Chapter 6 of 19

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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board (BSEAP) SSC Mathematics · ~6 marks · Board: AP Board

Arithmetic Progression (AP)

A sequence where the DIFFERENCE between consecutive terms is CONSTANT. Master the nth term formula and sum formulas for AP SSC Mathematics.

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By the end of this chapter you'll be able to…

1Identify an AP from a sequence by checking constant common difference
2Find the nth term using aₙ = a + (n−1)d
3Find the sum of n terms using Sₙ = n/2[2a+(n−1)d] or Sₙ = n/2(a+l)
4Set up and solve simultaneous equations to find a and d from given conditions
5Apply the 3-term trick: take three terms in AP as a−d, a, a+d

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Why this chapter matters

Arithmetic Progressions (AP) is a guaranteed question in AP SSC Mathematics — typically 4–6 marks. The nth term and sum formulas are simple to memorise and mechanically apply. Most AP exam questions test: find the nth term, find the sum, or set up equations to find a or d from given conditions. The '3-term trick' (taking terms as a−d, a, a+d) saves significant time on problems where sum is given. This is one of the easiest high-scoring chapters in Class 10 mathematics.

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Arithmetic Progression

"An AP grows by ADDING the same number each time. The mathematics is elegant — and highly tested."

Definition

A sequence where a₂−a₁ = a₃−a₂ = ... = d (COMMON DIFFERENCE). d can be positive, negative, or zero.

nth Term: aₙ = a + (n−1)d

a = first term. d = common difference.

Sum of First n Terms — Two Forms

Sₙ = n/2[2a + (n−1)d] — when you know a and d.
Sₙ = n/2(a + l) — when you know first AND last terms (l = aₙ).
'The sum is the AVERAGE of first and last, multiplied by n.'

Key Trick: Sₙ − Sₙ₋₁ = aₙ

Finding the nth term from sums.

Common Mistakes

Confusing AP (add d) with GP (multiply by r).
In Sₙ formula, the (n−1) factor — students often write n instead.

Worked Examples — nth Term

Example 1 — Find the 10th Term

AP: 2, 7, 12, 17, ... Find a₁₀. a = 2, d = 7−2 = 5. a₁₀ = 2 + (10−1)×5 = 2 + 45 = 47.

Example 2 — Find the Term Number

Which term of the AP: 21, 18, 15, ... is −81? a = 21, d = −3. aₙ = 21 + (n−1)(−3) = 24 − 3n. Set 24 − 3n = −81 → 3n = 105 → n = 35. The 35th term is −81.

Example 3 — Three Numbers in AP

Three numbers in AP have sum 15 and product 80. Find them. Let numbers be a−d, a, a+d. Sum: 3a = 15 → a = 5. Product: (5−d)×5×(5+d) = 80 → 5(25−d²) = 80 → 25−d² = 16 → d² = 9 → d = ±3. Numbers: 2, 5, 8 or 8, 5, 2.

Example 4 — Two APs with Same Common Difference

Two APs have the same common difference. The difference between their 100th terms is 100. Find the difference between their 1000th terms. Let first terms be a₁ and a₂. a₁₀₀ − b₁₀₀ = [a₁+99d] − [a₂+99d] = a₁−a₂ = 100. a₁₀₀₀ − b₁₀₀₀ = [a₁+999d] − [a₂+999d] = a₁−a₂ = 100. 'The difference between corresponding terms of two APs with the SAME common difference is CONSTANT.'

Sum of n Terms — Derivation and Examples

Derivation of Sₙ = n/2[2a + (n−1)d]

Sₙ = a + (a+d) + (a+2d) + ... + [a+(n−2)d] + [a+(n−1)d] Reverse: Sₙ = [a+(n−1)d] + [a+(n−2)d] + ... + (a+d) + a Add: 2Sₙ = [2a+(n−1)d] + [2a+(n−1)d] + ... + [2a+(n−1)d] (n terms) 2Sₙ = n[2a+(n−1)d] → Sₙ = n/2[2a+(n−1)d]

'This is the Gauss pairing method — each pair sums to the SAME value.'

Example 5 — Sum of First n Terms

Find S₂₀ for AP: 5, 8, 11, 14, ... a = 5, d = 3. S₂₀ = 20/2[2×5 + (20−1)×3] = 10[10 + 57] = 670.

Example 6 — Finding n Given Sum

How many terms of AP: 24, 21, 18, ... give sum 78? a = 24, d = −3. Sₙ = n/2[48 + (n−1)(−3)] = n/2[51 − 3n] = 78. n[51−3n] = 156 → 51n − 3n² = 156 → 3n² − 51n + 156 = 0 → n² − 17n + 52 = 0. (n−4)(n−13) = 0 → n = 4 or n = 13. Both valid: S₄ = 78 and S₁₃ = 78 (after n=4, terms become negative, sum first decreases then returns to 78).

