Chapter 8 · Sequences and Series

8.3 Arithmetic and Geometric Means

Two numbers have two natural averages: the arithmetic mean $A=\dfrac{a+b}{2}$, which sits midway in an AP, and the geometric mean $G=\sqrt{ab}$, which sits midway in a GP. For positive numbers $A\ge G$ always — a fact with surprising uses.

The price of a phone model rose by $10\%$ in one year and by $40\%$ the next. What was the average yearly rise? The tempting answer is $25\%$, the ordinary average of $10$ and $40$. But two rises of $25\%$ multiply the price by $1.25\times1.25=1.5625$, while the real rises multiplied it by $1.1\times1.4=1.54$. The right "average factor" is the number $g$ with $g\times g=1.1\times1.4$, that is $g=\sqrt{1.54}\approx1.241$ — an average rise of about $24.1\%$. When quantities are added, the ordinary average is the natural one; when they are multiplied, the natural average is a square root.

These two averages are the arithmetic mean and the geometric mean. In this section we define them, use them to insert any number of terms between two given numbers so that the whole list is an AP or a GP, and prove the inequality $A\ge G$ — the arithmetic mean of two positive numbers is never less than their geometric mean. That one inequality answers a whole family of "largest area" and "smallest cost" questions without any calculus.

The Arithmetic Mean and Inserting Arithmetic Means

📘 Definition 8.5 — Arithmetic mean

The arithmetic mean (A.M.) of two numbers $a$ and $b$ is $$A=\frac{a+b}{2}.$$ It is the number that makes $a,\ A,\ b$ an AP, since then $A-a=b-A$.

More generally, to insert $n$ arithmetic means between $a$ and $b$ means to find $n$ numbers $A_1,A_2,\dots,A_n$ such that $$a,\ A_1,\ A_2,\ \dots,\ A_n,\ b$$ is an AP. The list has $n+2$ terms, with $a$ first and $b$ last.

📐 Theorem 8.6 — Inserting $n$ arithmetic means

The $n$ arithmetic means between $a$ and $b$ are $$A_k=a+k\,d\quad(k=1,2,\dots,n),\qquad d=\frac{b-a}{n+1},$$ and their sum is $n\cdot\dfrac{a+b}{2}$, that is $n$ times the single A.M. of $a$ and $b$.

Proof. In the AP $a,A_1,\dots,A_n,b$ the number $b$ is the $(n+2)$th term, so $b=a+(n+1)d$, giving $d=\dfrac{b-a}{n+1}$, and the $(k+1)$th term is $A_k=a+kd$. The means themselves form an AP of $n$ terms with first term $A_1$ and last term $A_n$, so their sum is $\tfrac n2(A_1+A_n)$. Since $A_1=a+d$ and $A_n=b-d$, $A_1+A_n=a+b$. $\blacksquare$

✏️ Example 1 — Inserting arithmetic means
Insert $5$ arithmetic means between $8$ and $26$, and check that their sum is $5$ times the A.M. of $8$ and $26$.
Difference $n=5$, so the AP has $7$ terms and $d=\dfrac{26-8}{6}=3$.
Means $8,\ \mathbf{11},\ \mathbf{14},\ \mathbf{17},\ \mathbf{20},\ \mathbf{23},\ 26$. The means are $11,14,17,20,23$.
Check $11+14+17+20+23=85$ and $5\times\dfrac{8+26}{2}=5\times17=85$ ✓. Note that the middle mean, $17$, is itself the A.M. of $8$ and $26$.
✏️ Example 2 — How many means?
Some arithmetic means are inserted between $7$ and $71$, and the $5$th of them is $27$. How many means were inserted?
Set up With $n$ means, $d=\dfrac{71-7}{n+1}=\dfrac{64}{n+1}$, and $A_5=7+5d=27$.
Solve $5d=20$, $d=4$, so $\dfrac{64}{n+1}=4$ and $n+1=16$.
Answer $n=15$ means: $11,15,19,\dots,67$.
Check $A_{15}=7+15\times4=67$, and the next term $67+4=71$ ✓.

The Geometric Mean and Inserting Geometric Means

📘 Definition 8.6 — Geometric mean

The geometric mean (G.M.) of two positive numbers $a$ and $b$ is $$G=\sqrt{ab}.$$ It is the positive number that makes $a,\ G,\ b$ a GP, since then $\dfrac Ga=\dfrac bG$, i.e. $G^2=ab$.

