Chapter 8 · Sequences and Series

8.2 Geometric Progressions

Multiply by the same number $r$ again and again: $a,\ ar,\ ar^2,\dots$ . The $n$th term is $ar^{n-1}$, the first $n$ terms add to $\dfrac{a(r^n-1)}{r-1}$, and when $|r|\lt1$ all infinitely many terms add to $\dfrac{a}{1-r}$.

Put ₹10 000 in a bank account that pays $8\%$ interest per year, compounded annually. After one year you have $10\,000\times1.08=10\,800$; after two years $10\,800\times1.08=11\,664$; after three, $12\,597.12$. Each year's balance is the previous one multiplied by the same number, $1.08$. The same pattern describes a bouncing ball (each rebound is $\tfrac34$ of the last), a medicine leaving the bloodstream (each hour a fixed fraction remains), a city growing by $2\%$ a year, and the king's rice on the chessboard (each square doubles the last).

In an arithmetic progression you add the same number each time and the terms march along a straight line. Multiplying instead gives a geometric progression, and its behaviour is far richer: it can explode, shrink towards zero, or swing from positive to negative. In this section we find the $n$th term and the sum of the first $n$ terms of a geometric progression, and then meet one of the most surprising facts in school mathematics — that infinitely many positive numbers can add up to a finite total.

The Common Ratio and the $n$th Term

📘 Definition 8.4 — Geometric progression

A sequence is a geometric progression (GP) if its first term $a$ is non-zero and each later term is obtained by multiplying the term before it by the same non-zero constant $r$, the common ratio: $$a,\ ar,\ ar^2,\ ar^3,\dots,\qquad \frac{a_{n+1}}{a_n}=r\ \text{ for every } n.$$ No term of a GP is zero, so the ratio of consecutive terms always makes sense.

To test whether a list is a GP, divide each term by the one before it: $3,\,12,\,48,\,192$ gives $4,4,4$ — a GP with $r=4$. The list $2,6,12,20$ gives $3,2,\tfrac53$ — not a GP. The ratio may be a fraction ($81,27,9,3$ has $r=\tfrac13$) or negative ($5,-10,20,-40$ has $r=-2$).

📐 Theorem 8.3 — The $n$th term of a GP

The $n$th term of a GP with first term $a$ and common ratio $r$ is $$a_n=ar^{n-1}.$$ If the GP is finite with last term $l$, the $n$th term counted from the end is $l\left(\tfrac1r\right)^{n-1}$.

Proof. $a_1=a=ar^0$. To reach $a_n$ from $a_1$ we multiply by $r$ once for each step $a_1\to a_2\to\cdots\to a_n$, that is $n-1$ times, so $a_n=ar^{n-1}$. Read backwards, the terms form a GP with first term $l$ and ratio $\tfrac1r$, which gives the second formula. $\blacksquare$

The sign and size of $r$ decide how a GP behaves. If $r\gt1$ (and $a\gt0$) the terms grow ever faster; if $0\lt r\lt1$ they shrink towards $0$ but never reach it; if $r\lt0$ the signs alternate; if $r=1$ the GP is constant; and if $r=-1$ it flips between $a$ and $-a$. Interactive 1 lets you see all of these.

