Learn subjects / Grade 10 / Mathematics GRADE 10 · Mathematics · LEARNING COMPANION
Arithmetic Progressions Three guided lessons with explanations, visual steps, practice and a final check.
Original learning companion, not full textbook exercise solutions. Covers constant differences, term membership and finite sums, with selected contextual applications. It is not an exhaustive exercise bank.
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01 · Get ready 02 · Learn 03 · Practise 04 · Final check 05 · My learning You can revisit any stage. Opening a page does not count as passing. Read all lesson explanations Read all lessons These explanations remain available without JavaScript. The separate learner and solution pages can be printed on A4.
LESSON 1
Recognise a constant difference Distinguish an arithmetic progression from other patterns.
A sequence is an ordered list; the order matters.
Compare neighbours For 5, 9, 13, 17, subtract 5 from 9 to get 4.
Repeat the difference check 13 − 9 = 4 and 17 − 13 = 4, so the difference is constant.
Continue in the same way Add 4 to 17 to get 21. Moving backward from 5 subtracts 4 to give 1.
Check a non-example 2, 4, 8, 16 has differences 2, 4, 8. It is not an AP even though it follows a pattern.
Worked example: For 18, 13, 8, 3, the differences are all −5, so a = 18 and d = −5. The next term is −2.
Watch for this: An AP may decrease or remain constant; d can be negative or zero. A recognisable pattern alone does not make an AP.
Discuss: Create an increasing AP, a decreasing AP and a non-AP. Explain each using consecutive differences.
LESSON 2
Find an AP term without listing them all Use aₙ = a + (n − 1)d and check the index.
Going from term 1 to term 2 uses one jump; term 1 to term 5 uses four jumps.
Identify the start and jump A row pattern begins 6, 10, 14, …, so a = 6 and d = 4.
Count the jumps The eighth term is seven jumps from the first, not eight.
Calculate the term a₈ = 6 + 7 × 4 = 34.
Reverse-check a target To locate 50, solve 6 + (n − 1)4 = 50. This gives n − 1 = 11, so n = 12.
Worked example: In 9, 14, 19, …, term 15 is 9 + 14 × 5 = 79. To test 81, solve n − 1 = (81 − 9)/5 = 14.4; since n is not an integer, 81 is not a term.
Watch for this: Using a + nd shifts the answer by one term. A target belongs to an AP only if the resulting n is a positive integer.
Discuss: A seating row starts with 12 chairs and each next row adds 3. Find row 20, then decide whether a row can have 61 chairs.
LESSON 3
Add many AP terms efficiently Derive and apply the sum formula by pairing.
Reordering a finite addition does not change its sum.
Write a short AP both ways For 4 + 7 + 10 + 13 + 16, reverse the order underneath.
Add vertically Each vertical pair totals 20: 4 + 16, 7 + 13 and 10 + 10. There are five pairs across the two rows.
Remove the double count The two rows contain twice the original sum, so 2S = 5 × 20 and S = 50.
Generalise For first term a and last term l, Sₙ = n(a + l)/2 = n[2a + (n − 1)d]/2. This also works for odd n.
Worked example: Save 5, 8, 11, … units for 10 days. Last term = 5 + 9 × 3 = 32. Total = 10(5 + 32)/2 = 185 units.
Watch for this: The last term and the number of terms are different quantities. Do not forget to halve the total after pairing two rows.
Discuss: Design a 7-day saving plan that is an AP. Calculate the total by listing and by the formula, and compare.
Continue on paper Sources and coverage Original learning companion, not full textbook exercise solutions. Covers constant differences, term membership and finite sums, with selected contextual applications. It is not an exhaustive exercise bank.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
