Functions, Logarithms and Progressions

These three topics account for roughly three to five of the twenty-two QA questions, and CAT likes to combine them: a sequence whose terms are logarithms, or a function whose values form a progression. Each is small enough to learn completely — there is no long tail of special cases here, unlike arithmetic — which makes this the highest-return-per-hour block in the QA section. Learn the eight formulas in the progressions table and the five logarithm laws properly and you can attempt every question of these types that a paper sets, rather than the half you happen to recognise.

Arithmetic Progressions

An AP adds a fixed common difference d each step. With first term a, the nth term is a + (n − 1)d and the sum of n terms is (n/2)[2a + (n − 1)d], which is worth remembering in its more useful second form: n × (average of the first and last term).

The average form is the fast form. The sum of the integers from 17 to 63 is (number of terms) × (average) = 47 × 40 = 1880. Count the terms as 63 − 17 + 1 = 47, and the average of an AP is always the average of its two ends: (17 + 63)/2 = 40. No formula substitution required.
Choose symmetric variables. If three numbers in AP add to 24, write them a − d, a, a + d: the sum is 3a = 24, so a = 8 immediately. For four terms use a − 3d, a − d, a + d, a + 3d — the sum is 4a. This single habit removes most of the algebra from AP word problems.
Two terms give you everything. If the 7th term is 31 and the 15th is 63, then 8d = 63 − 31 = 32, so d = 4, and a = 31 − 6(4) = 7. Any two terms fix the whole progression, so a question giving three terms is giving you a check.
When the sum is given as a formula. If Sₙ = 3n² + 5n, then the nth term is Sₙ − Sₙ₋₁ = [3n² + 5n] − [3(n−1)² + 5(n−1)] = 6n + 2. Check n = 1: S₁ = 8 and 6(1) + 2 = 8 ✓. This subtraction trick works for any sequence, not just an AP.

Geometric and Harmonic Progressions

The formulas worth memorising
Quantity	Formula
nth term of a GP	arⁿ⁻¹
Sum of n terms of a GP (r ≠ 1)	a(rⁿ − 1)/(r − 1)
Sum to infinity of a GP	a/(1 − r), only when \|r\| < 1
Geometric mean of a and b	√(ab)
Harmonic mean of a and b	2ab/(a + b)
Sum of the first n natural numbers	n(n + 1)/2
Sum of the first n squares	n(n + 1)(2n + 1)/6
Sum of the first n cubes	[n(n + 1)/2]²

Worked example (infinite GP). The sum of 6 + 2 + 2/3 + … is a/(1 − r) with a = 6 and r = 1/3, giving 6/(2/3) = 9. The condition matters: the same formula applied to 2 + 4 + 8 + … would return −2, which is nonsense — the series diverges because r = 2 is not less than 1 in magnitude.
AM ≥ GM ≥ HM, for positive numbers, with all three equal only when the numbers are equal. And GM² = AM × HM for two numbers, which turns many "find the third mean" questions into one multiplication. Check with 4 and 9: AM = 6.5, GM = 6, HM = 2(36)/13 = 72/13 ≈ 5.54 — the order holds, and 6.5 × 72/13 = 36 = 6² ✓
An HP is an AP upside down. If a, b, c are in HP then 1/a, 1/b, 1/c are in AP. Never manipulate an HP directly — invert the terms, do the AP work, invert back.

Logarithms — Five Laws and One Habit

log_b(x) = y means b^y = x. That one line is the definition, and reading a logarithm back as a power resolves most confusion. The base must be positive and not 1, and x must be positive — so log of zero or of a negative number does not exist, which is the source of several CAT traps.