✏️ Example 1 — Finding terms
(a) Find the $10$th term and the $n$th term of the GP $5,10,20,\dots$ .
(b) Which term of the GP $2,\ 2\sqrt2,\ 4,\dots$ is $128$?
(c) Find the $8$th term of $3,\ -\tfrac32,\ \tfrac34,\dots$ .
(a) $a=5$, $r=\tfrac{10}{5}=2$: $a_n=5\cdot2^{n-1}$ and $a_{10}=5\cdot2^9=2560$.
(b) $r=\dfrac{2\sqrt2}{2}=\sqrt2$. Solve $2(\sqrt2)^{n-1}=128$: $(\sqrt2)^{n-1}=64=2^6=(\sqrt2)^{12}$, so $n-1=12$ and $n=13$. The $13$th term is $128$.
(c) $r=\dfrac{-3/2}{3}=-\tfrac12$: $a_8=3\left(-\tfrac12\right)^7=-\dfrac{3}{128}$.
Check (c) An odd power of a negative ratio is negative, and the even-numbered terms ($-\tfrac32,\dots$) are the negative ones ✓.
✏️ Example 2 — A GP from two of its terms
The $3$rd term of a GP is $12$ and the $6$th term is $-96$. Find the first term, the common ratio, the $n$th term and the $10$th term.
Set up $ar^2=12$ and $ar^5=-96$.
Divide $\dfrac{ar^5}{ar^2}=r^3=\dfrac{-96}{12}=-8$, so $r=-2$ (the only real cube root of $-8$). Then $a=\dfrac{12}{r^2}=\dfrac{12}{4}=3$.
Answer $a_n=3(-2)^{n-1}$ and $a_{10}=3(-2)^9=-1536$.
Check $3,-6,12,-24,48,-96$: the $3$rd term is $12$ and the $6$th is $-96$ ✓.

Three numbers in GP. If $a,b,c$ are consecutive terms of a GP then $\dfrac ba=\dfrac cb$, that is $$b^2=ac.$$ Conversely, non-zero $a,b,c$ with $b^2=ac$ are in GP. When a problem gives the product of three numbers in GP, call them $\dfrac{a}{r},\ a,\ ar$: the product is then simply $a^3$.

✏️ Example 3 — Three numbers in GP
Three numbers in GP have product $1$ and sum $\tfrac{39}{10}$. Find them.
Product Take $\tfrac ar,\ a,\ ar$. Product $a^3=1$, so $a=1$.
Sum $\tfrac1r+1+r=\tfrac{39}{10}$. Multiply by $10r$: $10+10r+10r^2=39r$, so $10r^2-29r+10=0$, $(5r-2)(2r-5)=0$, giving $r=\tfrac25$ or $r=\tfrac52$.
Answer Either value gives the same three numbers (in opposite orders): $\tfrac25,\ 1,\ \tfrac52$.
Check $\tfrac25\cdot1\cdot\tfrac52=1$ ✓ and $\tfrac{4+10+25}{10}=\tfrac{39}{10}$ ✓.
Interactive 1 — AP versus GP: Adding versus Multiplying

Both sequences start at the same $a$. The AP (blue) adds $d$ each step, so its dots lie on a straight line. The GP (red) multiplies by $r$ each step. Press Play to sweep $r$ from $-1.6$ to $1.6$ and watch the GP change character: oscillating and growing, oscillating and shrinking, decaying to $0$, constant at $r=1$, and exploding for $r\gt1$. Dots that leave the window are marked at its edge.

411.3
AP: $a_{12}=a+11d$—
GP: $a_{12}=ar^{11}$—
GP behaviour—

The Sum of the First $n$ Terms

For an AP we added the series to itself backwards. For a GP the trick is different: multiply the series by $r$. Every term then moves one place along, and subtracting makes almost everything cancel.

📐 Theorem 8.4 — Sum of $n$ terms of a GP

The sum of the first $n$ terms of a GP with first term $a$ and common ratio $r$ is $$S_n=\frac{a(r^n-1)}{r-1}=\frac{a(1-r^n)}{1-r}\qquad(r\ne1),$$ and $S_n=na$ if $r=1$. In terms of the last term $l=ar^{n-1}$: $S_n=\dfrac{lr-a}{r-1}$ ($r\ne1$).

