Arithmetic Progressions Class 10 NCERT Chapter Guide

Understanding the Core Concepts of an AP

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms remains constant. This constant difference is called the common difference, denoted by $d$. The first term of the sequence is denoted by $a$. Let's represent a general AP: $a, a+d, a+2d, a+3d, \dots$

If the common difference $d > 0$, the AP is increasing (e.g., $2, 5, 8, 11, \dots$ where $d=3$). If $d < 0$, the sequence decreases (e.g., $10, 8, 6, 4, \dots$ where $d=-2$). If $d = 0$, all terms are identical. To find the $n$-th term of an AP, we use the formula:
$a_n = a + (n - 1)d$
Here, $a_n$ represents the general term, $a$ is the first term, $n$ is the position of the term, and $d$ is the common difference. Mastering this formula helps us identify missing terms or determine whether a specific number belongs to a given AP.

Step-by-Step Derivation & Use of AP Formulas

1. Identifying the Variables — Before applying any formula, identify your key values from the given sequence: $a$ (first term), $d$ (common difference, calculated as $a_2 - a_1$), and $n$ (the term's position, which must always be a positive integer).
2. Calculating the $n$-th Term ($a_n$) — Use the general term formula $a_n = a + (n-1)d$. Substitute the known values to find the specific term, or use it to check if a given value exists in the series by solving for $n$.
3. Deriving the Sum of $n$ Terms ($S_n$) — To find the sum of the first $n$ terms, we use the formula: $S_n = \frac{n}{2} [2a + (n-1)d]$. If the last term ($l = a_n$) is known, we can simplify this to: $S_n = \frac{n}{2} [a + l]$.

Step-by-Step Worked Examples

• Example 1: Find the 15th term of the AP: 3, 8, 13, 18, ... 1. Identify parameters: Here, the first term $a = 3$, and the common difference $d = 8 - 3 = 5$. We need to find the 15th term ($n = 15$). 2. Apply the formula: $a_n = a + (n-1)d$ 3. Substitute values: $a_{15} = 3 + (15 - 1) \times 5 = 3 + 14 \times 5 = 3 + 70 = 73$. Result: The 15th term is 73.
• Example 2: Find the sum of the first 20 terms of the AP: 5, 9, 13, 17, ... 1. Identify parameters: First term $a = 5$, common difference $d = 9 - 5 = 4$, and number of terms $n = 20$. 2. Apply the sum formula: $S_n = \frac{n}{2} [2a + (n-1)d]$ 3. Substitute values: $S_{20} = \frac{20}{2} [2(5) + (20-1)4] = 10 [10 + 19 \times 4] = 10 [10 + 76] = 10 \times 86 = 860$. Result: The sum of the first 20 terms is 860.

Common Board Exam Mistakes & Tips

1. Fractional or Negative $n$ values: The number of terms ($n$) must always be a positive integer ($n \in \mathbb{Z}^+$). If you solve for $n$ and get a fraction or negative number, the given value is not a term of that AP.
2. Sign Mistakes with Common Difference ($d$): Remember that $d = a_{k+1} - a_k$. For a decreasing AP, $d$ will be negative. Don't omit the minus sign, or it will throw off all calculation steps.
3. Formula Confusion: Do not swap the $n$-th term formula with the sum formula. Read the question carefully: does it ask for 'the value of the 10th term' ($a_{10}$) or 'the sum of the first 10 terms' ($S_{10}$)?
4. Selecting terms for word problems: When asked to assume 3 numbers in an AP whose sum is known, use $(a-d)$, $a$, and $(a+d)$. Their sum is $3a$, which easily yields the value of $a$.

Practice Questions with Solutions

• Q: Which term of the AP: 21, 18, 15, ... is -81? A: Step 1: Identify the parameters of the given AP. The first term $a = 21$ and the common difference $d = 18 - 21 = -3$. We are given the $n$-th term $a_n = -81$. Step 2: Use the formula for the $n$-th term: $a_n = a + (n-1)d$. Substitute the known values: $-81 = 21 + (n-1)(-3)$. Step 3: Solve for $n$. Subtract 21 from both sides: $-102 = (n-1)(-3)$. Divide both sides by $-3$: $34 = n-1$. Therefore, $n = 35$. Final answer: The 35th term of the given AP is -81.
• Q: Find the sum of the first 15 multiples of 8. A: Step 1: Write down the AP formed by the first 15 multiples of 8. The sequence is: 8, 16, 24, 32, ... up to 15 terms. Here, $a = 8$, $d = 8$, and $n = 15$. Step 2: Apply the sum of $n$ terms formula: $S_n = \frac{n}{2} [2a + (n-1)d]$. Step 3: Substitute the values: $S_{15} = \frac{15}{2} [2(8) + (15-1)8] = \frac{15}{2} [16 + 14 \times 8] = \frac{15}{2} [16 + 112] = \frac{15}{2} \times 128 = 15 \times 64 = 960$. Final answer: The sum of the first 15 multiples of 8 is 960.
• Q: The 4th term of an AP is 14 and the 12th term is 70. Find the first term and the common difference. A: Step 1: Set up two algebraic equations using the formula $a_n = a + (n-1)d$. For the 4th term ($a_4 = 14$): $a + 3d = 14$ (Equation 1). For the 12th term ($a_{12} = 70$): $a + 11d = 70$ (Equation 2). Step 2: Subtract Equation 1 from Equation 2 to eliminate $a$: $(a + 11d) - (a + 3d) = 70 - 14 \implies 8d = 56 \implies d = 7$. Step 3: Substitute the value of $d$ back into Equation 1: $a + 3(7) = 14 \implies a + 21 = 14 \implies a = -7$. Final answer: The first term $a$ is -7 and the common difference $d$ is 7.
• Q: Find the sum of all two-digit odd positive numbers. A: Step 1: Identify the first and last two-digit odd positive integers. The first is 11 and the last is 99. The sequence is 11, 13, 15, ..., 99, which forms an AP with $a = 11$, $d = 2$, and the last term $l = 99$. Step 2: Find the number of terms ($n$) using $l = a + (n-1)d$: $99 = 11 + (n-1)2 \implies 88 = (n-1)2 \implies 44 = n-1 \implies n = 45$. Step 3: Apply the simplified sum formula $S_n = \frac{n}{2}[a + l]$: $S_{45} = \frac{45}{2}[11 + 99] = \frac{45}{2} \times 110 = 45 \times 55 = 2475$. Final answer: The sum of all two-digit odd positive numbers is 2475.

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