JEE Mains Maths · Sequences and Series

AP and GP Conditions, Means and AM–GM

Turning 'in AP', 'in GP' and 'in HP' into equations, the arithmetic, geometric and harmonic means of two numbers, and the AM–GM inequality for maxima and minima.

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Why this matters

Twenty-five PYQs. Two thirds state that numbers are in AP, GP or HP, often one set in AP and a changed set in GP, and ask for the numbers. The rest use the means of two numbers or find a least or greatest value by AM–GM. Three ideas cover the page.

Concept 1 of 3: Turning 'in AP', 'in GP' and 'in HP' into equations

Three numbers

a,b,c

are in AP when

2b=a+c

, in GP when

b^2=ac

, and in HP when their reciprocals are in AP,

b=\frac{2ac}{a+c}

. Logarithms in AP mean the numbers are in GP. When one set is in AP and a shifted set is in GP, write the AP with one unknown

d

and put the shifted terms into

b^2=ac

: one quadratic in

d

Definition

AP: $2b=a+c$ . GP: $b^2=ac$ . HP: $\frac2b=\frac1a+\frac1c$ .
$\log a,\log b,\log c$ in AP $\iff a,b,c$ in GP.
AP then shifted GP: write $a,a+d,a+2d$ , shift, and apply $b^2=ac$ .
Reject a root that makes a GP term zero.

Three conditions

2b=a+c,\qquad b^2=ac,\qquad b=\frac{2ac}{a+c}

Worked example

1,a,b

are in AP and

1,a-2,b-3

are in GP. Find

a

and

b

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q77Moderate

Example 1 · Sequences and Series · AP and GP Conditions, Means and AM-GM

Let

3,a,b,c

be in A.P. and

3,a- 1,b+ 1,c+ 9

be in G.P. Then, the arithmetic mean of

a,b

and

c

is:

A zero term is not a GP

A root that makes any GP term zero, or the ratio undefined, must be rejected even though it satisfies

b^2=ac

Drill 16 more on turning 'in ap', 'in gp' and 'in hp' into equations

Concept 2 of 3: AM, GM and HM of two numbers, and means inserted between them

For positive

a,b

A=\frac{a+b}2

G=\sqrt{ab}

H=\frac{2ab}{a+b}

. Always

A\ge G\ge H

, and

G^2=AH

. Inserting

n

arithmetic means makes an AP of

n+2

terms, so the means add to

n

times the single AM. Inserting

n

geometric means makes a GP, so their product is

G^n

Definition

$A=\frac{a+b}2$ , $G=\sqrt{ab}$ , $H=\frac{2ab}{a+b}$ ; $A\ge G\ge H$ , $G^2=AH$ .
$a,b$ are the roots of $x^2-2Ax+G^2=0$ .
$n$ AMs add to $n\cdot\frac{a+b}2$ . $n$ GMs multiply to $(\sqrt{ab})^n$ ; their ratio is $\left(\frac ba\right)^{1/(n+1)}$ .

The three means

A\ge G\ge H,\qquad G^2=A\,H

Worked example

Find the three means of 4 and 16.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 15 Apr 2023 · Q70Moderate

Example 2 · Sequences and Series · AP and GP Conditions, Means and AM-GM

Let

A_{1}

and

A_{2}

be two arithmetic means and

G_{1},G_{2}

G_{3}

be three geometric means of two distinct positive numbers. The

G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2}G_{3}^{2}

is equal to

The larger root is the AM

When the AM and GM are given as the roots of a quadratic, the larger root is the AM, because

A\ge G

Drill 1 more on am, gm and hm of two numbers, and means inserted between them

Concept 3 of 3: Least and greatest values by AM–GM

For positive numbers, the average is at least the geometric mean, with equality only when all are equal. To make a sum such as

px+qy

least while

x^ay^b

is fixed, split

px

into

a

equal pieces and

qy

into

b

equal pieces, so the product of the pieces is a fixed number. The same split finds the greatest product under a fixed sum. Comparisons of powers, like

11^n

against

10^n+9^n

, go by dividing through and checking small

n

Definition

$\frac{x_1+\dots+x_n}{n}\ge\sqrt[n]{x_1x_2\cdots x_n}$ , all $x_i>0$ .
Equality when all $x_i$ are equal: this gives the point where the extreme occurs.
Split to match the powers: for $x^ay^b$ fixed, use $a$ copies of $\frac{px}{a}$ and $b$ copies of $\frac{qy}{b}$ .

AM–GM

\frac{x_1+x_2+\dots+x_n}{n}\ge\left(x_1x_2\cdots x_n\right)^{1/n}

Worked example

x,y>0

and

xy^2=32

. Find the least value of

x+y

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q154Moderate

Example 3 · Sequences and Series · AP and GP Conditions, Means and AM-GM

Let

x,y> 0

. If

x^{3}y^{2}=2^{15}

, then the least value of

3x + 2y

Check the equality case

AM–GM gives a bound, and the bound is the answer only if the equal-pieces point is allowed: positive, and inside any stated range.

Drill 5 more on least and greatest values by am–gm

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 (3)

Turning 'in AP', 'in GP' and 'in HP' into equations
Three conditions
$2b=a+c,\qquad b^2=ac,\qquad b=\frac{2ac}{a+c}$
AM, GM and HM of two numbers, and means inserted between them
The three means
$A\ge G\ge H,\qquad G^2=A\,H$
Least and greatest values by AM–GM
AM–GM
$\frac{x_1+x_2+\dots+x_n}{n}\ge\left(x_1x_2\cdots x_n\right)^{1/n}$

Watch out for (3)

A zero term is not a GP
→ Turning 'in AP', 'in GP' and 'in HP' into equations
The larger root is the AM
→ AM, GM and HM of two numbers, and means inserted between them
Check the equality case
→ Least and greatest values by AM–GM

Test yourself on Sequences and Series

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