Sequences and Series

Two job offers, both starting at five lakh a year.

Offer A adds fifty thousand to the salary every year: an arithmetic progression. Offer B raises the salary by $8$ per cent every year: a geometric progression.

Fifty thousand is $10$ per cent of the starting salary, and $10$ beats $8$. So Offer A pays more. That is the reflex, and over the first decade it is very nearly right.

Over ten years A wins, by about seven thousand rupees in seventy-two lakh. Over twenty, B wins by roughly thirty-four lakh.

Nothing changed except the number of years. The AP's increments stay at fifty thousand for ever, while the GP's increments are $8$ per cent of a salary that keeps growing, so they start smaller and end enormous.

$\text{repeated addition}\Rightarrow \text{linear},\text{repeated multiplication}\Rightarrow \text{exponential}$

The crossing was never in doubt. Multiplication eventually beats addition however small the ratio, and the only question is when.

One structure, two operations. Take the logarithm of every term of a GP and the constant ratio becomes a constant difference: the GP has become an AP. Every GP result has an AP result standing behind it, and the two means, arithmetic and geometric, are the same average computed through the two operations.

$a,ar,a{r}^2,\dots \stackrel{\log }{\to }\log a,\log a+\log r,\log a+2\log r,\dots$

1. Sequences, Series and Notation

Sequence. An ordered list of numbers. Series. The sum of a sequence's terms. The distinction matters because $T_n$ and $S_n$ answer different questions.

$T_n=S_n-S_{n-1}$

That relation is worth more than it looks: given a formula for $S_n$, it recovers every term, and it is the fastest way to test whether a given $S_n$ belongs to an AP or a GP at all.

Illustration 1

The sum of the first $n$ terms of a sequence is $S_n=3n^2+5n$. Find the $n$th term and identify the sequence.

Subtract consecutive sums, which is the only tool needed.

$T_n=S_n-S_{n-1}=\left(3n^2+5n\right)-\left(3(n-1{)}^2+5(n-1)\right)$

$=3\left(n^2-(n-1{)}^2\right)+5=3(2n-1)+5=6n+2$

$T_1=8,T_2=14,T_3=20\Rightarrow \text{an AP with }a=8,d=6$

Check the edge case, which the subtraction cannot see: $S_1$ must equal $T_1$, and $3+5=8$ agrees.

A quadratic $S_n$ always signals an AP, because subtracting two quadratics that differ by one in $n$ leaves something linear. Likewise an $S_n$ built from a power of a constant signals a GP.

2. Arithmetic Progressions

AP. Each term exceeds the previous by a fixed common difference $d$.

$T_n=a+(n-1)d$

$S_n=\frac{n}{2}\left[2a+(n-1)d\right]=\frac{n}{2}\left(a+\ell \right)$

The second form, with $\ell$ the last term, is the faster one whenever both ends are known. It also says what an AP sum really is: the number of terms times the average of the first and last.

Properties worth using

Trap. For four terms in AP the symmetric choice has common difference $2d$, not $d$. Using $a-2d,a-d,a+d,a+2d$ is not an AP at all.

Illustration 2

The $7$th term of an AP is $34$ and the $13$th is $64$. Find the sum of the first $20$ terms.

Two facts give two equations, and subtracting them removes $a$ immediately.

$a+6d=34,a+12d=64\Rightarrow 6d=30\Rightarrow d=5,a=4$

$S_{20}=\frac{20}{2}\left[2(4)+19(5)\right]=10(8+95)=1030$

There is a shorter route worth seeing. The gap between the $7$th and $13$th terms spans six common differences, so $d$ is the difference of the values divided by the difference of the positions, $30/6$. That reading works for any two terms and skips the simultaneous equations entirely.

Illustration 3

For the AP $25,22,19,\dots$, find the sum of the first $20$ terms and the greatest sum the progression can reach.

Here $a=25$ and $d=-3$, so the terms fall and eventually go negative. The greatest sum is reached just before that happens.

