Progressions (AP & GP)

Progressions (AP & GP)

🟢 Lite

Key Formula/Rule

AP: nth term = a + (n–1)d, sum = n/2 × [2a + (n–1)d]. GP: nth term = ar^(n–1), sum = a(r^n – 1)/(r – 1) for r > 1.

Quick Memory Trick

**“A” for “Always Add” in AP (constant difference “d” added each step). **“G” for “Grow by Multiply” in GP (constant ratio “r” multiplied each step).

1-Sentence Summary

AP numbers walk up stairs (keep adding the same step size), GP numbers grow like rabbits (each term is the last one multiplied by the same factor).

Quick Example

Q: Find the 10th term of the AP: 3, 7, 11, 15, … A: a=3, d=4, n=10 → T₁₀ = 3 + (10–1)×4 = 3 + 36 = 39

Must Remember

• AP: constant difference (d); GP: constant ratio (r)
• Sum of AP = n/2 × (first + last) = n/2 × [2a + (n–1)d]
• GP sum = a(r^n – 1)/(r – 1) when r ≠ 1; = n×a when r = 1
• Arithmetic mean of two numbers = (a + b)/2; Geometric mean = √(ab)
• AP + AP = AP; GP × GP = GP (useful for combining)

🟡 Standard

Concept Explanation

A sequence is just a list of numbers that follow some rule. The two most important types you’ll encounter are Arithmetic Progressions (AP) and Geometric Progressions (GP). They show up everywhere — from calculating loan repayments to modeling population growth — which is why they’re exam favorites.

An Arithmetic Progression is a sequence where you add the same number every time to get the next term. That constant number is called the “common difference” (denoted d). So if you start at 5 and add 3 each time, you get 5, 8, 11, 14, 17… The beauty of AP is that any term can be found instantly if you know the first term (a), the step size (d), and which term you want (n). You just plug into: Tₙ = a + (n–1)d.

A Geometric Progression works differently. Instead of adding, you multiply by the same factor every time. That factor is the “common ratio” (denoted r). So if you start at 3 and multiply by 2 each time, you get 3, 6, 12, 24, 48… The ratio between consecutive terms never changes — that’s the defining property. GP gets interesting because the numbers grow (or shrink) much faster than AP. A GP with r > 1 grows exponentially, while one with 0 < r < 1 shrinks toward zero.

Key Formulas

Symbol	Meaning
a	First term of the sequence
d	Common difference (AP)
r	Common ratio (GP)
n	Number of terms
l	Last term
Sₙ	Sum of first n terms

Step-by-Step Example

Q: The 3rd term of an AP is 16 and the 10th term is 58. Find the sum of the first 20 terms.

Step 1: Express terms in terms of a and d. T₃ = a + 2d = 16 T₁₀ = a + 9d = 58

Step 2: Subtract to find d. (a + 9d) – (a + 2d) = 58 – 16 7d = 42 → d = 6

Step 3: Find a. a + 2(6) = 16 → a = 4

Step 4: Find sum of first 20 terms. S₂₀ = n/2 × [2a + (n–1)d] = 20/2 × [2(4) + 19(6)] = 10 × [8 + 114] = 10 × 122 = 1220

Answer: 1220

Common Mistakes

• Confusing the (n–1) multiplier → In Tₙ = a + (n–1)d, when n = 1, the term is just a (makes sense). If you accidentally use n instead of (n–1), you’d get a + d as the first term — wrong.
• Forgetting the GP sum formula changes when r < 1 → When r < 1, the formula a(r^n – 1)/(r – 1) still works if you flip the sign correctly. Write it as a(1 – r^n)/(1 – r) for r < 1 to avoid sign confusion.
• Mixing up AP and GP formulas → AP involves addition, GP involves multiplication. Check which one you’re solving.

Quick Test (2 Qs)

1. Q: The 2nd term of a GP is 12 and the 5th term is 96. Find the common ratio. Options: A) 2 B) 3 C) 4 D) 8. Ans: A (Reason: T₂ = ar = 12, T₅ = ar⁴ = 96. Divide: ar⁴/ar = r³ = 96/12 = 8 → r = 2)

2. Q: Three numbers are in AP with sum 21 and product 280. Find them. Options: A) 7, 8, 9 B) 5, 7, 9 C) 4, 7, 10 D) 6, 7, 8. Ans: B (Reason: Let numbers be (a–d), a, (a+d). Sum = 3a = 21 → a = 7. Product = (7–d)(7)(7+d) = 280 → 49 – d² = 40 → d² = 9 → d = 3. Numbers: 4, 7, 9)

🔴 Extended

Concept Deep Dive

Let’s dig into why these formulas exist and what they really tell us about the sequences they describe.

The Nature of AP:

An arithmetic progression is fundamentally a linear model. Each step adds the same amount, so the terms grow (or shrink) at a constant rate. Graphically, if you plot AP terms against their position numbers, you get a straight line. The slope of that line is exactly the common difference d.