We keep $a$ and $b$ positive throughout: for numbers of opposite sign $ab\lt0$ and there is no real G.M. at all, and for two negative numbers the convention would need a negative root. To insert $n$ geometric means between $a$ and $b$ means to find $G_1,\dots,G_n$ such that $a,\ G_1,\ \dots,\ G_n,\ b$ is a GP; we take the positive common ratio, so that all the means are positive.

📐 Theorem 8.7 — Inserting $n$ geometric means

For positive $a$ and $b$, the $n$ geometric means between $a$ and $b$ are $$G_k=a\,r^k\quad(k=1,2,\dots,n),\qquad r=\left(\frac ba\right)^{\frac{1}{n+1}},$$ and their product is $\left(\sqrt{ab}\right)^n=G^n$, the $n$th power of the single G.M. of $a$ and $b$.

Proof. The number $b$ is the $(n+2)$th term of the GP, so $b=ar^{n+1}$, $r^{n+1}=\dfrac ba$, and $r$ is the positive $(n+1)$th root of $\tfrac ba$. For the product, pair the means from the two ends: $G_k\,G_{n+1-k}=ar^k\cdot ar^{n+1-k}=a^2r^{n+1}=ab$. Multiplying all $n$ means and pairing them in this way, $(G_1G_2\cdots G_n)^2=(ab)^n$, so $G_1G_2\cdots G_n=(ab)^{n/2}=\left(\sqrt{ab}\right)^n$, the positive root since all the means are positive. $\blacksquare$

✏️ Example 3 — Inserting geometric means
(a) Insert $3$ geometric means between $1$ and $256$, and check their product.
(b) Insert $4$ geometric means between $3$ and $96$.
(a) $r^4=\dfrac{256}{1}=256$, so $r=4$: $1,\ \mathbf{4},\ \mathbf{16},\ \mathbf{64},\ 256$.
(a) Check $4\times16\times64=4096$ and $G=\sqrt{256}=16$, $G^3=4096$ ✓.
(b) $r^5=\dfrac{96}{3}=32$, so $r=2$: $3,\ \mathbf{6},\ \mathbf{12},\ \mathbf{24},\ \mathbf{48},\ 96$.
Compare In (a) the A.M. of $1$ and $256$ is $128.5$, while the G.M. is $16$ — far smaller. The geometric means crowd towards the smaller number; the arithmetic means are evenly spread.
Interactive 1 — Inserting Means: Even Steps versus Even Ratios

Choose $a$, $b$ and the number of means $n$. The top line shows the $n$ arithmetic means — equal steps $d$. The middle line shows the $n$ geometric means on the same ordinary scale — they bunch towards the smaller end. The bottom line uses a ratio scale, where each tick is a fixed multiple of the one before: there the geometric means are evenly spaced. Press Play to add means one by one.

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step $d$—
ratio $r$—
sum of A.M.s $=nA$—
product of G.M.s $=G^n$—

The Inequality A.M. $\ge$ G.M.

Try a few pairs: for $4$ and $9$, $A=6.5$ and $G=6$; for $1$ and $256$, $A=128.5$ and $G=16$; for $5$ and $5$, $A=G=5$. The arithmetic mean always seems to win, and the two tie only when the numbers are equal. This is true for every pair of positive numbers.

📐 Theorem 8.8 — The A.M.–G.M. inequality

For any two positive numbers $a$ and $b$, $$\frac{a+b}{2}\ \ge\ \sqrt{ab},$$ that is $A\ge G$, with equality if and only if $a=b$.

Proof. Since $a,b\gt0$ they have real square roots, with $a=(\sqrt a)^2$ and $b=(\sqrt b)^2$. Then $$A-G=\frac{a+b}{2}-\sqrt{ab}=\frac{(\sqrt a)^2-2\sqrt a\sqrt b+(\sqrt b)^2}{2}=\frac{\left(\sqrt a-\sqrt b\right)^2}{2}.$$ A square of a real number is never negative, so $A-G\ge0$. It is zero exactly when $\sqrt a=\sqrt b$, i.e. when $a=b$. $\blacksquare$

The inequality has a beautiful picture (Interactive 2). Draw a semicircle on a diameter of length $a+b$ and mark the point $C$ that splits the diameter into lengths $a$ and $b$. The radius is $\tfrac{a+b}{2}=A$. The perpendicular from $C$ up to the semicircle has length $\sqrt{ab}=G$ (by similar triangles, its square is $a\times b$). A half-chord can never be longer than the radius — so $G\le A$, with equality when $C$ is the centre, i.e. $a=b$.

