4.3 Arithmetic Series and Sum

An arithmetic series is the sum of the terms of an arithmetic sequence. If the sequence is $a_1,a_2,a_3,\dots ,a_n$ with common difference $d$, the corresponding series is:

$a_1+a_2+a_3+\cdots +a_n$

---

Key Formulas

Formula 1 — When $a_1$, $a_n$, and $n$ are known:

$S_n=\frac{n}{2}(a_1+a_n)$

This is the most efficient formula when the first and last terms are both known.

Formula 2 — When $a_1$, $d$, and $n$ are known:

$S_n=\frac{n}{2}[2a_1+(n-1)d]$

This is derived by substituting $a_n=a_1+(n-1)d$ into Formula 1.

---

Derivation of the Sum Formula

Write the sum forwards and backwards:

$S_n=a_1+(a_1+d)+(a_1+2d)+\cdots +a_n$ $S_n=a_n+(a_n-d)+(a_n-2d)+\cdots +a_1$

Adding both rows term by term:

$2S_n=n(a_1+a_n)⟹S_n=\frac{n}{2}(a_1+a_n)$

---

Finding $n$ When It Is Unknown

If the first term $a_1$, last term $a_n$, and common difference $d$ are known but $n$ is not, use:

$n=\frac{a_n-a_1}{d}+1$

Example: Find the number of terms in $5+8+11+\cdots +50$.

$n=\frac{50-5}{3}+1=15+1=16$

---

Finding $n$ When $S_n$, $a_1$, and $d$ Are Given

Substitute into $S_n=\frac{n}{2}[2a_1+(n-1)d]$ and solve the resulting quadratic equation in $n$:

$d{n}^2+(2a_1-d)n-2S_n=0$

---

Finding the First Three Terms Given $n$, $a_n$, and $S_n$

Step 1: Use $S_n=\frac{n}{2}(a_1+a_n)$ to find $a_1$.

Step 2: Use $a_n=a_1+(n-1)d$ to find $d$.

Step 3: The first three terms are $a_1,a_1+d,a_1+2d$.

Example: Given $n=10$, $a_{10}=30$, $S_{10}=165$, find the first three terms.

• $165=\frac{10}{2}(a_1+30)\Rightarrow 33=a_1+30\Rightarrow a_1=3$
• $30=3+9d\Rightarrow d=3$
• First three terms: $3,6,9$

---

Sum of Multiples Between Two Bounds

To find the sum of all multiples of $k$ between $L$ and $U$:

1. Find $a_1$ = smallest multiple of $k$ that is $\ge L$.
2. Find $a_n$ = largest multiple of $k$ that is $\le U$.
3. Find $n=\frac{a_n-a_1}{k}+1$.
4. Apply $S_n=\frac{n}{2}(a_1+a_n)$.

Example: Sum of multiples of 4 between 14 and 523.

• $a_1=16$, $a_n=520$, $d=4$
• $n=\frac{520-16}{4}+1=127$
• $S_{127}=\frac{127}{2}(16+520)=\frac{127\times 536}{2}=34036$

---

Real-World Applications

In word problems where a quantity increases by a fixed amount each period (e.g., daily savings, monthly salary increments):

• The total number of periods = $n$ (number of terms)
• The fixed increase per period = $d$ (common difference)
• The initial value = $a_1$ (first term)

Example: A person saves Rs. 100 on day 1 and increases savings by Rs. 20 each day. Total savings in 30 days:

$S_{30}=\frac{30}{2}[2(100)+(30-1)(20)]=15[200+580]=15\times 780=11700$

---