The sum formula for AP, Sₙ = n/2 × [2a + (n–1)d], has a beautiful intuitive proof: write the sum forward and backward, then pair terms. The nth term plus the 1st term always equals (a + l), where l is the last term. There are n such pairs, so Sₙ = n(a + l)/2 = n/2 × [2a + (n–1)d]. This pairing trick works because of the symmetry built into every AP.

A practical analogy: think of AP like walking up a staircase with equal steps. The “height” after n steps is your nth term. The total vertical distance climbed after n steps is your sum. And yes, the staircase metaphor even works for negative steps (walking down) or fractional steps.

The Nature of GP:

A geometric progression is an exponential model. Each step multiplies by the same factor, so the terms grow (or shrink) at a rate proportional to their current value. Plot this on a graph and you get a curve that bends upward (for r > 1) or gradually descends toward zero (for 0 < r < 1). No straight line can capture this — that’s what makes GP fundamentally different from AP.

The GP sum formula, Sₙ = a(r^n – 1)/(r – 1), comes from a clever trick: multiply the sum by r, then subtract the original sum. Nearly everything cancels out, leaving a simple expression. Try it yourself with a 3-term GP and you’ll see exactly why it works.

The most important thing to understand about GP is the ratio r. When r > 1, numbers explode toward infinity (think compound interest, population growth, viral spread). When r = 1, the GP is constant — every term equals a. When 0 < r < 1, numbers shrink toward zero asymptotically (think radioactive decay, drug absorption). When r < 0, you get alternating signs — the terms flip between positive and negative.

Advanced Formula Derivation

Infinite GP Sum:

When |r| < 1 and n → ∞, the term r^n becomes vanishingly small. So S∞ = a/(1 – r). This formula describes anything that decays toward a limit — the total distance a bouncing ball travels as it gradually stops, or the total time a leaking bucket takes to empty.

Example: 1, 1/2, 1/4, 1/8, … Here a = 1, r = 1/2. S∞ = 1/(1 – 1/2) = 2. No matter how many terms you add, you’ll never exceed 2.

GATE-Level Numerical Problems

Q1 (GATE 2020 — style): The sum of three numbers in GP is 26 and their product is 216. Find the numbers.

• Working: Let numbers be a/r, a, ar. Product = (a/r) × a × ar = a³ = 216 → a = 6. Sum = 6/r + 6 + 6r = 26 → 6/r + 6r = 20 → Multiply by r: 6 + 6r² = 20r → 6r² – 20r + 6 = 0 → Divide by 2: 3r² – 10r + 3 = 0 → (3r – 1)(r – 3) = 0 → r = 3 or r = 1/3.
• Answer: If r = 3: numbers are 2, 6, 18. If r = 1/3: numbers are 18, 6, 2. So the set is {2, 6, 18}.
• Common error: Forgetting the symmetric case — if a/r, a, ar works, then the reciprocals also work.

Q2 (GATE 2019 — style): The 12th term of an AP is 8 more than the 6th term. If the 4th term is 14, find the first term and common difference.

• Working: T₁₂ = a + 11d, T₆ = a + 5d. Given T₁₂ – T₆ = 8 → (a + 11d) – (a + 5d) = 8 → 6d = 8 → d = 4/3. T₄ = a + 3d = 14 → a + 3(4/3) = 14 → a + 4 = 14 → a = 10.
• Answer: a = 10, d = 4/3
• Common error: Trying to find the actual terms when only the difference was asked. Here, the question only needed a and d.

Q3: A person saves ₹1000 in the first month, ₹1200 in the second month, ₹1400 in the third month, and so on (increasing by ₹200 each month). How much will they save in 24 months total?

• Working: This is an AP with a = 1000, d = 200, n = 24. S₂₄ = 24/2 × [2(1000) + 23(200)] = 12 × [2000 + 4600] = 12 × 6600 = ₹79,200.
• Answer: ₹79,200

Multiple Approaches

Method A (Standard formula): Use Sₙ = n/2 × [2a + (n–1)d] directly. Safe, reliable, always works.

Method B (Average × count): The average of the first and last term = (a + l)/2. Multiply by n gives the same answer. Useful when you know the last term but not d explicitly.

When to use: Method A when you have a and d. Method B when you can quickly find the last term (l = a + (n-1)d) and want a faster calculation.

Tricky Cases

• AP of consecutive integers: Numbers like 5, 6, 7, 8 are an AP with a = 5, d = 1. Their sum is n(a + l)/2 — same formula but simpler because d = 1.
• GP with negative ratio: If r = –2, terms alternate signs: a, –2a, 4a, –8a… The absolute values grow, but the sign keeps flipping. Sum formula still works.
• Harmonic Progression (HP): When the reciprocals of terms form an AP. No special “HP formulas” needed — just convert to AP by taking reciprocals, solve, then convert back.

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