Read in two directions, the inequality solves optimisation problems:

Finally, the mean values pin down the numbers themselves. If $a$ and $b$ have A.M. $A$ and G.M. $G$, then $a+b=2A$ and $ab=G^2$, so $a$ and $b$ are the roots of a quadratic.

📐 Theorem 8.9 — Recovering two numbers from their means

If two positive numbers have arithmetic mean $A$ and geometric mean $G$, they are the roots of $$x^2-2Ax+G^2=0,$$ namely $$A\pm\sqrt{A^2-G^2}=A\pm\sqrt{(A+G)(A-G)}.$$

Proof. A quadratic with roots $a$ and $b$ is $x^2-(a+b)x+ab=0$; substitute $a+b=2A$ and $ab=G^2$ and use the quadratic formula. The roots are real because $A^2-G^2=(A+G)(A-G)\ge0$ by Theorem 8.8. $\blacksquare$

✏️ Example 4 — Numbers from their means
(a) The A.M. and G.M. of two positive numbers are $10$ and $8$. Find the numbers.
(b) Can two positive numbers have A.M. $5$ and G.M. $6$?
(a) By Theorem 8.9 the numbers satisfy $x^2-20x+64=0$, i.e. $(x-4)(x-16)=0$: they are $4$ and $16$.
(a) Check $\tfrac{4+16}{2}=10$ ✓ and $\sqrt{64}=8$ ✓.
(b) No: Theorem 8.8 says $A\ge G$, but here $A=5\lt6=G$. (The quadratic $x^2-10x+36=0$ has discriminant $100-144\lt0$ — no real roots.)
✏️ Example 5 — Smallest values
(a) Find the least value of $x+\dfrac9x$ for $x\gt0$, and the $x$ where it occurs.
(b) Show that $(a+b)\left(\dfrac1a+\dfrac1b\right)\ge4$ for all positive $a,b$.
(a) $x$ and $\tfrac9x$ are positive with fixed product $9$. By A.M. $\ge$ G.M., $\dfrac{x+9/x}{2}\ge\sqrt{x\cdot\tfrac9x}=3$, so $x+\dfrac9x\ge6$.
(a) Equality Equality needs $x=\tfrac9x$, i.e. $x^2=9$, $x=3$ (positive). So the least value is $6$, at $x=3$. Check: $x=2$ gives $6.5$, $x=4$ gives $6.25$ — both larger ✓.
(b) $a+b\ge2\sqrt{ab}$ and $\dfrac1a+\dfrac1b\ge2\sqrt{\dfrac1{ab}}$ (A.M. $\ge$ G.M. for the positive numbers $\tfrac1a,\tfrac1b$). Both sides are positive, so we may multiply: $(a+b)\left(\tfrac1a+\tfrac1b\right)\ge4\sqrt{ab}\cdot\dfrac{1}{\sqrt{ab}}=4$.
(b) Equality Only when $a=b$. (Expanding gives $2+\tfrac ab+\tfrac ba$, and $\tfrac ab+\tfrac ba\ge2$ is the same fact.)
✏️ Example 6 — The largest field
A farmer has $200\ \text{m}$ of fencing to enclose a rectangular field. Show that the area cannot exceed $2500\ \text{m}^2$, and find the dimensions that give this area. What if the field lies along a straight river, so that only three sides need fencing?
Four sides Sides $x$ and $y$ with $2x+2y=200$, so $x+y=100$. By A.M. $\ge$ G.M., $\sqrt{xy}\le\dfrac{x+y}{2}=50$, so the area $xy\le2500\ \text{m}^2$, with equality when $x=y=50$: a $50\ \text{m}\times50\ \text{m}$ square.
Three sides Let the two sides perpendicular to the river be $x$ and the side parallel to it $y$: $2x+y=200$. Apply the inequality to the two positive numbers $2x$ and $y$, whose sum is fixed: $\sqrt{2x\cdot y}\le\dfrac{2x+y}{2}=100$, so $2xy\le10\,000$ and $xy\le5000\ \text{m}^2$.
Answer Equality when $2x=y=100$: the field is $50\ \text{m}$ by $100\ \text{m}$ (long side along the river), area $5000\ \text{m}^2$ — twice as much, because the river does a side's work.
Trap Applying A.M. $\ge$ G.M. to $x$ and $y$ in the river case gives $xy\le\left(\tfrac{x+y}{2}\right)^2$, but $x+y$ is not fixed there, so this bound is useless. Choose the two numbers so that their sum is fixed.
Interactive 2 — Seeing A.M. $\ge$ G.M.

