Chapter 17 of 19

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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board (BSEAP) SSC Mathematics · ~8 marks · Board: AP Board

Progressions — Arithmetic and Geometric

ARITHMETIC PROGRESSION (common difference d) and GEOMETRIC PROGRESSION (common ratio r). nth term and sum formulas for both. AP Board Class 10 Mathematics.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

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

1Distinguish AP from GP; identify the common difference (d) and common ratio (r)
2Find the nth term and sum of n terms of both AP and GP
3Apply the distance formula between two points
4Apply the section formula (internal division) to find a point dividing a segment in a given ratio
5Find the area of a triangle using coordinate geometry; test three points for collinearity

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Why this chapter matters

The Progressions chapter in AP SSC covers BOTH Arithmetic Progressions (AP) and Geometric Progressions (GP), PLUS Coordinate Geometry (distance formula, section formula, area of triangle). Together these account for approximately 8–10 marks in the AP SSC paper. GP is harder than AP — the ratio (not difference) connects the terms. Coordinate geometry (distance, section formula, collinearity) is highly formula-driven and reliable for full marks once the formulas are memorised.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Progressions — Arithmetic and Geometric

Arithmetic Progression (AP)

A sequence where the DIFFERENCE between consecutive terms is CONSTANT (= d).

nth Term: aₙ = a + (n−1)d

Sum: Sₙ = n/2[2a + (n−1)d] = n/2(a + l)

Geometric Progression (GP)

A sequence where the RATIO between consecutive terms is CONSTANT (= r).

nth Term: aₙ = arⁿ⁻¹

Sum: Sₙ = a(rⁿ−1)/(r−1) [r>1] or Sₙ = a(1−rⁿ)/(1−r) [r<1]

Coordinate Geometry

Distance Formula: d = √[(x₂−x₁)² + (y₂−y₁)²]

Section Formula (Internal): ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))

Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2)

Area of Triangle: ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|. Area = 0 → COLLINEAR.

Detailed Formulas for Arithmetic Progression

General Form of an AP

An arithmetic progression is represented as: a, a+d, a+2d, a+3d, ...

nth Term — Deep Dive

The nth term formula aₙ = a + (n−1)d gives the value of ANY term without listing all previous terms.

Example 1: Find the 10th term of the AP: 2, 5, 8, 11, ... Here a = 2, d = 3, n = 10. a₁₀ = 2 + (10−1)×3 = 2 + 27 = 29.

Example 2: The 5th term of an AP is 17 and the 10th term is 32. Find the AP. Given: a₅ = a + 4d = 17 ... (i) a₁₀ = a + 9d = 32 ... (ii) Subtract (i) from (ii): 5d = 15 → d = 3 Substitute in (i): a + 12 = 17 → a = 5 So the AP is: 5, 8, 11, 14, ... ✓

Sum of n Terms — Derivation

Sₙ = a + (a+d) + (a+2d) + ... + [a+(n−2)d] + [a+(n−1)d] Write in REVERSE order: Sₙ = [a+(n−1)d] + [a+(n−2)d] + ... + (a+d) + a Add the two equations. Each pair sums to 2a + (n−1)d. There are n pairs. 2Sₙ = n[2a + (n−1)d] Sₙ = n/2[2a + (n−1)d] ✓

When the LAST TERM l = a + (n−1)d is known: Sₙ = n/2(a + l)

Example 3: Find the sum of the first 20 terms of the AP: 1, 3, 5, 7, ... a = 1, d = 2, n = 20 S₂₀ = 20/2[2×1 + (20−1)×2] = 10[2 + 38] = 10 × 40 = 400 ✓

Example 4: How many terms of the AP: 24, 21, 18, ... must be taken for the sum to be 78? a = 24, d = −3, Sₙ = 78 78 = n/2[48 + (n−1)(−3)] 78 = n/2[48 − 3n + 3] 78 = n/2[51 − 3n] 156 = n(51 − 3n) 156 = 51n − 3n² 3n² − 51n + 156 = 0 n² − 17n + 52 = 0 (n − 4)(n − 13) = 0 n = 4 or n = 13 Both are valid! For n = 4, sum = 78. For n = 13, sum is ALSO 78 because later terms are NEGATIVE and cancel out. 'Understand this — sometimes more terms don't mean MORE sum.'