Example 7 — Sum of First n Natural Numbers

Find S₁₀₀ for natural numbers 1, 2, 3, ..., 100. a = 1, d = 1, n = 100. S₁₀₀ = 100/2[2×1 + (100−1)×1] = 50[2 + 99] = 5050. 'Shortcut: Sum of first n natural numbers = n(n+1)/2.'

Example 8 — Real-life Savings

A man saves ₹5 in week 1, ₹7 in week 2, ₹9 in week 3, and so on. Find total savings after 52 weeks. a = 5, d = 2, n = 52. S₅₂ = 52/2[10 + 51×2] = 26[10 + 102] = 26×112 = ₹2912.

Example 9 — Real-life Penalty

A contractor pays ₹200 penalty for day 1, ₹250 for day 2, ₹300 for day 3, and so on. Find total penalty for 30 days. a = 200, d = 50, n = 30. S₃₀ = 30/2[400 + 29×50] = 15[400 + 1450] = 15×1850 = ₹27,750.

Key Derivations

nth term from sum: aₙ = Sₙ − Sₙ₋₁. This is useful when sums are given but terms are not directly known.

Sum of n terms of an AP with first term a and last term l: Sₙ = n/2(a+l). 'This is the AVERAGE of first and last terms multiplied by the number of terms.'

AP Exam Focus

Topic	Marks	Frequency
nth term identification	2-3	Very Common
Sum of n terms	4-5	Very Common
Word problems on AP	4-5	Common
Finding n or d given conditions	3-4	Moderate
Real-life applications	4	Common

Self-Test Questions

Find the 15th term of AP: −3, 1, 5, 9, ... (Answer: a = −3, d = 4, a₁₅ = −3+14×4 = 53)
Which term of AP: 3, 8, 13, 18, ... is 78? (Answer: 3+(n−1)5 = 78 → n = 16)
Find S₂₅ for AP: 7, 11, 15, 19, ... (Answer: S₂₅ = 25/2[14+24×4] = 25/2×110 = 1375)
How many terms of AP: 9, 17, 25, ... sum to 450? (Answer: n/2[18+8(n−1)]=450 → 4n²+5n−450=0 → n = 10)
Sum of three numbers in AP = 21, sum of squares = 165. Find numbers. (Answer: a=7, d=±2 → 5,7,9)
A theatre has 20 rows. Row 1 has 30 seats, each subsequent row has 2 more seats. Find total seats. (Answer: S₂₀ = 20/2[60+19×2] = 980 seats)
Sum of all two-digit natural numbers divisible by 3. (Answer: 12,15,...,99. n=30. S₃₀ = 15[24+29×3] = 1665)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Arithmetic Progression

GENERAL TERM: aₙ = a + (n−1)d (a = first term, d = common difference). SUM OF n TERMS: Sₙ = n/2[2a+(n−1)d] = n/2(a+l) where l = last term. FINDING d: d = a₂−a₁ (any two consecutive terms). COMMON DIFFERENCE: negative for decreasing AP, positive for increasing. THREE TERMS IN AP: Take as (a−d), a, (a+d). Sum = 3a. FOUR TERMS: (a−3d), (a−d), (a+d), (a+3d). Sum = 4a. CHECKING: Three numbers p, q, r are in AP iff 2q = p+r.

AP SSC TRICK: If question says 'find 3 numbers in AP with sum S and product P': Let terms = a−d, a, a+d. Sum = 3a = S → a = S/3. Then use product condition to find d. This always works cleanly. AP Board frequently uses this setup.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing the sum formula — using n instead of n/2