Left: a semicircle on a diameter split into $a$ and $b$. The radius (purple) is the A.M.; the half-chord standing at the split point (green) is the G.M. Right: the $a\times b$ rectangle (area $ab=G^2$) and the square with the same perimeter, of side $A$ (area $A^2$). Press Play to slide the split point with $a+b$ fixed: the green segment and the rectangle's area are largest exactly when $a=b$.

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A.M. $A$—
G.M. $G$—
$A-G=\tfrac{(\sqrt a-\sqrt b)^2}{2}$—
areas $ab$ vs $A^2$—
⚠️ Common mistakes with means

1. Counting the wrong number of terms. Inserting $n$ means makes an AP or GP of $n+2$ terms, so $d=\dfrac{b-a}{n+1}$ and $r^{n+1}=\dfrac ba$ — not $n$ or $n+2$.

2. Forgetting positivity. A.M. $\ge$ G.M. is for positive numbers. For $x\lt0$, $x+\tfrac1x\le-2$, so "$x+\tfrac1x\ge2$" is false there.

3. Claiming a minimum that is never reached. The bound from A.M. $\ge$ G.M. is the least value only if the equality case $a=b$ can actually happen. Always find the $x$ that gives equality.

4. Applying the inequality when nothing is fixed. Use it on two numbers whose sum (to bound the product) or product (to bound the sum) is constant, as in Example 6.

5. Averaging growth rates arithmetically. Percentage changes multiply; the average growth factor is a G.M., as in the phone-price example.

