NDA Maths · Sequence & Series

Interrelating AP, GP and HP — the bridge tricks

NDA's favourite hard genre: take logs to turn a GP into an AP, take reciprocals to turn an HP into an AP, and translate "roots of an equation" conditions into progression conditions.

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Why this matters

Fifteen PYQs and the chapter's signature difficulty — six of these are HARD. The wins come from three reflexes: take logs to turn a GP into an AP, take reciprocals to turn an HP into an AP, and write down the three standard three-term conditions before doing anything else. Almost every "if … are in AP and … are in GP" chain falls to writing those conditions and eliminating. Master the five moves below and the hard band of this chapter opens up.

Concept 1 of 5

The three three-term conditions

Intuition

Every progression has a one-line test for three numbers. AP: the middle is the average. GP: the middle squared is the product. HP: flip to reciprocals and test for AP. A neat unifier is the ratio

\tfrac{a-b}{b-c}

: it equals 1 for an AP,

\tfrac{a}{b}

for a GP, and

\tfrac{a}{c}

for an HP — a quick way to tell them apart.

Definition

For three numbers $a, b, c$ :

AP $\iff 2b = a + c$ , and then $\dfrac{a-b}{b-c} = 1$ .
GP $\iff b^2 = ac$ , and then $\dfrac{a-b}{b-c} = \dfrac{a}{b}$ .
HP $\iff b = \dfrac{2ac}{a+c}$ , and then $\dfrac{a-b}{b-c} = \dfrac{a}{c}$ .

These three are the workhorses of the entire subtopic — write them down first, always.

The unifying ratio

\frac{a-b}{b-c} = \begin{cases} 1 & \text{AP} \\[2pt] a/b & \text{GP} \\[2pt] a/c & \text{HP} \end{cases}

Worked example

Are

3, 6, 12

in AP, GP, or HP? Compute

\dfrac{a-b}{b-c}

and confirm it matches.

Test GP: $b^2 = 6^2 = 36$ and $ac = 3 \times 12 = 36$ — equal, so they are in GP (ratio 2).
Compute the ratio: $\dfrac{3 - 6}{6 - 12} = \dfrac{-3}{-6} = \dfrac12$ .
For a GP this should equal $\dfrac{a}{b} = \dfrac{3}{6} = \dfrac12$ — it matches.

Answer:GP;

\dfrac{a-b}{b-c} = \dfrac12 = \dfrac{a}{b}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Sequence & SeriesHARD

If a, b, c are in AP or GP or HP, then

\frac{a-b}{b-c}

is equal to

[Q47 · Sep · 2018]

Concept 2 of 5

The log bridge: a GP becomes an AP

Intuition

Logarithms turn multiplication into addition. So taking the log of every term of a GP turns the constant ratio into a constant difference — the logs are in AP. This single move handles questions that mix "in GP" with "

\ln

… in AP", and questions with exponents like

p^x = q^y = r^z

Definition

If $x, y, z$ are in GP ( $y^2 = xz$ ), then $\log x, \log y, \log z$ are in AP (since $2\log y = \log x + \log z$ ) — for any base. Conversely, if logs are in AP the originals are in GP. For $p^x = q^y = r^z = k$ : take logs to get $x\log p = y\log q = z\log r$ , then $\log p, \log q, \log r$ relate through the progression of $x, y, z$ .

The bridge

x, y, z \text{ in GP} \iff \log x, \log y, \log z \text{ in AP}

Worked example

The numbers

2, 4, 8

are in GP. Show that

\log 2, \log 4, \log 8

are in AP.

Write each: $\log 2,\ \log 4 = 2\log 2,\ \log 8 = 3\log 2$ .
These are $\log 2 \cdot (1, 2, 3)$ — equally spaced, common difference $\log 2$ .
Check the AP condition: $2\log 4 = 4\log 2 = \log 2 + \log 8$ . ✓

Answer:

\log 2, \log 4, \log 8

are in AP with common difference

\log 2

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Sequence & SeriesMODERATE

p^2

q^2

and

r^2

(where

p,q,r>0

) are in GP, then which of the following is/are correct? 1.

p

q

and

r

are in GP. 2.

\ln p

\ln q

and

\ln r

are in AP. Select the correct answer using the code given below:

[Q42 · Apr · 2020]

Drill 5 more on the log bridge: a gp becomes an ap

Concept 3 of 5

The reciprocal bridge: an HP becomes an AP

Intuition

An HP is defined by its reciprocals being in AP — so any HP condition is best handled by flipping. The classic NDA shape wraps the reciprocals in expressions like

\tfrac{1}{b+c}, \tfrac{1}{c+a}, \tfrac{1}{a+b}

: "in HP" means the denominators

b+c, c+a, a+b

are in AP, which collapses to a simple relation among

a, b, c

Definition

$\tfrac{1}{u}, \tfrac{1}{v}, \tfrac{1}{w}$ are in HP $\iff u, v, w$ are in AP $\iff 2v = u + w$ . So whenever you see reciprocals "in HP", drop to the denominators and impose the AP condition on them.