Applications of AP

Simple Interest: Equal amounts added each year — forms an AP of interest amounts. 'If you deposit Rs. 1000 at 5% simple interest, you earn Rs. 50 EVERY year — it never changes.'
Stacking: Bricks in a triangular stack — each row has a fixed number fewer than the row below.
Depreciation: Value decreases by a CONSTANT amount each year.

Harmonic Progression (HP)

A sequence whose RECIPROCALS form an AP. General form: 1/a, 1/(a+d), 1/(a+2d), ...

Example: 1/2, 1/5, 1/8, 1/11, ... is an HP because 2, 5, 8, 11, ... is an AP.

There is NO direct formula for the sum of an HP. To solve HP problems, convert to AP, solve, then convert BACK.

Geometric Progression — Deep Dive

General Form of a GP

a, ar, ar², ar³, ...

Types of GP

Type	Condition	Example
Finite GP	Limited number of terms	1, 2, 4, 8
Infinite GP	Unlimited terms	1, 1/2, 1/4, 1/8, ...
Increasing GP	r > 1	2, 6, 18, 54, ...
Decreasing GP	0 < r < 1	81, 27, 9, 3, ...
Constant GP	r = 1	5, 5, 5, 5, ...
Alternating GP	r < 0	1, −3, 9, −27, ...

nth Term — Examples

Example 5: Find the 8th term of the GP: 3, 6, 12, 24, ... a = 3, r = 2, n = 8 a₈ = 3 × 2⁷ = 3 × 128 = 384 ✓

Example 6: The 3rd term of a GP is 12 and the 6th term is 96. Find the GP. a₃ = ar² = 12 ... (i) a₆ = ar⁵ = 96 ... (ii) Divide (ii) by (i): r³ = 96/12 = 8 → r = 2 Substitute in (i): a × 4 = 12 → a = 3 The GP is: 3, 6, 12, 24, 48, 96, ... ✓

Sum of n Terms — Derivation

Sₙ = a + ar + ar² + ... + arⁿ⁻¹ Multiply by r: rSₙ = ar + ar² + ar³ + ... + arⁿ Subtract: Sₙ − rSₙ = a − arⁿ Sₙ(1 − r) = a(1 − rⁿ) Sₙ = a(1 − rⁿ)/(1 − r) [for r ≠ 1]

For r > 1, it is convenient to write: Sₙ = a(rⁿ − 1)/(r − 1)

Example 7: Find the sum of the first 6 terms of the GP: 2, 6, 18, ... a = 2, r = 3, n = 6 S₆ = 2(3⁶ − 1)/(3 − 1) = 2(729 − 1)/2 = 728 ✓

Sum of Infinite GP

When |r| < 1, as n → ∞, rⁿ → 0 S∞ = a/(1 − r) [for |r| < 1]

Example 8: Find the sum of the infinite GP: 1, 1/2, 1/4, 1/8, ... a = 1, r = 1/2 S∞ = 1/(1 − 1/2) = 1/(1/2) = 2 ✓

Applications of GP

Compound Interest: The amount at the end of each year forms a GP. 'Compound interest is GP in action — your money grows by a RATIO, not a constant amount.'
Population growth: If population grows at 2% per year, it forms a GP.
Radioactive decay: The amount remaining halves every half-life — a GP with r = 1/2.
Installment payments: Many loan repayment schedules follow GP patterns.

Common Mistakes — AP and GP

Mistake	Correction
Confusing AP and GP formulas	'AP uses ADDITION (d). GP uses MULTIPLICATION (r). Remember: A for ADDITION.'
Using r = 1 formula without checking	Direct formula gives division by zero. Treat r = 1 as Sₙ = n×a
Forgetting n−1 in nth term	aₙ = a + (n−1)d, NOT a + nd
Wrong ratio in GP	r = a₂/a₁, always check with a₃/a₂ too
Using AP sum for GP sum	Never mix Sₙ formulas

AP SSC Board Exam Focus

Topic	Marks	Frequency
nth term of AP/GP	2-3	Very High
Sum of n terms of AP	3-4	Very High
Sum of n terms of GP	3-4	High
Application problems	4-5	Moderate
Finding a and d/r from conditions	4	High
Infinite GP sum	2	Moderate

Self-Test Questions

Find the 15th term of the AP: 7, 11, 15, 19, ...
The 7th term of an AP is 34 and the 13th term is 64. Find the AP.
Find the sum of the first 25 terms of the AP: 3, 7, 11, 15, ...
How many terms of the AP: 63, 60, 57, ... give a sum of 693?
Find the 10th term of the GP: 2, 6, 18, 54, ...
The 2nd term of a GP is 6 and the 5th term is 162. Find the GP.
Find the sum of the first 8 terms of the GP: 1, 3, 9, 27, ...
Find the sum of the infinite GP: 4, 2, 1, 1/2, ...
A man saves Rs. 1250 in the first year, Rs. 1350 in the second year, Rs. 1450 in the third year, and so on. How much does he save in 15 years?
A bacteria culture doubles every hour. If there are 100 bacteria now, how many will there be after 8 hours?