Exercises

I
Arithmetic means.
  • 1 Find the A.M. of $7$ and $25$.
  • 2 Insert $4$ arithmetic means between $3$ and $23$.
  • 3 Insert $6$ arithmetic means between $-5$ and $16$.
  • 4 Find the sum of the $10$ arithmetic means inserted between $4$ and $30$.
  • 5 Between $2$ and $50$ some arithmetic means are inserted, with common difference $4$. How many?
  • 6 Find $x$ if $2x+1$ is the A.M. of $x$ and $13$.
  • 7 $n$ arithmetic means are inserted between $1$ and $31$ so that the $7$th mean and the $(n-1)$th mean are in the ratio $5:9$. Find $n$.
1 $\tfrac{7+25}{2}=16$.
2 $d=\tfrac{23-3}{5}=4$: $7,11,15,19$.
3 $d=\tfrac{16-(-5)}{7}=3$: $-2,1,4,7,10,13$.
4 $10\times\tfrac{4+30}{2}=170$ (Theorem 8.6).
5 $\tfrac{50-2}{n+1}=4$, so $n+1=12$, $n=11$.
6 $\tfrac{x+13}{2}=2x+1$, $x+13=4x+2$, $x=\tfrac{11}{3}$.
7 $d=\tfrac{30}{n+1}$, $A_7=1+7d$, $A_{n-1}=31-2d$. $9(1+7d)=5(31-2d)$ gives $73d=146$, $d=2$, so $n+1=15$, $n=14$. Check: $A_7=15$, $A_{13}=27$, $15:27=5:9$ ✓.
II
Geometric means.
  • 1 Find the G.M. of $4$ and $25$.
  • 2 Find the G.M. of $\tfrac23$ and $\tfrac{27}{2}$.
  • 3 Insert $2$ geometric means between $2$ and $54$.
  • 4 Insert $3$ geometric means between $\tfrac19$ and $9$.
  • 5 Insert $4$ geometric means between $160$ and $5$.
  • 6 Find the product of the $5$ geometric means inserted between $2$ and $18$.
  • 7 Find two positive numbers whose A.M. is $25$ and G.M. is $7$.
Check: the means of two positive numbers
A.M.—
G.M.—
quadratic $x^2-2Ax+G^2=0$—
1 $\sqrt{100}=10$.
2 $\sqrt{\tfrac23\cdot\tfrac{27}{2}}=\sqrt9=3$.
3 $r^3=27$, $r=3$: $6,18$.
4 $r^4=81$, $r=3$: $\tfrac13,1,3$.
5 $r^5=\tfrac{5}{160}=\tfrac1{32}$, $r=\tfrac12$: $80,40,20,10$.
6 $G=\sqrt{36}=6$; product $6^5=7776$ (Theorem 8.7).
7 $x^2-50x+49=0$, $(x-1)(x-49)=0$: $1$ and $49$.
III
The inequality A.M. $\ge$ G.M.
  • 1 Verify the inequality for the numbers $4$ and $9$, and for $10$ and $0.1$.
  • 2 The A.M. and G.M. of two positive numbers are $34$ and $16$. Find the numbers.
  • 3 Prove that $x+\dfrac1x\ge2$ for every $x\gt0$. When does equality hold?
  • 4 Find the least value of $4x+\dfrac{25}{x}$ for $x\gt0$, and where it occurs.
  • 5 If $a,b\gt0$ and $a+b=12$, find the greatest possible value of $ab$.
  • 6 Prove that $(a+b)(b+c)(c+a)\ge8abc$ for all positive $a,b,c$.
  • 7 Prove that $\dfrac{a}{b}+\dfrac{b}{a}\ge2$ for all positive $a,b$.
  • 8 The sum of two positive numbers is $10$ times their geometric mean. Show that the numbers are in the ratio $(5+2\sqrt6):(5-2\sqrt6)$.
1 $4,9$: $A=6.5\ge G=6$. $10,0.1$: $A=5.05\ge G=1$.
2 $x^2-68x+256=0$; discriminant $4624-1024=3600$, so $x=\tfrac{68\pm60}{2}$: $64$ and $4$.
3 A.M. $\ge$ G.M. for $x$ and $\tfrac1x$: $\tfrac{x+1/x}{2}\ge\sqrt{1}=1$. Equality iff $x=\tfrac1x$, i.e. $x=1$.
4 $4x+\tfrac{25}{x}\ge2\sqrt{100}=20$, with equality when $4x=\tfrac{25}{x}$, $x^2=\tfrac{25}{4}$, $x=\tfrac52$. Least value $20$.
5 $\sqrt{ab}\le6$, so $ab\le36$, reached at $a=b=6$.
6 $a+b\ge2\sqrt{ab}$, $b+c\ge2\sqrt{bc}$, $c+a\ge2\sqrt{ca}$; all sides are positive, so multiply: $\ge8\sqrt{a^2b^2c^2}=8abc$.
7 $\tfrac ab$ and $\tfrac ba$ are positive with product $1$: their A.M. is at least $1$, so the sum is at least $2$ (equality iff $a=b$).
8 $a+b=10G$ means $A=5G$. By Theorem 8.9 the numbers are $A\pm\sqrt{A^2-G^2}=5G\pm\sqrt{24}\,G=G(5\pm2\sqrt6)$, so their ratio is $(5+2\sqrt6):(5-2\sqrt6)$.
IV
Applications.
  • 1 A city's water tariff rose by $10\%$ one year and by $40\%$ the next. Find the average yearly percentage rise, correct to one decimal place, and explain why it is not $25\%$.
  • 2 A rectangle has area $64\ \text{cm}^2$. Find its least possible perimeter.
  • 3 A rectangular pen is built against a long wall using $120\ \text{m}$ of fencing for the other three sides. Find the largest possible area.
  • 4 A culture had $1000$ bacteria at 8 a.m. and $16\,000$ at noon, multiplying by the same factor every hour. Find the counts at 9, 10 and 11 a.m.
  • 5 Nine lamp posts are to be placed between two poles $2\ \text{km}$ apart so that all eleven posts are equally spaced. How far from the first pole is each lamp post?
  • 6 On a piano, the frequencies of the $13$ notes from A ($440\ \text{Hz}$) up to the next A ($880\ \text{Hz}$) form a GP: $11$ geometric means are inserted between $440$ and $880$. Find the common ratio, and the frequency of C, three notes above the first A, to the nearest $0.01\ \text{Hz}$.
1 Average factor $\sqrt{1.1\times1.4}=\sqrt{1.54}\approx1.2410$: about $24.1\%$ a year. Two rises of $25\%$ would give $1.5625$, more than the actual $1.54$; the G.M. of the factors is the rate that reproduces the true total.
2 Sides $x$, $\tfrac{64}{x}$: $x+\tfrac{64}{x}\ge2\cdot8=16$, so the perimeter $\ge32\ \text{cm}$, for the $8\ \text{cm}\times8\ \text{cm}$ square.
3 Sides $x,x$ (perpendicular to the wall) and $y$: $2x+y=120$. $\sqrt{2x\cdot y}\le60$, so $xy\le1800\ \text{m}^2$, at $2x=y=60$: $30\ \text{m}\times60\ \text{m}$.
4 $3$ G.M.s between $1000$ and $16\,000$: $r^4=16$, $r=2$: $2000$, $4000$, $8000$ bacteria.
5 $9$ A.M.s between $0$ and $2000\ \text{m}$: $d=\tfrac{2000}{10}=200\ \text{m}$; the lamp posts are $200,400,\dots,1800\ \text{m}$ from the first pole.
6 $r^{12}=2$, $r=2^{1/12}\approx1.0595$. C: $440\,r^3=440\cdot2^{1/4}\approx523.25\ \text{Hz}$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask you to insert means (2 marks), to find two numbers from their A.M. and G.M. or prove a result like $a,b=A\pm\sqrt{(A+G)(A-G)}$ (3 marks), and set case studies on largest areas. JEE uses A.M. $\ge$ G.M. to find least values quickly and combines it with properties of means.