Proof. If $r=1$ every term is $a$ and $S_n=na$. Otherwise write $$S_n=a+ar+ar^2+\cdots+ar^{n-1}.$$ Multiplying by $r$, $$rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n.$$ The two right-hand sides share the terms $ar,\dots,ar^{n-1}$; subtracting, $rS_n-S_n=ar^n-a$, i.e. $(r-1)S_n=a(r^n-1)$. Since $r\ne1$ we may divide by $r-1$. Multiplying top and bottom by $-1$ gives the second form, and $ar^n=lr$ gives the last. $\blacksquare$

Both forms are correct for every $r\ne1$; use $\dfrac{a(r^n-1)}{r-1}$ when $r\gt1$ and $\dfrac{a(1-r^n)}{1-r}$ when $r\lt1$, so that the numbers stay positive. For the chessboard, $a=1$, $r=2$, $n=64$: $S_{64}=2^{64}-1$ grains, as in the chapter introduction.

✏️ Example 4 — Using the sum formula
(a) Find the sum of the first $8$ terms of $3+6+12+\cdots$ .
(b) Find the sum of the first $10$ terms of $1+\tfrac12+\tfrac14+\cdots$ .
(c) How many terms of $3+3^2+3^3+\cdots$ are needed to give the sum $120$?
(a) $a=3$, $r=2$: $S_8=\dfrac{3(2^8-1)}{2-1}=3\times255=765$.
(b) $a=1$, $r=\tfrac12$: $S_{10}=\dfrac{1-(1/2)^{10}}{1-1/2}=2\left(1-\tfrac{1}{1024}\right)=\dfrac{1023}{512}$ — just less than $2$.
(c) $a=3$, $r=3$: $\dfrac{3(3^n-1)}{2}=120$, so $3^n-1=80$, $3^n=81=3^4$, $n=4$.
Check (c) $3+9+27+81=120$ ✓.
✏️ Example 5 — Money that grows by a fixed percentage
(a) ₹20 000 is invested at $10\%$ per year, compounded annually. Find the amount after $5$ years.
(b) A family deposits ₹5000 at the start of every year for $5$ years in an account paying $8\%$ per year, compounded annually. How much is in the account at the end of the fifth year?
(a) Each year the amount is multiplied by $1+\tfrac{10}{100}=1.1$. After $5$ years: $20\,000\times1.1^5=20\,000\times1.61051=\text{₹}\,32\,210.20$. (The amounts at the end of years $0,1,2,\dots$ form a GP with ratio $1.1$; this is the compound interest formula $A=P\left(1+\tfrac{R}{100}\right)^n$.)
(b) Each deposit The deposit made at the start of year $1$ earns interest for $5$ years and grows to $5000(1.08)^5$; the one at the start of year $2$ grows to $5000(1.08)^4$; … ; the last one, made at the start of year $5$, to $5000(1.08)$.
(b) Add Total $=5000\,[1.08+1.08^2+\cdots+1.08^5]$, a GP with $a=1.08$, $r=1.08$, $n=5$: $$5000\times\frac{1.08\,(1.08^5-1)}{0.08}\approx5000\times6.33593=\text{₹}\,31\,679.65.$$
Check The family paid in ₹25 000; the extra ₹6679.65 is interest. Year by year: $5400\to11\,232\to17\,530.56\to24\,333.00\to31\,679.65$ (add ₹5000 and multiply by $1.08$ each year) ✓.
⚠️ Common mistakes with GPs

1. Using $r^n$ for the $n$th term. The $n$th term is $ar^{n-1}$: the first term has no factor $r$ at all.

2. Dividing the wrong way. $r=\dfrac{a_2}{a_1}$ (a later term over the earlier one). For $16,8,4$, $r=\tfrac12$, not $2$.

3. Writing $S_n=\dfrac{a(r^n-1)}{r-1}$ when $r=1$. The formula divides by zero; the sum is simply $na$.

4. Losing the sign of a negative ratio. $(-2)^{9}=-512$ but $(-2)^{10}=1024$; and $-2^{10}$ (without brackets) means $-1024$.

5. Percent growth as an AP. "Grows by $10\%$ a year" multiplies by $1.1$ each year (a GP); "grows by ₹1000 a year" adds a fixed amount (an AP).