✓ The sum formula is Sₙ = n/2 × [2a + (n−1)d] — the n/2 is essential. A common error is writing Sₙ = n[2a + (n−1)d] (forgetting to divide by 2). Alternatively, use Sₙ = n/2(first term + last term). Quick check: sum of first 10 natural numbers = 10/2 × (1+10) = 5 × 11 = 55 ✓.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Arithmetic Progression (AP)?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•DEFINITION OF AP: A sequence where each term after the first is obtained by adding a CONSTANT (d = common difference). General form: a, a+d, a+2d, a+3d, ... The difference between any two consecutive terms is always d. Positive d → increasing AP. Negative d → decreasing AP. d=0 → constant sequence (still an AP).
•CHECKING IF AP: Subtract consecutive terms: if (a₂−a₁) = (a₃−a₂) = (a₄−a₃) = constant, it is an AP. Equivalently, three numbers p, q, r are in AP if and only if 2q = p+r (the middle term is the arithmetic mean of its neighbours).
•GENERAL TERM FORMULA: aₙ = a + (n−1)d. Here a = first term, d = common difference, n = position (1st, 2nd, 3rd...). To find WHICH position a value occurs: set aₙ = given value, solve for n. If n comes out as a positive integer, it is in the AP; otherwise it is not.
•SUM FORMULA: Sₙ = n/2 × [2a + (n−1)d]. Alternative form: Sₙ = n/2 × (a + l) where l = last term. NOTE: The difference between consecutive sums gives the terms: Sₙ − Sₙ₋₁ = aₙ. This relationship is tested in AP SSC.
•THREE-TERM TRICK: If sum of three terms in AP is given, take them as (a−d), a, (a+d). Sum = 3a → find a immediately. Then use product or other condition to find d. FOUR-TERM TRICK: Take as (a−3d), (a−d), (a+d), (a+3d). Sum = 4a.
•FINDING a AND d FROM TWO CONDITIONS: Typically given two terms: 'p-th term = X and q-th term = Y.' Write two equations from aₙ = a+(n−1)d: a+(p−1)d = X and a+(q−1)d = Y. Subtract to find d, then back-substitute to find a. This is the standard 4-mark question setup.
•SUM OF NATURAL NUMBERS: Sum of first n natural numbers = n(n+1)/2 (special case: a=1, d=1). Sum of first n odd numbers = n². Sum of first n even numbers = n(n+1). These are derived from the AP sum formula and appear in AP SSC questions.
•WORD PROBLEMS: 'A factory produces 100 items on day 1, increasing by 5 items per day. How many items in first 30 days?' → AP with a=100, d=5, n=30. S₃₀ = 30/2 × [200 + 29×5] = 15 × [200+145] = 15 × 345 = 5175 items. Identify a, d, n from the word problem context.
•ARITHMETIC MEAN: The arithmetic mean (AM) of two numbers p and q is (p+q)/2. A number inserted between p and q to form an AP must be (p+q)/2. The AM is always between the two numbers. If n arithmetic means are inserted between a and b, the common difference = (b−a)/(n+1).
•AP NOTATION WARNING: The problem chapter is called 'Arithmetic Progressions' but the subject board is also 'AP Board.' Do not confuse 'AP' meaning the mathematical sequence with 'AP' meaning Andhra Pradesh. In math context, AP = Arithmetic Progression.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

EMI (Equated Monthly Instalments) and loan repayment

Some loan repayment schemes (simple interest loans) follow an AP where each payment is the same amount — constant common difference = 0. More interestingly, when a borrower makes extra payments that reduce principal by a fixed amount each month, the interest component decreases by a fixed amount each month — forming a decreasing AP. Understanding AP is foundational to financial mathematics and the concept of amortisation (how loan balances decrease over time with regular payments).

Stadium seating and architecture

Stadium seating is often designed in an AP — the first row has a certain number of seats, and each subsequent row has a few more. Architects use the AP sum formula to calculate total seating capacity. Example: a cricket stadium with 20 rows, first row 50 seats, each row adding 3 more → total seats = S₂₀ = 20/2 × [100 + 19×3] = 10 × 157 = 1570 seats. Similarly, theatre seating, amphitheatres, and tiered parking garages all involve AP-based counting.