$T_n=25+(n-1)(-3)=28-3n$

$T_n\ge 0⟺n\le \frac{28}{3}=9.33\Rightarrow T_9=1,T_{10}=-2$

Adding $T_{10}$ would reduce the total, so the maximum sum occurs at $n=9$.

$S_9=\frac{9}{2}\left[50+8(-3)\right]=\frac{9}{2}(26)=117$

$S_{20}=\frac{20}{2}\left[50+19(-3)\right]=10(-7)=-70$

The two answers differ in sign, which is the point: with a negative common difference, "sum of more terms" and "larger sum" are different requests.

Illustration 4

Four numbers in AP have sum $20$ and sum of squares $120$. Find them.

Choose the symmetric form, which makes the sum condition collapse instantly.

$a-3d,a-d,a+d,a+3d\Rightarrow \text{sum}=4a=20\Rightarrow a=5$

In the sum of squares, every cross term cancels in pairs, which is exactly why the symmetric form is chosen.

$4a^2+20d^2=120\Rightarrow 100+20d^2=120\Rightarrow d^2=1$

$d=\pm 1\Rightarrow 2,4,6,8\text{(or the same numbers reversed)}$

Both signs of $d$ give the same set, listed forwards or backwards, so the answer is one progression and not two.

3. Arithmetic Means

Arithmetic mean of $a$ and $b$: $\frac{a+b}{2}$, the single number making $a$, $A$, $b$ an AP.

Inserting $n$ arithmetic means between $a$ and $b$ builds an AP with $n+2$ terms in total.

$d=\frac{b-a}{n+1}$

The denominator is $n+1$ because inserting $n$ means creates $n+1$ gaps, not $n$. Their sum is $n$ times the single arithmetic mean of $a$ and $b$.

4. Geometric Progressions

GP. Each term is the previous multiplied by a fixed common ratio $r$.

$T_n=a{r}^{n-1},S_n=a\frac{r^n-1}{r-1}(r\ne 1)$

For $r=1$ every term equals $a$ and the sum is simply $na$; the formula divides by zero there and must not be used.

Properties worth using

Illustration 5

In a GP the sum of the first two terms is $12$ and the sum of the third and fourth is $48$. Find the progression.

Write both conditions and notice that the second is the first multiplied by $r^2$.

$a(1+r)=12,a{r}^2(1+r)=48$

Dividing removes $a$ and the whole bracket at once.

$r^2=4\Rightarrow r=2\text{ or }r=-2$

$r=2\Rightarrow a=4:4,8,16,32$

$r=-2\Rightarrow a(1-2)=12\Rightarrow a=-12:-12,24,-48,96$

Check the second: $-48+96=48$. Both progressions satisfy every condition given.

Trap. A square root in a GP question almost always admits both signs, and a negative ratio produces a legitimate alternating progression. Discarding it because it "looks wrong" costs the mark.

Illustration 6

Show that if $a$, $b$, $c$ are in GP then $\log a$, $\log b$, $\log c$ are in AP, and use this on $3,6,12$.

GP means the ratio of consecutive terms is constant, so $b/a=c/b$.

$\log \frac{b}{a}=\log \frac{c}{b}\Rightarrow \log b-\log a=\log c-\log b$

which is exactly the statement that the three logarithms have a constant difference.

For $3,6,12$ taken in base $2$: ${\log }_{2}3\approx 1.585$, then $2.585$, then $3.585$. The common difference is $1$, which is ${\log }_{2}2={\log }_{2}r$.

$\text{constant ratio }r\stackrel{\log }{\to }\text{constant difference }\log r$

Every GP identity has an AP identity behind it, and this is the translation. The geometric mean becoming the arithmetic mean of the logarithms is the same fact applied to three terms.

The infinite sum

When $|r|<1$ the powers of $r$ shrink towards zero and the partial sums settle on a limit.