Answers: 1) 63, 2) AP: 4, 9, 14, ..., 3) 1275, 4) 14 terms, 5) 39366, 6) GP: 2, 6, 18, 54, ..., 7) 3280, 8) 8, 9) Rs. 28,500, 10) 25,600

Key formulas & results

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

AP and GP Formulas

AP: General term: aₙ = a+(n−1)d. Sum: Sₙ = n/2[2a+(n−1)d]. GP: General term: aₙ = arⁿ⁻¹. Sum: Sₙ = a(rⁿ−1)/(r−1) for r>1, or a(1−rⁿ)/(1−r) for r<1. COORDINATE GEOMETRY: Distance: d = √[(x₂−x₁)²+(y₂−y₁)²]. Section Formula (internal, ratio m:n): P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)). Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2). Area of triangle: ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|. Area = 0 → COLLINEAR.

GP DISTINCTION FROM AP: AP has CONSTANT DIFFERENCE (add d each time). GP has CONSTANT RATIO (multiply r each time). Examples: 2, 4, 8, 16... → GP (r=2). 2, 5, 8, 11... → AP (d=3). THREE TERMS IN GP: Take as a/r, a, ar. Product = a³. THREE TERMS IN AP: Take as a−d, a, a+d. Sum = 3a.

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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

✗ Using the AP sum formula for GP

✓ AP sum: Sₙ = n/2[2a+(n−1)d]. GP sum: Sₙ = a(rⁿ−1)/(r−1). These are COMPLETELY DIFFERENT. The key distinguishing check: Is the pattern adding a constant (AP) or multiplying by a constant (GP)? Dividing any term by the previous term: constant ratio → GP. Subtracting: constant difference → AP.

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 Progressions — Arithmetic and Geometric?

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.

•AP vs GP IDENTIFICATION: AP = constant DIFFERENCE (subtract consecutive terms). GP = constant RATIO (divide consecutive terms). Test: 2, 4, 6, 8 → differences: 2,2,2 → AP (d=2). Test: 2, 4, 8, 16 → ratios: 2,2,2 → GP (r=2). Test: 2, 4, 12 → differences: 2,8 (not constant) → ratios: 2,3 (not constant) → NEITHER.
•GP FORMULAS: nth term: aₙ = arⁿ⁻¹. Sum of n terms: Sₙ = a(rⁿ−1)/(r−1) for r≠1. For r=1: Sₙ = na. Sum of infinite GP (only when |r|<1): S∞ = a/(1−r). THREE TERMS IN GP: Take as a/r, a, ar. Product = a³. Note: GP uses MULTIPLICATION not addition.
•COORDINATE GEOMETRY — DISTANCE FORMULA: Distance between (x₁,y₁) and (x₂,y₂) = √[(x₂−x₁)²+(y₂−y₁)²]. A point equidistant from two fixed points lies on the perpendicular bisector of the segment joining them.
•COORDINATE GEOMETRY — SECTION FORMULA: Point dividing (x₁,y₁) and (x₂,y₂) in ratio m:n internally: P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)). MIDPOINT (m=n=1): M = ((x₁+x₂)/2, (y₁+y₂)/2). For external division: P = ((mx₂−nx₁)/(m−n), (my₂−ny₁)/(m−n)).
•COORDINATE GEOMETRY — AREA OF TRIANGLE: Area = ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|. COLLINEARITY TEST: Area = 0 means three points are collinear (on the same line). Always take the ABSOLUTE VALUE — area cannot be negative.
•GP RATIO CONDITION: If three numbers p, q, r are in GP, then q² = pr (the square of the middle term = product of extremes). This is the GP equivalent of 2q = p+r for AP.
•APPLICATIONS OF GP: Compound interest: A = P(1+r/100)ⁿ follows a GP — each year's amount is r obtained by multiplying by (1+r/100). Population growth with constant growth rate. Radioactive decay. Bank FD calculations. All exponential growth/decay processes follow GP.
•CENTROID OF TRIANGLE: The point dividing each median in ratio 2:1 from vertex. Coordinates: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). The centroid always lies inside the triangle. This is derived from the section formula (the centroid divides each median 2:1 from vertex).
•DISTANCE SPECIAL CASES: If two points have the same y-coordinate, distance = |x₂−x₁| (horizontal distance). If same x-coordinate, distance = |y₂−y₁| (vertical distance). These are special cases of the distance formula but faster to compute.
•SIGN CONVENTION: In coordinate geometry, always keep signs carefully. (−2, 3) means x = −2 (left of origin), y = 3 (above origin). When computing (x₂−x₁)², the result is always positive (square of any number ≥ 0). Double-check sign before squaring — error in sign before squaring is the most common mistake.