The Sum of an Infinite GP

Stand $1$ metre from a wall. Walk half the distance to it, then half of what remains, then half of that, and so on for ever. You walk $\tfrac12+\tfrac14+\tfrac18+\cdots$ metres. After $n$ steps the distance walked is $S_n=1-\left(\tfrac12\right)^n$ (Theorem 8.4 with $a=r=\tfrac12$), and the distance left is $\left(\tfrac12\right)^n$. That gap halves at every step — after $10$ steps it is under a millimetre, after $40$ steps less than the width of an atom — so the partial sums get as close to $1$ as we please. We therefore say that the infinite series has sum $1$: $$\tfrac12+\tfrac14+\tfrac18+\cdots=1.$$

The same thing happens for any GP with $|r|\lt1$: the powers $r^n$ get closer and closer to $0$ as $n$ grows (for example $0.9^{50}\approx0.005$ and $0.9^{100}\approx0.00003$). Then in $$S_n=\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}-\frac{a\,r^n}{1-r}$$ the last part shrinks to nothing, and $S_n$ settles on $\dfrac{a}{1-r}$. (Limits are made precise in Chapter 12; here the idea of "getting as close as we like" is enough.)

📐 Theorem 8.5 — Sum of an infinite GP

If $-1\lt r\lt1$ (that is, $|r|\lt1$), the partial sums of the infinite GP $a+ar+ar^2+\cdots$ approach $$S_\infty=\frac{a}{1-r},$$ called the sum to infinity of the GP. If $|r|\ge1$ (and $a\ne0$), the partial sums do not settle on any number, and the infinite GP has no sum.

For $|r|\ge1$ the terms themselves do not shrink: $1+2+4+\cdots$ grows without bound, $1+1+1+\cdots$ too, and $1-1+1-1+\cdots$ has partial sums $1,0,1,0,\dots$ that never settle. Interactive 2 shows all three behaviours.

✏️ Example 6 — Sums to infinity
(a) Find the sum to infinity of $8-4+2-1+\cdots$ .
(b) Write the recurring decimals $0.\overline{45}$ and $0.1\overline{6}$ as fractions.
(c) A ball is dropped from a height of $16\ \text{m}$. Each time it hits the ground it rebounds to $\tfrac34$ of the height it fell from. Find the total distance it travels before coming to rest.
(a) $a=8$, $r=-\tfrac12$, and $|r|\lt1$: $S_\infty=\dfrac{8}{1-(-1/2)}=\dfrac{8}{3/2}=\dfrac{16}{3}$.
(b) $0.\overline{45}=0.45+0.0045+0.000045+\cdots$, a GP with $a=0.45$, $r=0.01$: $\dfrac{0.45}{0.99}=\dfrac{45}{99}=\dfrac{5}{11}$. And $0.1\overline{6}=0.1+(0.06+0.006+\cdots)=\dfrac1{10}+\dfrac{0.06}{0.9}=\dfrac1{10}+\dfrac1{15}=\dfrac16$.
(c) Set up The ball falls $16$ m, then rises and falls $12$ m, then $9$ m, then $6.75$ m, … . Every rebound height is travelled twice (up and down).
(c) Answer Total $=16+2\,(12+9+6.75+\cdots)=16+2\cdot\dfrac{12}{1-3/4}=16+2\cdot48=112\ \text{m}$.
Interactive 2 — Partial Sums Settling (or Not)

Each bar is a partial sum $S_n=a+ar+\cdots+ar^{n-1}$; the small mark on it shows the term just added. When $|r|\lt1$ the dashed line is $S_\infty=\dfrac{a}{1-r}$ and the bars close in on it — from one side if $r\gt0$, zig-zagging round it if $r\lt0$. Push $|r|$ up to $1$ or beyond and the bars never settle.

40.5
$n$—
$S_n$—
$S_\infty=\tfrac{a}{1-r}$—
gap $|S_\infty-S_n|$—
🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026, as an extension of the sums of progressions and infinite geometric series it lists; not in JEE Main 2026 or the CBSE year-end exam. Arithmetico-geometric series are a JEE Advanced favourite. In such a series each term is a term of an AP times the matching term of a GP, e.g. $1+2x+3x^2+4x^3+\cdots$ .