$S_{\infty }=\frac{a}{1-r},|r|<1$

Illustration 7

Sum $1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\cdots$, then apply the same formula to $1+2+4+8+\cdots$ and explain the result.

The first has $a=1$ and $r=2/3$, comfortably inside the range.

$S_{\infty }=\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3$

The partial sums $1,1.667,2.111,2.407,\dots$ climb towards $3$ and never reach it, which is what a limit looks like.

The second has $r=2$. Applying the formula regardless gives

$\frac{1}{1-2}=-1$

A sum of positive numbers cannot be negative, so the formula has been used where it says nothing. Its derivation divides by $1-r$ after letting $r^n$ vanish, and $2^n$ does not vanish.

Trap. $S_{\infty }=a/(1-r)$ is a statement about a limit, and the limit exists only for $|r|<1$. Quote the condition alongside the formula.

5. Geometric Means

Geometric mean of positive $a$ and $b$: $\sqrt{ab}$, the single number making $a$, $G$, $b$ a GP.

Inserting $n$ geometric means between $a$ and $b$ builds a GP with $n+2$ terms.

$r={\left(\frac{b}{a}\right)}^{1/(n+1)}$

Again $n+1$, for the same reason as before, and their product is $G^n$ where $G=\sqrt{ab}$. Every statement here is the arithmetic-means statement with sums replaced by products, which is the logarithm at work again.

6. The Relation Between AM and GM

For positive numbers, the arithmetic mean is never below the geometric mean.

$A\ge G,\text{equality if and only if all the numbers are equal}$

For two numbers the proof is one line, because a square is never negative.

${\left(\sqrt{a}-\sqrt{b}\right)}^2\ge 0\Rightarrow a+b\ge 2\sqrt{ab}\Rightarrow \frac{a+b}{2}\ge \sqrt{ab}$

The equality condition does real work, and it is where marks are lost. Equality holds exactly when the numbers coincide, so any minimum obtained this way is attained only at that point.

Using it to find minimum values

Whenever a positive expression is a sum whose corresponding product is constant, its minimum follows without calculus.

$x+\frac{4}{x}(x>0):\text{product}=4\Rightarrow \frac{x+4/x}{2}\ge 2\Rightarrow x+\frac{4}{x}\ge 4$

Equality at $x=2$, where the two parts are equal.

Trap. Check that the product is constant before using the inequality. If the product still contains the variable, the bound moves with the variable and proves nothing.

Illustration 8

Find the minimum of $x^2+\frac{2}{x}$ for $x>0$.

Applied to two parts the product is $2x$, which is not constant, so the inequality gives nothing useful. The repair is to split until the product is constant.

$x^2+\frac{1}{x}+\frac{1}{x}\text{has product}{x}^2\cdot \frac{1}{x}\cdot \frac{1}{x}=1$

Now apply AM-GM to those three positive parts.

$\frac{x^2+\frac{1}{x}+\frac{1}{x}}{3}\ge \sqrt[3]{1}=1\Rightarrow x^2+\frac{2}{x}\ge 3$

Equality needs all three parts equal: $x^2=1/x$, so $x^3=1$ and $x=1$.

$\text{minimum }3,\text{ at }x=1$

Check by substitution: $1+2=3$. Differentiating gives $2x-2/x^2=0$, so $x^3=1$, the same answer through several more lines.

Illustration 9

A student writes $\tan \theta +\cot \theta \ge 2$ because the product is $1$. Where does this fail?

The product is indeed the constant $1$, so that condition is satisfied. The other condition is not.

AM-GM requires the terms to be positive. In the first quadrant both are positive and the conclusion holds, with minimum $2$ at $\theta =\pi /4$.

In the second quadrant $\tan \theta$ and $\cot \theta$ are both negative, so the inequality does not apply. Substituting $\theta =3\pi /4$ gives $-1+(-1)=-2$, well below the claimed bound of $2$.

$\text{constant product is not enough; the terms must be positive}$

Writing $u=-\tan \theta >0$ recovers the truth: $u+1/u\ge 2$, so $\tan \theta +\cot \theta \le -2$ there. The bound flipped from a minimum to a maximum.

