="mord">1</span><span class="mord"><span class="mpunct">,</span></span><span class="mord">800</span><span class="mrel">=</span></span><span class="base"><span class="mord">49</span><span class="mord"><span class="mpunct">,</span></span><span class="mord">200</span></span></span></span>. </li>  	<li data-start="8053" data-end="8289"> Total earnings over 5 years. <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>S</mi><mn>5</mn></msub><mo>=</mo><mn>5</mn><mi mathvariant="normal">/</mi><mn>2</mn><mo>×</mo><mo stretchy="false">(</mo><mn>42,000</mn><mo>+</mo><mn>49,200</mn><mo stretchy="false">)</mo><mo>=</mo><mn>2.5</mn><mo>×</mo><mn>91,200</mn><mo>=</mo><mn>228,000</mn></mrow><annotation encoding="application/x-tex">S_5 = 5/2 × (42{,}000 + 49{,}200) = 2.5 × 91{,}200 = 228{,}000</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord"><span class="mord mathnormal">S</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mrel">=</span></span><span class="base"><span class="mord">5/2</span><span class="mbin">×</span></span><span class="base"><span class="mopen">(</span><span class="mord">42</span><span class="mord"><span class="mpunct">,</span></span><span class="mord">000</span><span class="mbin">+</span></span><span class="base"><span class="mord">49</span><span class="mord"><span class="mpunct">,</span></span><span class="mord">200</span><span class="mclose">)</span><span class="mrel">=</span></span><span class="base"><span class="mord">2.5</span><span class="mbin">×</span></span><span class="base"><span class="mord">91</span><span class="mord"><span class="mpunct">,</span></span><span class="mord">200</span><span class="mrel">=</span></span><span class="base"><span class="mord">228</span><span class="mord"><span class="mpunct">,</span></span><span class="mord">000</span></span></span></span>.If the company adds a bonus only in year 5, that is not arithmetic; it's a one-off adjustment after the arithmetic pattern. </li> </ul> Example 5: Finding a missing term in context.Suppose runway lights are placed along a straight taxiway so that distances between consecutive lights are equal. The first light is at the start, the 10th light is at 135 m, and the spacing is constant. What is the distance to the first few lights? Treat positions as an arithmetic sequence where a₁ = 0 (start) and a₁₀ = 135. Solve for d using <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>a</mi><mi>n</mi></msub><mo>=</mo><msub><mi>a</mi><mn>1</mn></msub><mo>+</mo><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mi>d</mi></mrow><annotation encoding="application/x-tex">aₙ = a₁ + (n−1)d</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord"><span class="mord mathnormal">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathnormal mtight">n</span></span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mrel">=</span></span><span class="base"><span class="mord"><span class="mord mathnormal">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist"><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span><span class="vlist-s">​</span></span></span></span></span><span class="mbin">+</span></span><span class="base"><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mbin">−</span></span><span class="base"><span class="mord">1</span><span class="mclose">)</span><span class="mord mathnormal">d</span></span></span></span>:<span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>135</mn><mo>=</mo><mn>0</mn><mo>+</mo><mn>9</mn><mi>d</mi><mo>⇒</mo><mi>d</mi><mo>=</mo><mn>15</mn></mrow><annotation encoding="application/x-tex">135 = 0 + 9d \Rightarrow d = 15</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="mord">135</span><span class="mrel">=</span></span><span class="base"><span class="mord">0</span><span class="mbin">+</span></span><span class="base"><span class="mord">9</span><span class="mord mathnormal">d</span><span class="mrel">⇒</span></span><span class="base"><span class="mord mathnormal">d</span><span class="mrel">=</span></span><span class="base"><span class="mord">15</span></span></span></span> meters.So light positions are 0 m, 15 m, 30 m, … This calculation is common in layout and safety design. Example 6: Inverse problems—how many terms do we need?You want at least3,000 in a project budget by making regular 125 contributions, starting now and adding the same amount each month. How many months until the sum of contributions first reaches or exceeds3,000? Here a₁=125, d=125, and you need the smallest n such that
Sn=n2[2×125+(n1)×125]3,000.Sₙ = \dfrac{n}{2} [2×125 + (n−1)×125] \ge 3{,}000.
That simplifies to Sn=125n2(n+1)3,000Sₙ = \dfrac{125n}{2}(n+1) \ge 3{,}000.
1252(n2+n)3,000n2+n48.\dfrac{125}{2}(n^2 + n) \ge 3{,}000 \Rightarrow n^2 + n \ge 48.
Try n = 6: 36+6=4236 + 6 = 42 (too small). n = 7: 49+7=5649 + 7 = 56 (enough).
It will take 7 months of contributions to meet or exceed the goal under a pure arithmetic scheme.

Example 7: Deriving the sum formula quickly.
Write the first n terms forward and backward:
Sn=a1+(a1+d)+(a1+2d)++anSₙ = a₁ + (a₁ + d) + (a₁ + 2d) + \cdots + aₙ
Sn=an+(and)+(an2d)++a1Sₙ = aₙ + (aₙ − d) + (aₙ − 2d) + \cdots + a₁
Add vertically to pair terms: each column sums to a1+ana₁ + aₙ, and there are n columns. Hence 2Sn=n(a1+an)2Sₙ = n(a₁ + aₙ) so Sn=n2(a1+an)Sₙ = \dfrac{n}{2}(a₁ + aₙ). This classic trick also explains why the average of first and last term governs the total.

5) Common Pitfalls, Extensions, and Teaching Tips

Mistaking geometric for arithmetic. In a geometric sequence, you multiply by a constant ratio rr; in an arithmetic sequence, you add by a constant difference d. If your terms double or halve, that’s not arithmetic. Check one or two gaps to be sure.

Indexing off by one. Many errors stem from plugging n with the wrong starting index. If a schedule labels the first payment as “Month 0,” the explicit formula needs adjusting, or you rewrite the scenario so that the first counted term is a₁.

Sign of the difference. A decreasing pattern means d < 0. The formulas still hold. For totals, decreasing sequences can still have positive sums if the early terms are large enough; just be clear about how many terms are being added.

Forgetting units and interpretation. Be explicit about what each term represents—a deposit, a row’s seat count, a tile count per step, or a position along a line. The same algebra gives different real-world meanings depending on context.

Connecting to linear functions. Writing aₙ = a₁ + (n−1)d as aₙ = dn + (a₁ − d) shows the line form y = mx + b with m = d. This unlocks graphing, quick predictions, and comparison to other linear models. In teaching, switching between the sequence view and the graph view deepens understanding.

Practical extension—spreadsheets. The recursive form an=an1+daₙ = aₙ₋₁ + d is perfect for a quick spreadsheet: put a₁ in the first cell, then copy a formula that adds d down the column. For totals, use the average-of-first-and-last insight or the closed form Sₙ to verify the spreadsheet sum.

Designing assessments. Mix direct formula questions (find a37a_{37}) with context problems (tiles, seating, budgets). Include at least one inverse problem (find n given a target total) to ensure students can work backward from the sum formula.

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