Andhra Pradesh (BIEAP) marks blueprint

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Where this shows up in the real world

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

Compound interest and exponential growth (GP)

Every bank fixed deposit, mutual fund NAV, and loan with compound interest follows a geometric progression. A principal of ₹10,000 at 10% per year compound interest: Year 1: ₹11,000, Year 2: ₹12,100, Year 3: ₹13,310 — each year multiplied by 1.1 (GP with r=1.1). The GP formula aₙ = arⁿ⁻¹ directly gives you the amount after n years. Understanding GP is essential for all personal finance decisions — comparing FD rates, understanding doubling time (rule of 72), and comparing simple vs compound interest.

GPS and smartphone location (distance formula)

Your phone's map app calculates your distance to any destination using the 3D generalisation of the distance formula: distance = √[(x₂−x₁)²+(y₂−y₁)²+(z₂−z₁)²]. Route optimisation algorithms repeatedly compute distances between points to find the shortest path (Dijkstra's algorithm). Even 'How far is this restaurant?' in Google Maps is ultimately a distance formula calculation applied to GPS coordinates. The Class 10 formula is the exact foundation — extended to 3D and spherical geometry for real Earth distances.

Seismic waves and earthquake epicentre location

Seismologists use the distance formula (in 3D) with data from multiple seismometers to locate the epicentre of an earthquake. Each seismometer gives a distance to the epicentre (from the wave arrival time). The intersection of multiple distance circles (in 3D, spheres) gives the exact location — this is called triangulation. The section formula is also used to interpolate where the epicentre lies between known stations. Class 10 coordinate geometry is the literal mathematical tool used for earthquake location.

Exam strategy

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

GP identification (2 marks): explicitly compute a₂/a₁ and a₃/a₂ — if equal, state 'common ratio r = ...' This explicit check earns the identification mark. Don't assume a sequence is GP without computing the ratio.
Section formula (4 marks): label which is point A (with x₁,y₁) and which is B (x₂,y₂), and which is m and n in the ratio m:n. Write the formula explicitly before substituting. The most common error is swapping m and n — labelling carefully prevents this.
Coordinate geometry area (4 marks): write the three vertices with their coordinates clearly. Apply the formula in the standard format: ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|. Calculate each bracket term separately before adding — avoids sign errors.
GP sum formula (4 marks): first identify a (first term) and r (common ratio). If r>1: use Sₙ = a(rⁿ−1)/(r−1). If r<1: use Sₙ = a(1−rⁿ)/(1−r). Both forms are algebraically equivalent but avoiding negative denominators reduces errors.
Collinearity check: use the area formula and check if result = 0. Write 'Since area = 0, the three points are collinear' — this explicit conclusion earns the final mark.

Going beyond the textbook

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

Research the Geometric Mean and its applications — the geometric mean of n positive numbers is the nth root of their product: GM = (a₁×a₂×...×aₙ)^(1/n). For a GP, the GM of all n terms equals the middle term (for odd n). The AM-GM inequality states that for positive numbers, the arithmetic mean is always ≥ geometric mean, with equality only when all numbers are equal. Research how AM-GM is used to prove optimization results (e.g., a rectangle with fixed perimeter has maximum area when it is a square).
Investigate Zeno's Paradoxes and the sum of infinite GP — Zeno argued that Achilles can never catch a tortoise because before catching up, he must reach where the tortoise was, but by then it has moved, and so on infinitely. The sum 1/2 + 1/4 + 1/8 + ... = 1 (infinite GP with a=1/2, r=1/2, S∞=a/(1−r)=1) shows that infinitely many steps can add up to a finite total. This resolved Zeno's paradox and was a key step toward the development of calculus and limits.
Explore Taxicab geometry (Manhattan distance) — on a grid (like city streets), the distance between two points is not the straight-line Euclidean distance but |x₂−x₁| + |y₂−y₁| (taxicab or Manhattan metric). A 'circle' in taxicab geometry is a square tilted 45°. Research how different distance metrics change the geometry of circles, triangles, and the triangle inequality — and how taxicab geometry is used in GPS routing and network analysis.
Research the 4-point and n-point Geometry problems — given n points in a plane, how many distinct distances are there? The Erdős distinct distances problem (solved in 2010 by Guth and Katz) asks: what is the minimum number of distinct distances formed by n points? This seemingly simple question used advanced algebraic geometry (polynomial method) in its proof. The Class 10 distance formula is the starting point for this deep research area.