Recovering the numbers from the means

If $A$ and $G$ are the two means of a pair of numbers, those numbers are the roots of a quadratic you can write immediately, using sum $2A$ and product $G^2$.

$x^2-2Ax+G^2=0$

Illustration 10

Two positive numbers are said to have arithmetic mean $5$ and geometric mean $6$. Show that no such numbers exist.

Form the quadratic and look at its discriminant.

$x^2-10x+36=0\Rightarrow D=100-144=-44<0$

The roots are non-real, so no pair of real numbers has these means.

The general statement is the same calculation with letters: $D=4A^2-4G^2$, which is non-negative exactly when $A\ge G$. Claiming $A<G$ is claiming a negative discriminant.

This is a second and independent proof of AM-GM, arriving through quadratics rather than through squares, and it also explains the equality case: $A=G$ makes the discriminant zero, so the two numbers coincide.

7. A Note on Syllabus Emphasis

The official unit text names arithmetic and geometric progressions, the insertion of arithmetic and geometric means between two given numbers, and the relation between AM and GM. That is the whole unit.

Those unnamed topics remain examinable in JEE Advanced, so do not discard them if you are sitting that paper. For JEE Main the marks live in the four items the unit text lists, and AM-GM in particular repays far more attention than its single line of syllabus suggests.

Illustration 11

Beyond JEE Main, for Advanced sitters: sum $S=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\cdots$ to infinity.

This is an arithmetico-geometric series: an AP $1,2,3,\dots$ multiplied term by term by a GP with ratio $\frac{1}{2}$. The method is to multiply by the ratio and subtract.

$S=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\cdots ,\frac{S}{2}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\cdots$

$S-\frac{S}{2}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$

Subtracting has turned the difficult series into an ordinary infinite GP, because consecutive numerators differ by exactly $1$.

$\frac{S}{2}=\frac{1}{1-\frac{1}{2}}=2\Rightarrow S=4$

The shift-and-subtract idea is the whole technique, and it is worth knowing even for Main, because it occasionally rescues a question that looks like an ordinary GP and is not.

Summary

An AP repeats addition and a GP repeats multiplication; logarithms turn one into the other, so every GP result has an AP result behind it.

Multiplication eventually overtakes addition however small the ratio, which is why an $8$ per cent raise beats a fixed increment worth $10$ per cent, given enough years.

$T_n=S_n-S_{n-1}$ recovers every term from a sum formula; a quadratic $S_n$ always means an AP, and the case $n=1$ must be checked separately.

For an AP, $T_n=a+(n-1)d$ and $S_n=\frac{n}{2}(a+\ell )$: the number of terms times the average of the ends.

Terms equidistant from the ends of an AP have constant sum; in a GP they have constant product.

Symmetric choices simplify: $a-d,a,a+d$ for three, and $a-3d,a-d,a+d,a+3d$ for four, where the common difference is $2d$.

With negative $d$, the greatest sum stops at the last non-negative term, so more terms can mean a smaller total.

For a GP, $S_n=a(r^n-1)/(r-1)$, and $r=1$ must be handled separately as $na$.

A square root in a GP question usually admits both signs, and a negative ratio gives a legitimate alternating progression.

$S_{\infty }=a/(1-r)$ holds only for $|r|<1$; applied to $r=2$ it returns $-1$ for a sum of positives.

Inserting $n$ means creates $n+1$ gaps, so the difference is $(b-a)/(n+1)$ and the ratio is $(b/a{)}^{1/(n+1)}$.

$A\ge G$ for positive numbers, with equality only when they are all equal, and the semicircle picture shows why: a half chord cannot exceed a radius.

For minimum values, check that the product is constant and the terms positive. If the product is not constant, split the expression until it is.

$x^2-2Ax+G^2=0$ recovers a pair from its means, and its discriminant is a second proof of $A\ge G$.