E1. Board · 1 mark The geometric mean of $3$ and $27$ is (A) $15$   (B) $9$   (C) $\pm9$   (D) $81$

Idea $G=\sqrt{3\times27}=\sqrt{81}=9$.
Trap (A) is the A.M.; (C) forgets that the G.M. of two positive numbers is the positive root; (D) is $G^2$.
Answer (B).

E2. Board · 2 marks Insert three arithmetic means between $6$ and $-10$.

Difference $5$ terms in all: $d=\dfrac{-10-6}{4}=-4$. (1 mark)
Answer $6,\ 2,\ -2,\ -6,\ -10$: the means are $2,-2,-6$. (1 mark)

E3. Board · 3 marks If $A$ and $G$ are the A.M. and G.M. of two positive numbers $a$ and $b$, prove that the numbers are $A\pm\sqrt{(A+G)(A-G)}$. Hence find two numbers whose A.M. is $13$ and G.M. is $12$.

Sum and product $a+b=2A$ and $ab=G^2$, so $(a-b)^2=(a+b)^2-4ab=4A^2-4G^2$ and $a-b=\pm2\sqrt{(A+G)(A-G)}$. (1 mark)
Solve Adding and subtracting with $a+b=2A$: $a,b=A\pm\sqrt{(A+G)(A-G)}$. (1 mark)
Answer $13\pm\sqrt{25\times1}=13\pm5$: the numbers are $18$ and $8$. Check: $\tfrac{18+8}{2}=13$, $\sqrt{144}=12$ ✓. (1 mark)

E4. Board · Case study A school wants a rectangular garden plot fenced with exactly $80\ \text{m}$ of fencing. Let one side be $x$ metres. (i) Write the area in terms of $x$. (1 mark) (ii) Use A.M. $\ge$ G.M. to show that the area is at most $400\ \text{m}^2$. (1 mark) (iii) The plan finally needs an area of exactly $300\ \text{m}^2$. Find the sides. (2 marks)