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 — coordinate geometry (distance, section, area) and progressions (AP/GP) together account for 6–10 marks in AP SSC Mathematics

JEE Main and AdvancedVery High — coordinate geometry (straight lines, circles, conics) and sequences/series are among the most heavily tested topics in JEE; Class 10 is the foundation

NTSE (Mathematics)High — coordinate geometry and number sequences appear in NTSE Stage I and II

AP EAPCETVery High — coordinate geometry (straight lines, circles, conics) is a major portion of the EAPCET Mathematics syllabus

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Questions students ask

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ARITHMETIC PROGRESSION (AP): Terms increase (or decrease) by ADDING a constant d. Each term = previous term + d. Example: 3, 7, 11, 15, ... (add 4 each time). GEOMETRIC PROGRESSION (GP): Terms increase (or decrease) by MULTIPLYING by a constant r. Each term = previous term × r. Example: 3, 6, 12, 24, ... (multiply by 2 each time). QUICK TEST: Subtract any two consecutive terms — if the difference is constant → AP. Divide any term by the previous one — if the ratio is constant → GP. Warning: the same sequence of numbers like 1, 4, 16 could look like either — check: differences are 3, 12 (not constant, NOT AP), ratios are 4, 4 (constant → GP with r=4).

Use the Section Formula. Let A = (x₁, y₁) and B = (x₂, y₂). Point P dividing AB in ratio m:n = 2:1 internally: x = (m×x₂ + n×x₁)/(m+n) = (2x₂ + 1×x₁)/3. y = (2y₂ + 1×y₁)/3. MNEMONIC: 'm' goes with the SECOND point (B), 'n' with the FIRST point (A). Example: A(1,2), B(7,5), ratio 2:1. x = (2×7 + 1×1)/3 = 15/3 = 5. y = (2×5 + 1×2)/3 = 12/3 = 4. P = (5,4). Verify: AP distance = √[(5−1)²+(4−2)²] = √20 = 2√5. PB = √[(7−5)²+(5−4)²] = √5. Ratio AP:PB = 2:1 ✓.

Three points are COLLINEAR if they all lie on the SAME STRAIGHT LINE — meaning no triangle can be formed between them (area = 0). TEST 1: Area formula: if ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)| = 0, the points are collinear. TEST 2: Check if the slope from point 1 to point 2 equals the slope from point 2 to point 3: (y₂−y₁)/(x₂−x₁) = (y₃−y₂)/(x₃−x₂). TEST 3: Use the distance approach: if AB + BC = AC (total distance = direct distance), then B lies on segment AC. The area test is fastest for AP SSC.

Yes — a constant sequence (all terms equal) is BOTH AP and GP. Example: 5, 5, 5, 5, ... is AP with d=0 AND GP with r=1. These are degenerate cases. Any non-constant sequence cannot be both — because AP requires constant difference (d≠0) while GP requires constant ratio (r≠1), and a non-constant AP cannot have a constant ratio. For AP SSC, all practical problems have sequences that are clearly either AP or GP (not both).

The distance formula computes (x₂−x₁)² + (y₂−y₁)², which is ALWAYS non-negative (sum of squares). You CANNOT get a negative value inside the square root if the formula is applied correctly. Possible errors: (1) You computed (x₂−x₁)² as x₂²−x₁² (wrong — this is a factoring identity, not squaring a difference: (a−b)² = a²−2ab+b²). (2) You used a subtraction symbol instead of addition between the two squared terms. (3) Sign error in x₂−x₁ BEFORE squaring — but since squaring removes the sign, this doesn't affect the result. Check: recompute (x₂−x₁) and (y₂−y₁) separately, square each, then add.

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