The laws
Law	Use it when
log(mn) = log m + log n	A product appears inside a log
log(m/n) = log m − log n	A quotient appears inside a log
log(mᵏ) = k log m	A power appears inside — the most used law of the five
log_b(x) = log(x)/log(b)	Two different bases must be compared (change of base)
log_b(b) = 1 and log_b(1) = 0	Always — these two collapse whole expressions

Worked example 1. log₂8 + log₃9 = 3 + 2 = 5, because 2³ = 8 and 3² = 9. Read each as "what power gives this?" and no method is needed.
Worked example 2 (change of base). log₄8 = log8/log4 = 3log2/2log2 = 3/2. The log2 cancels, which is why change of base is worth applying even when the numbers look awkward.
Worked example 3 (an equation). Solve log₂(x) + log₂(x − 2) = 3. Combine: log₂[x(x − 2)] = 3, so x(x − 2) = 8, x² − 2x − 8 = 0, giving x = 4 or x = −2. But x = −2 makes both logs undefined, so the only answer is x = 4. Discarding the invalid root is not optional here — it is the whole question.
Counting digits. The number of digits in a positive integer N is ⌊log₁₀N⌋ + 1. Since log₁₀2 ≈ 0.301, the number 2¹⁰⁰ has ⌊30.1⌋ + 1 = 31 digits. CAT has asked this exact form more than once.

⚠️ The domain trap

Every logarithmic equation must be solved twice: once algebraically, and once by asking whether each answer keeps every log's argument positive. Questions in this topic are frequently written so that one root fails that test — the "extra" root is the distractor option.

Functions — Domain, Composition and Graphs

Domain is where the rule makes sense. Three restrictions cover nearly every CAT question: a denominator cannot be zero, the inside of an even root cannot be negative, and the argument of a log must be positive. For f(x) = √(x − 3)/(x − 5) the domain is x ≥ 3 except x = 5.
Composition runs right to left. f(g(x)) means apply g first. With f(x) = 2x + 1 and g(x) = x², f(g(3)) = f(9) = 19 while g(f(3)) = g(7) = 49. They are different, and a question that swaps the order is testing exactly that.
Odd and even. f(−x) = f(x) makes f even (symmetric about the y-axis); f(−x) = −f(x) makes it odd (symmetric about the origin). x², |x| and cos-type shapes are even; x, x³ and 1/x are odd. Most functions are neither, and knowing the symmetry sometimes answers a question with no computation at all — the sum f(a) + f(−a) is 0 for any odd function.
Graph transformations, in the order they confuse people. f(x) + k moves the graph up by k. f(x + k) moves it left by k — not right, which is the near-universal error. −f(x) flips it vertically; f(−x) flips it horizontally. |f(x)| reflects everything below the x-axis upward.
Worked example (a functional equation). If f(x) + 2f(1/x) = 3x for all x ≠ 0, find f(2). Put x = 2: f(2) + 2f(1/2) = 6. Put x = 1/2: f(1/2) + 2f(2) = 3/2. Multiply the second by 2 and subtract the first: 4f(2) − f(2) = 3 − 6, so 3f(2) = −3 and f(2) = −1. Substituting the reciprocal is the standard move for this whole question type.

🧠 Test with a number before trusting a rule

Unsure whether f(x + 2) shifts left or right? Take f(x) = x² and ask where the new graph bottoms out. f(x + 2) = (x + 2)², which is zero at x = −2 — to the left. Ten seconds of substitution beats a memorised rule you half-recall.