Reciprocal flip

\frac{1}{u}, \frac{1}{v}, \frac{1}{w} \text{ in HP} \iff u, v, w \text{ in AP}

Worked example

\dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}

are in HP, show that

a, b, c

are in AP.

"Reciprocals in HP" means the denominators $b+c,\ c+a,\ a+b$ are in AP.
AP condition: $2(c+a) = (b+c) + (a+b)$ .
Expand: $2c + 2a = a + 2b + c \Rightarrow a + c = 2b$ .
That is exactly the AP condition for $a, b, c$ .

Answer:

a, b, c

are in AP.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Sequence & SeriesHARD

\frac{1}{b+c}

\frac{1}{c+a}

\frac{1}{a+b}

are in HP, then which of the following is/are correct? 1.

a

b

c

are in AP 2.

(b+c)^{2}

(c+a)^{2}

(a+b)^{2}

are in GP Select the correct answer using the code given below.

[Q10 · Sep · 2021]

Concept 4 of 5

Mixed and chained progression problems

Intuition

When a problem strings several conditions together — some terms in AP, others in GP, others in HP — the method never changes: write each condition algebraically (

2b=a+c

c^2=bd

\tfrac{2}{d}=\tfrac1c+\tfrac1e

), then eliminate to get the asked relation. A famous special case: three numbers that are simultaneously in AP and GP must be equal.

Definition

Translate every clause into its three-term condition, then substitute and eliminate. Useful facts that drop out: numbers in both AP and GP are all equal ( $a = b = c$ ); and chains like $a,b,c$ in AP, $b,c,d$ in GP, $c,d,e$ in HP force $a, c, e$ into GP. There is no shortcut beyond careful elimination — set it up cleanly.

Worked example

a, b, c

are in AP and also in GP, prove that

a = b = c

AP: $a + c = 2b$ . GP: $b^2 = ac$ .
Treat $a$ and $c$ as roots of $t^2 - (a+c)t + ac = 0$ , i.e. $t^2 - 2bt + b^2 = 0$ .
That factors as $(t - b)^2 = 0$ , so both roots equal $b$ : $a = c = b$ .

Answer:

a = b = c

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Sequence & SeriesHARD

a,b,c

are in AP;

b,c,d

are in GP;

c,d,e

are in HP, then which of the following is/are correct? (A)

a,c

and

e

are in GP. (B)

\dfrac{1}{a},\dfrac{1}{c},\dfrac{1}{e}

are in GP. Select the correct answer using the code given below:

[Q26 · Apr · 2024]

Drill 4 more on mixed and chained progression problems

Concept 5 of 5

Roots, coefficients, and progression conditions

Intuition

Some questions hide a progression inside a quadratic. The link is Vieta's relations: for

ax^2 + bx + c = 0

, the roots sum to

-b/a

and multiply to

c/a

. A condition on the roots (their ratio, or a symmetric expression in them) becomes an equation in the coefficients — which often turns out to say the coefficients are in AP, GP, or HP.

Definition

For $ax^2 + bx + c = 0$ with roots $\alpha, \beta$ :

\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.

Symmetric functions reduce to these:

\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta

\tfrac{1}{\alpha} + \tfrac{1}{\beta} = \tfrac{\alpha+\beta}{\alpha\beta}

, etc. Plug a given root-condition into these and simplify to read off the progression among the coefficients.

Vieta's relations (monic-friendly)

\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}

Worked example

For

x^2 + px + q = 0

with roots

\alpha, \beta

, express

\alpha^2 + \beta^2

in terms of

p

and

q

Vieta: $\alpha + \beta = -p$ and $\alpha\beta = q$ .
Use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ .
Substitute: $(-p)^2 - 2q = p^2 - 2q$ .

Answer:

\alpha^2 + \beta^2 = p^2 - 2q

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Sequence & SeriesHARD

The sum of the roots of the equation

ax^2 + x + c = 0

(where

a

and

c

are non-zero) is equal to the sum of the reciprocals of their squares. Then

a,\, ca^2,\, c^2

are in

[Q17 · Apr · 2017]

Drill 1 more on roots, coefficients, and progression conditions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The three three-term conditions
The unifying ratio
$\frac{a-b}{b-c} = \begin{cases} 1 & \text{AP} \\[2pt] a/b & \text{GP} \\[2pt] a/c & \text{HP} \end{cases}$
The log bridge: a GP becomes an AP
The bridge
$x, y, z \text{ in GP} \iff \log x, \log y, \log z \text{ in AP}$
The reciprocal bridge: an HP becomes an AP
Reciprocal flip
$\frac{1}{u}, \frac{1}{v}, \frac{1}{w} \text{ in HP} \iff u, v, w \text{ in AP}$
Roots, coefficients, and progression conditions
Vieta's relations (monic-friendly)
$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$

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