(i) The sides add to $40$, so the other side is $40-x$ and the area is $x(40-x)\ \text{m}^2$, $0\lt x\lt40$. (1 mark)
(ii) $\sqrt{x(40-x)}\le\dfrac{x+(40-x)}{2}=20$, so the area $\le400\ \text{m}^2$ (a $20\ \text{m}$ square). (1 mark)
(iii) set-up $x(40-x)=300$, $x^2-40x+300=0$. (1 mark)
Answer $(x-10)(x-30)=0$: the plot is $10\ \text{m}\times30\ \text{m}$. (The numbers $10,30$ have A.M. $20$ and G.M. $\sqrt{300}$.) (1 mark)

E5. JEE Main The least value of $4^x+4^{1-x}$, $x\in\mathbb{R}$, is (A) $2$   (B) $4$   (C) $5$   (D) $1$

Idea Both terms are positive and their product is $4^x\cdot4^{1-x}=4$, a constant. By A.M. $\ge$ G.M., $4^x+4^{1-x}\ge2\sqrt4=4$.
Equality $4^x=4^{1-x}$ gives $x=\tfrac12$, and then $2+2=4$, so the bound is reached.
Trap (A) uses "$t+\tfrac1t\ge2$" without noticing that the product is $4$, not $1$; (C) is the value at $x=0$, not the least value.
Answer (B).

E6. JEE Advanced One or more options correct. Let $a,b$ be distinct positive numbers with A.M. $A$ and G.M. $G$, and let $n\ge1$ arithmetic means and $n$ geometric means be inserted between $a$ and $b$. Then (A) $A\gt G$   (B) $A^2-G^2=\left(\dfrac{a-b}{2}\right)^2$   (C) the product of the $n$ geometric means is $G^n$   (D) the sum of the $n$ arithmetic means is $nG$

(A) True: $A-G=\tfrac12(\sqrt a-\sqrt b)^2\gt0$ because $a\ne b$.
(B) True: $\dfrac{(a+b)^2}{4}-ab=\dfrac{a^2-2ab+b^2}{4}=\left(\dfrac{a-b}{2}\right)^2$.
(C), (D) (C) True by Theorem 8.7. (D) False: the sum is $nA$ (Theorem 8.6), and $nA\gt nG$ by (A). E.g. $a=1$, $b=9$, $n=1$: the A.M. is $5$, not $G=3$.
Answer (A), (B), (C).
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. What is the "average yearly rise" in the price of something your family buys regularly — and does the arithmetic mean of the yearly rises over- or under-state it?

Why it matters. Economists, banks and governments quote average growth rates for prices, salaries, GDP and investments all the time. The correct average of percentage changes is a geometric mean (the "compound annual growth rate"); averaging the percentages arithmetically always gives a figure at least as big, by A.M. $\ge$ G.M. — a gap that can mislead.

What to do.

  1. Choose an item — milk, petrol, a bus or metro fare, a school fee, a cinema ticket — and collect its price in each of the last $5$–$8$ years (old bills, family memory, newspaper archives or official price data).
  2. For each year compute the growth factor $g_k=\dfrac{\text{price this year}}{\text{price last year}}$.
  3. Compute the arithmetic mean of the factors and their geometric mean $\left(g_1g_2\cdots g_n\right)^{1/n}$ (a spreadsheet's AVERAGE and GEOMEAN).
  4. Starting from the first price, apply each mean for $n$ years. Which one lands exactly on today's price?

Tools: A spreadsheet; price records

Hand in: Your price table, the yearly factors, both means, the two "predicted" final prices, and a paragraph explaining which average a fair report should use, and why the other one is always at least as large.

Going further. A fund advertises "average returns of $25\%$ a year" after gaining $100\%$ one year and losing $50\%$ the next. What happened to an investment of ₹10 000? Compute both means of the factors $2$ and $0.5$.

Instructor note: the G.M. of the factors always reproduces the final price exactly (it is defined that way), while the A.M. overshoots unless every year had the same rise. In "going further", the A.M. of $2$ and $0.5$ is $1.25$ (the advertised $25\%$) but the G.M. is $1$: the investor is back to exactly ₹10 000. This makes a memorable class discussion.

🧪 Self-check quiz

0/5 correct

Q1. The arithmetic mean of $6$ and $14$ is

Q2. The geometric mean of $4$ and $9$ is

Q3. Four arithmetic means are inserted between $2$ and $22$. Their common difference is

Q4. The product of the three geometric means inserted between $2$ and $32$ is

Q5. For $x\gt0$, the least value of $x+\dfrac4x$ is

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