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.

All Sequence & Series notes NDA Maths strategy

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

Try it yourself

For the HP

6, 3, 2

, verify that

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

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Three-term test for GP?
2.
$\tfrac{a-b}{b-c}$ for an AP equals?
3.
Are $2, 4, 6$ AP, GP, or HP?
4.
$\tfrac{a-b}{b-c}$ for an HP equals?

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

Try it yourself

p^2, q^2, r^2

are in GP (with

p, q, r > 0

), show

\ln p, \ln q, \ln r

are in AP.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Taking logs turns a GP into a(n)?
2.
If $\log a, \log b, \log c$ are in AP, then $a, b, c$ are in?
3.
Are $\log 3, \log 9, \log 27$ in AP?
4.
If $x, y, z$ in GP, what is the common difference of $\ln x, \ln y, \ln z$ ?

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

Try it yourself

a, b, c

are in HP, find

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

in a simpler form.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\tfrac1u, \tfrac1v, \tfrac1w$ in HP means $u, v, w$ in?
2.
To solve an HP problem, first take?
3.
If $\tfrac1x, \tfrac1y, \tfrac1z$ are in AP, then $x, y, z$ are in?
4.
Denominators $b+c, c+a, a+b$ in AP give which relation?

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

Try it yourself

p, 1, q

are in AP and

p, 2, q

are in GP, show that

p, 4, q

are in HP.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Three numbers in both AP and GP must be?
2.
AM, GM, HM of two numbers are themselves in which progression?
3.
If $a, b, c$ in AP, what equals $2b$ ?
4.
If $b, c, d$ in GP, write the condition.

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

Try it yourself

The roots of

x^2 - 18x + k = 0

are in the ratio

4 : 5

. Find

k

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Sum of roots of $x^2 - 7x + 12 = 0$ ?
2.
Product of roots of $2x^2 + 3x + 5 = 0$ ?
3.
$\tfrac1\alpha + \tfrac1\beta$ equals?
4.
If roots of $x^2 - bx + c$ are equal, then?

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

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Sequence & SeriesHARD

p^x = q^y = r^z

, where

x

y

and

z

are in GP, then consider the following statements: I.

p

q

and

r

are in AP. II.

\ln p

\ln q

and

\ln r

are in GP. Which of the statements given above is/are correct?

[Q1 · Sep · 2025]

Example 2Sequence & SeriesMODERATE

p, 1, q

are in AP and

p, 2, q

are in GP, then which of the following statements is/are correct? I.

p, 4, q

are in HP. II.

\frac{1}{p}, \frac{1}{4}, \frac{1}{q}

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

[Q6 · Apr · 2025]

Example 3Sequence & SeriesHARD

The sum of the roots of the equation

x^2 + bx + c = 0

(where

b

and

c

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

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

are in

[Q16 · Apr · 2017]

Example 4Sequence & SeriesMODERATE

For the following two (02) items: Let

\alpha

and

\beta

be the roots of the quadratic equation

x^2 + (\log_{0.5}(a^2))x + (\log_{0.5}(a^2))^4 = 0

where

a^2 \neq 1

and

\log_{0.5}(a^2) > 0

. Further,

\beta^2 = \alpha(\log_{a^2}(0.5))

(10 + \log_{10} x)

(10 + \log_{10} y)

and

(10 + \log_{10} z)

are in AP, then consider the following statements: I. The GM of

x

and

z

y^2

. II. The AM of

\log_{10} x

and

\log_{10} z

\log_{10} y

. Which of the statements given above is/are correct?

[Q5 · Sep · 2025]

Example 5Sequence & SeriesMODERATE

If x, 3/2, z are in AP; x, 3, z are in GP; then which is in HP?

[Q34 · Apr · 2018]

Drill every past-year question on this subtopic

15 questions from the bank — paginated, with cart and Word-export support.

Drill the 15 questions

Drill every past-year question on this subtopic

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