Key takeaways

The sum of any AP is (number of terms) × (average of the first and last term). Use that form, not the two-line formula.
Write three terms in AP as a − d, a, a + d. The sum becomes 3a and the word problem collapses to one equation.
An infinite GP sums to a/(1 − r) only when |r| < 1. If the formula returns something impossible, the series diverges and the answer is "no finite sum".
log_b(x) = y is just b^y = x. Reading a logarithm back as a power answers most questions without any law at all.
Solve every log equation twice: algebraically, then by checking that each root keeps every argument positive. The rejected root is usually an option.
f(x + k) shifts the graph left, not right. When in doubt, substitute a number into a simple function instead of trusting the memory.
Functional equations in f(x) and f(1/x) are solved by substituting x and then 1/x, and treating the two results as simultaneous equations.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The sum of all integers from 17 to 63, both inclusive, is:
1. 1840
2. 1880
3. 1920
4. 1800
Show answer
Answer: B — 1880
Number of terms = 63 − 17 + 1 = 47 (the "+1" is what candidates forget). Average of an AP is the average of its ends: (17 + 63)/2 = 40. Sum = 47 × 40 = 1880. Dropping the +1 gives 46 × 40 = 1840, which is option A.
In an AP, the 7th term is 31 and the 15th term is 63. The first term is:
1. 7
2. 4
3. 3
4. 11
Show answer
Answer: A — 7
The two terms are 8 steps apart, so 8d = 63 − 31 = 32 and d = 4. Then a = 31 − 6d = 31 − 24 = 7. Verify: the 7th term is 7 + 6(4) = 31 ✓ and the 15th is 7 + 14(4) = 63 ✓. Option B is the common difference, offered as a distractor.
The sum to infinity of the series 6 + 2 + 2/3 + 2/9 + … is:
1. 8
2. 9
3. 12
4. The series has no finite sum
Show answer
Answer: B — 9
Each term is one third of the one before, so r = 1/3 and a = 6. Since |r| < 1 the infinite sum exists and equals a/(1 − r) = 6/(1 − 1/3) = 6/(2/3) = 9. Sanity check by adding the first four terms: 6 + 2 + 0.667 + 0.222 = 8.889, already close to 9 and rising.
If Sₙ = 3n² + 5n is the sum of the first n terms of a sequence, the 10th term is:
1. 62
2. 65
3. 350
4. 56
Show answer
Answer: A — 62
The nth term is Sₙ − Sₙ₋₁. Here S₁₀ = 300 + 50 = 350 and S₉ = 243 + 45 = 288, so the 10th term is 350 − 288 = 62. Option C is S₁₀ itself — the sum of ten terms, offered to anyone who stops one step early. (The general term is 6n + 2, which at n = 10 gives 62 ✓)
The value of log₂8 + log₃9 + log₅125 is:
1. 6
2. 8
3. 9
4. 10
Show answer
Answer: B — 8
Read each as a power: 2³ = 8 so the first is 3; 3² = 9 so the second is 2; 5³ = 125 so the third is 3. The total is 3 + 2 + 3 = 8. No law of logarithms is needed — only the definition.
The number of real values of x satisfying log₂(x) + log₂(x − 2) = 3 is:
1. 2
2. 1
3. 0
4. Infinitely many
Show answer
Answer: B — 1
Combining gives log₂[x(x − 2)] = 3, so x(x − 2) = 8 and x² − 2x − 8 = 0, whose roots are 4 and −2. Now apply the domain test: the original equation needs x > 0 and x − 2 > 0, so x must exceed 2. x = 4 qualifies; x = −2 makes both logarithms undefined and is rejected. Exactly one solution. Answering 2 means the domain test was skipped.
If f(x) = 2x + 1 and g(x) = x², then f(g(3)) − g(f(3)) equals:
1. 0
2. −30
3. 30
4. 68
Show answer
Answer: B — −30
f(g(3)): apply g first, g(3) = 9, then f(9) = 19. g(f(3)): apply f first, f(3) = 7, then g(7) = 49. The difference is 19 − 49 = −30. Option A is what you get by assuming composition commutes, which it does not.
A function satisfies f(x) + 2f(1/x) = 3x for every non-zero x. The value of f(2) is:
1. 2
2. −1
3. 1
4. −2
Show answer
Answer: B — −1
Substituting x = 2 gives f(2) + 2f(1/2) = 6. Substituting x = 1/2 gives f(1/2) + 2f(2) = 3/2. Double the second equation: 2f(1/2) + 4f(2) = 3. Subtract the first from it: 4f(2) − f(2) = 3 − 6, so 3f(2) = −3 and f(2) = −1. The reciprocal substitution is the whole method for this question type.