Computer science — array access and algorithms

In programming, iterating through an array with a fixed step (e.g., every 3rd element: index 0, 3, 6, 9, ...) produces an AP. The time complexity of such loops is O(n/step) — which is directly the number of terms in the AP. Linear search takes 1, 2, 3, ..., n comparisons in worst case — the average is n/2, derived from the AP sum formula. Understanding AP is foundational to analysing algorithm efficiency.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Identify a, d, n FIRST: before writing any formula, list these three values at the top. Every AP problem is solved once you have two of three values (a, d, n) and a target (term value or sum). This structured start prevents setting up the wrong formula.
Two-condition problems (4 marks): write TWO equations from the general term formula (one for each given term). Label them (1) and (2). Subtract to find d. Substitute back to find a. Write the AP. Show all steps — each step earns a mark.
3-term trick: if any question says 'three numbers in AP with sum S', immediately write them as (a−d), a, (a+d) and note sum = 3a → a = S/3. This saves setup time.
Sum formula: always write Sₙ = n/2[2a+(n−1)d], then substitute. The n/2 factor is most commonly forgotten. Check your answer: the last term should approximately equal 2Sₙ/n − a₁ (average term ≈ middle term).
Word problem checklist: read carefully to identify what is a, what is d, what is n, and what is being asked (sum or nth term). Write these before starting the calculation. 'First day', 'each subsequent day', 'total in N days' map directly to a, d, Sₙ.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research the connection between AP and arithmetic mean — in a sequence of n numbers in AP, the arithmetic mean of ALL n numbers equals the MIDDLE term (for odd n) or the average of the two middle terms (for even n). This is why the mean and median coincide for AP distributions. Explore how this relates to the symmetry of normal distributions in statistics.
Investigate Gauss's childhood sum: young Carl Friedrich Gauss (age ~10, 1787) reportedly solved the sum of 1+2+3+...+100 by noticing that pairing first and last terms gives 101 each, with 50 pairs → sum = 5050. This is exactly the AP sum formula derived geometrically. Research how Gauss went on to prove the prime number theorem and revolutionised mathematics — starting from this childhood insight.
Explore the connection between AP and averages in cricket: a batsman's run total over n innings follows an AP if they score a constant difference more each innings. The batting average (total runs / innings) equals the MEAN of the AP, which equals (a_n + a_1)/2 for an AP. Research how statisticians use sequences and averages to analyse sports performance — Duckworth-Lewis method in cricket involves similar mathematical structures.
Research arithmetic progressions in music theory — equal temperament in Western music divides an octave into 12 equal semitones. The frequencies do NOT follow an AP (they follow a geometric progression — each semitone multiplies frequency by 2^(1/12)). But the NAMES of notes (C, C#, D, D#, ...) are equally spaced — an AP of semitone steps. This explains why tuning by ear (using simple frequency ratios) versus tuning by equal temperament gives slightly different results — a deep connection between AP, GP, and music.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10)High — AP/GP problems are standard 4–6 mark questions in every AP SSC Mathematics paper

JEE Main and Advanced (Sequences and Series)Very High — AP, GP, and their sums are among the most frequently tested topics in JEE Mathematics; Class 10 AP is the foundation

NTSE (Mathematics)High — Number sequences including AP appear in NTSE Stage I mathematics

AP EAPCET (Mathematics)High — Sequences and series including AP, GP, and HP are a dedicated chapter in EAPCET syllabus

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Questions students ask

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Set Sₙ = given value and solve for n. This gives a quadratic equation in n (since Sₙ involves n²). For example: how many terms of AP 3, 5, 7, ... give sum 120? a=3, d=2, Sₙ = n/2[2(3)+(n−1)(2)] = n/2[6+2n−2] = n/2[4+2n] = n(2+n) = 2n+n². Set 2n+n² = 120 → n²+2n−120 = 0 → (n+12)(n−10) = 0 → n = 10 or n = −12. Since n must be a positive integer, n = 10. Always reject negative and fractional values of n.

When d is NEGATIVE, the AP is decreasing. After some point, the terms become negative. The sum increases until the last positive term, then decreases as negative terms are added. The sum is maximum when we include all non-negative terms. Example: AP 10, 7, 4, 1, −2, −5... (a=10, d=−3). Terms become negative at n=5 (a₅=−2). Sum is maximum at n=4 (S₄ = 22) or n=3 (S₃ = 21 — check both when the boundary term is 0 or close to 0). This type of 'find n for maximum sum' question appears in AP SSC.

Yes, absolutely. A DECREASING AP has a negative common difference. Example: 100, 97, 94, 91, ... has d = −3. The sequence is still an AP — the requirement is only that the common difference is CONSTANT. Temperature decreasing by 2°C each hour: 20, 18, 16, 14, ... (d=−2). A ball thrown upward loses 2 m/s speed each second: 20, 18, 16, 14, ..., 0 m/s — this is a decreasing AP (though the ball eventually stops, at which point the model no longer applies). AP SSC questions sometimes use decreasing APs in word problems about salary reductions, temperature drops, or decreasing production.

TERM: The actual VALUE at a position — e.g., in AP 3, 7, 11, 15, ..., the 4th term is 15 (the value). TERM NUMBER (n): The POSITION of the term — e.g., 15 is at position 4 (n=4). When a problem asks 'which term is 99?', it is asking for the TERM NUMBER n such that aₙ = 99. When it asks 'find the 10th term', it is asking for the TERM VALUE at n=10. This distinction is critical for setting up equations correctly. 'Find the term number' → solve aₙ = value for n. 'Find the term' → substitute n into aₙ = a+(n−1)d.

AP with a=5, d=2, n=12. Total trees = S₁₂ = 12/2 × [2(5) + (12−1)(2)] = 6 × [10 + 22] = 6 × 32 = 192 trees. This is a standard word problem format. Note: the 12th row alone has a₁₂ = 5 + 11(2) = 5 + 22 = 27 trees. The total (192) is always larger than the last row count × number of rows because early rows have fewer trees.

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