1. **Stating the Problem** We will learn about arithmetic sequences and series, understand their formulas, and then solve 5 example problems of varying difficulty with detailed steps and explanations. 2. **Understanding Arithmetic Sequence** An arithmetic sequence is a list of numbers where the difference between consecutive terms is constant. This difference is called the "common difference" and is denoted by $d$. If the first term is $a_1$, the second term is $a_2 = a_1 + d$, the third term is $a_3 = a_2 + d = a_1 + 2d$, and so on. The general formula for the $n$-th term is: $$ a_n = a_1 + (n - 1)d $$ 3. **Understanding Arithmetic Series** An arithmetic series is the sum of the terms of an arithmetic sequence. If we want to sum the first $n$ terms, the series sum $S_n$ is: $$ S_n = a_1 + a_2 + a_3 + \cdots + a_n $$ Using the formula: $$ S_n = \frac{n}{2} (a_1 + a_n) $$ or equivalently substituting $a_n$: $$ S_n = \frac{n}{2} [2a_1 + (n - 1)d] $$ This formula is very useful for finding the sum of a known number of terms quickly. --- ### Example Problems **Problem 1 (Easy):** Find the 10th term of an arithmetic sequence where $a_1 = 3$ and $d = 4$. **Solution:** - Step 1: Use the formula $a_n = a_1 + (n - 1)d$. - Step 2: Substitute $n=10$, $a_1=3$, $d=4$. $$ a_{10} = 3 + (10 - 1) \times 4 = 3 + 9 \times 4 = 3 + 36 = 39 $$ **Answer:** The 10th term is 39. --- **Problem 2 (Easy):** Calculate the sum of the first 5 terms of an arithmetic sequence with $a_1 = 7$ and $d = 2$. **Solution:** - Step 1: Find the 5th term using $a_n = a_1 + (n - 1)d$. $$ a_5 = 7 + (5 - 1) \times 2 = 7 + 8 = 15 $$ - Step 2: Use $S_n = \frac{n}{2} (a_1 + a_n)$: $$ S_5 = \frac{5}{2} (7 + 15) = \frac{5}{2} \times 22 = 5 \times 11 = 55 $$ **Answer:** The sum of the first 5 terms is 55. --- **Problem 3 (Medium):** An arithmetic sequence has the 3rd term $a_3 = 14$ and the 7th term $a_7 = 26$. Find $a_1$ and $d$. **Solution:** - Step 1: Write equations from the formula $a_n = a_1 + (n - 1)d$: $$ a_3 = a_1 + 2d = 14 $$ $$ a_7 = a_1 + 6d = 26 $$ - Step 2: Subtract the first from the second: $$ (a_1 + 6d) - (a_1 + 2d) = 26 - 14 \Rightarrow 4d = 12 \Rightarrow d = 3 $$ - Step 3: Substitute $d=3$ into the first equation: $$ a_1 + 2 \times 3 = 14 \Rightarrow a_1 + 6 = 14 \Rightarrow a_1 = 8 $$ **Answer:** $a_1 = 8$, common difference $d = 3$. --- **Problem 4 (Medium):** Find the sum of the first 20 terms if $a_1 = 5$ and $d = -2$. **Solution:** - Step 1: Find the 20th term: $$ a_{20} = 5 + (20 - 1)(-2) = 5 + 19 \times (-2) = 5 - 38 = -33 $$ - Step 2: Use the sum formula: $$ S_{20} = \frac{20}{2} (5 + (-33)) = 10 \times (-28) = -280 $$ **Answer:** The sum of the first 20 terms is -280. --- **Problem 5 (Hard):** The sum of the first 15 terms of an arithmetic sequence is 270 and the sum of the next 15 terms (terms 16 to 30) is 525. Find $a_1$ and $d$. **Solution:** - Step 1: Write formulas for sums: $$ S_{15} = \frac{15}{2} [2a_1 + (15 - 1)d] = 270 $$ $$ S_{30} = \frac{30}{2} [2a_1 + (30 - 1)d] $$ - Step 2: The sum of terms 16 to 30 is $S_{30} - S_{15} = 525$: $$ S_{30} = S_{15} + 525 = 270 + 525 = 795 $$ - Step 3: Write two equations: $$ \frac{15}{2} [2a_1 + 14d] = 270 \Rightarrow 15[2a_1 + 14d] = 540 \Rightarrow 2a_1 + 14d = 36 $$ $$ \frac{30}{2} [2a_1 + 29d] = 795 \Rightarrow 15[2a_1 + 29d] = 795 \Rightarrow 2a_1 + 29d = 53 $$ - Step 4: Subtract equations: $$ (2a_1 + 29d) - (2a_1 + 14d) = 53 - 36 \Rightarrow 15d = 17 \Rightarrow d = \frac{17}{15} $$ - Step 5: Substitute $d$ back: $$ 2a_1 + 14 \times \frac{17}{15} = 36 \Rightarrow 2a_1 + \frac{238}{15} = 36 \Rightarrow 2a_1 = 36 - \frac{238}{15} = \frac{540}{15} - \frac{238}{15} = \frac{302}{15} $$ $$ a_1 = \frac{302}{30} = \frac{151}{15} $$ **Answer:** $a_1 = \frac{151}{15}$, $d = \frac{17}{15}$.