Arithmetic Progressions (Class 10) — Study Notes

Overview

An Arithmetic Progression (AP) is a sequence of numbers where each term after the first is obtained by adding a fixed constant (called the common difference) to the previous term. This topic is critical for SOF IMO Class 10, appearing regularly in both Mathematical Reasoning and Achievers sections.

AP questions test three core skills: finding the nth term when the first term and common difference are known, computing the sum of n terms, and translating word problems into AP formulas. Mastery requires recognizing AP patterns quickly, applying formulas correctly without algebraic errors, and solving multi-step word problems that embed APs in real-world contexts like savings, seating arrangements, or construction projects.

Strong performance on this topic directly impacts your Mathematical Reasoning score and lays the foundation for series and sequences in higher mathematics. Focus on memorizing the two main formulas, practicing substitution carefully, and building speed in word-problem translation.

Key Concepts

• **Arithmetic Progression definition**: A sequence where the difference between consecutive terms is constant. Example: 3, 7, 11, 15, … has common difference d = 4.
• **Common difference (d)**: Found by subtracting any term from the next term: d = a₂ − a₁. The sign of d determines if the AP is increasing (d > 0), decreasing (d < 0), or constant (d = 0).
• **First term (a)**: The starting value of the sequence, often denoted as a or a₁. Every AP formula requires knowing the first term.
• **nth term formula**: Gives the value of any term in the sequence directly without listing all previous terms. Saves time in competitive exams.
• **Sum of n terms formula**: Computes the total of the first n terms efficiently. Two versions exist—use the one that fits the given information.
• **Word problems**: Typically describe patterns (salary increments, seating rows, stacking objects) that form an AP. Identify a, d, and n from the context, then apply formulas.
• **Finite AP**: Has a specific number of terms. Problems often ask for the last term or the sum of all terms in such sequences.

Formulas / Key Facts

**nth term formula**: aₙ = a + (n − 1)d where a = first term, d = common difference, n = term number. Use this to find any specific term or to find n when aₙ is given.

**Sum of n terms (version 1)**: Sₙ = (n/2)[2a + (n − 1)d] Use when you know a, d, and n but not the last term directly.

**Sum of n terms (version 2)**: Sₙ = (n/2)[a + l] where l = last term. Use when you know the first term, last term, and number of terms.

**Common difference from terms**: d = aₙ − aₙ₋₁ for any consecutive terms.

**Three terms in AP**: If a − d, a, a + d are three consecutive terms in AP, their sum is 3a and middle term is the average of the outer two.

**Last term**: l = a + (n − 1)d, just another name for aₙ when n is the total count.

**Checking if a sequence is AP**: Compute differences between consecutive terms. If all differences equal d, it's an AP.

Worked Examples

**Example 1: Find the 20th term** AP: 5, 9, 13, 17, … Given: a = 5, d = 9 − 5 = 4, n = 20 Apply: a₂₀ = a + (n − 1)d = 5 + (20 − 1) × 4 = 5 + 19 × 4 = 5 + 76 = 81 Answer: The 20th term is 81.

**Example 2: Sum of first 15 terms** AP: 7, 10, 13, … Given: a = 7, d = 3, n = 15 Apply: S₁₅ = (n/2)[2a + (n − 1)d] = (15/2)[2 × 7 + 14 × 3] = (15/2)[14 + 42] = (15/2) × 56 = 15 × 28 = 420 Answer: Sum is 420.

**Example 3: Word problem—stacking bricks** A construction worker stacks bricks in rows. The bottom row has 25 bricks, and each successive row has 2 fewer bricks. How many bricks are in the 10th row? What is the total number of bricks in the first 10 rows? Solution: First row a = 25, common difference d = −2 (decreasing by 2), n = 10. 10th term: a₁₀ = 25 + (10 − 1)(−2) = 25 − 18 = 7 bricks. Sum of 10 rows: S₁₀ = (10/2)[2 × 25 + 9 × (−2)] = 5[50 − 18] = 5 × 32 = 160 bricks. Answer: 10th row has 7 bricks; total is 160 bricks.

Common Mistakes

**Mistake 1: Confusing n and aₙ** Wrong thinking: Using the value of a term as the term number. Correct fix: n is the position (1st, 2nd, 3rd…), aₙ is the value at that position. Always identify which is given and which is required.

**Mistake 2: Sign errors with negative d** Wrong thinking: Writing d = −3 as +3 in calculations or forgetting the minus sign in (n − 1)d. Correct fix: Keep d with its sign throughout. If d = −3, write a + (n − 1)(−3) = a − 3(n − 1) carefully.

**Mistake 3: Using wrong sum formula** Wrong thinking: Applying Sₙ = (n/2)[a + l] when l is not given, or vice versa. Correct fix: Choose the formula that matches the known quantities. If you know a, d, n, use Sₙ = (n/2)[2a + (n − 1)d]. If you know a, l, n, use Sₙ = (n/2)[a + l].

**Mistake 4: Off-by-one errors in word problems** Wrong thinking: Counting rows or terms incorrectly, like treating "from 5th to 10th term" as 5 terms instead of 6. Correct fix: Carefully count the number of terms. From term p to term q inclusive, there are (q − p + 1) terms.

**Mistake 5: Forgetting to simplify before substitution** Wrong thinking: Substituting large numbers into (n − 1)d without simplifying (n/2) first. Correct fix: Simplify algebraically before plugging in values. Cancel common factors to reduce arithmetic errors.

Quick Reference

• **nth term**: aₙ = a + (n − 1)d — find any term directly.
• **Sum of n terms**: Sₙ = (n/2)[2a + (n − 1)d] or Sₙ = (n/2)[a + l].
• **Common difference**: d = second term − first term. Constant for all pairs.
• **Three-term shortcut**: If x, y, z in AP then y = (x + z)/2.
• **Check AP**: Verify a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = constant.
• **Word problems**: Extract a (starting value), d (increment/decrement), n (number of terms) from the problem statement, then apply formulas methodically.