Method. Multiply by the ratio $x$ and subtract, exactly as in the proof of Theorem 8.4. With $S=1+2x+3x^2+\cdots+nx^{n-1}$: $$S-xS=1+x+x^2+\cdots+x^{n-1}-nx^n=\frac{1-x^n}{1-x}-nx^n.$$ For $|x|\lt1$ both $x^n$ and $nx^n$ shrink to $0$ as $n$ grows, so the infinite series has sum $$1+2x+3x^2+\cdots=\frac{1}{(1-x)^2}.$$

Worked example. Find $\dfrac12+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\cdots$ .
Solution. $S=\dfrac12\left(1+2\cdot\tfrac12+3\cdot\tfrac14+\cdots\right)=\dfrac12\cdot\dfrac{1}{(1-\frac12)^2}=\dfrac12\cdot4=2$. Check: the partial sums $0.5,\ 1,\ 1.375,\ 1.625,\ 1.78,\dots$ creep up towards $2$ ✓.

Exercises

I
Terms of a GP.
  • 1 Find the $20$th term of the GP $\tfrac52,\tfrac54,\tfrac58,\dots$ .
  • 2 The $8$th term of a GP is $192$ and its common ratio is $2$. Find its $12$th term.
  • 3 Which term of the GP $2,8,32,\dots$ is $131\,072$?
  • 4 Which term of the GP $\sqrt3,\ 3,\ 3\sqrt3,\dots$ is $729$?
  • 5 The $2$nd term of a GP is $6$ and the $5$th term is $162$. Find its $n$th term.
  • 6 Find $x$ if $-\tfrac27,\ x,\ -\tfrac72$ are in GP.
  • 7 The $4$th, $10$th and $16$th terms of a GP are $x$, $y$ and $z$. Show that $y^2=xz$.
  • 8 Three numbers in GP have sum $21$ and product $216$. Find them.
1 $a=\tfrac52$, $r=\tfrac12$: $a_{20}=\tfrac52\left(\tfrac12\right)^{19}=\dfrac{5}{2^{20}}=\dfrac{5}{1\,048\,576}$.
2 $a_{12}=a_8\,r^4=192\times16=3072$.
3 $2\cdot4^{n-1}=131\,072$, $4^{n-1}=65\,536=4^8$, $n=9$.
4 $a=r=\sqrt3$, so $a_n=(\sqrt3)^n$. $729=3^6=(\sqrt3)^{12}$, so $n=12$.
5 $r^3=\tfrac{162}{6}=27$, $r=3$, $a=2$: $a_n=2\cdot3^{n-1}$.
6 $x^2=\left(-\tfrac27\right)\left(-\tfrac72\right)=1$, so $x=1$ or $x=-1$ (both give a GP: ratios $-\tfrac72$ or $\tfrac72$).
7 $x=ar^3$, $y=ar^9$, $z=ar^{15}$: $xz=a^2r^{18}=(ar^9)^2=y^2$.
8 $\tfrac ar,a,ar$ with $a^3=216$, $a=6$. $\tfrac6r+6+6r=21$ gives $2r^2-5r+2=0$, $r=2$ or $\tfrac12$: the numbers are $3,6,12$.
II
Sums of $n$ terms.
  • 1 Find the sum of the first $20$ terms of $0.15+0.015+0.0015+\cdots$ .
  • 2 Find the sum of the first $n$ terms of $\sqrt7+\sqrt{21}+3\sqrt7+\cdots$ .
  • 3 Find the sum of the first $n$ terms of $1-a+a^2-a^3+\cdots$ ($a\ne-1$).
  • 4 Evaluate $\displaystyle\sum_{k=1}^{11}\left(2+3^k\right)$.
  • 5 How many terms of $1+2+4+\cdots$ give the sum $1023$?
  • 6 A GP has first term $2$, last term $486$ and common ratio $3$. Find the number of terms and their sum.
  • 7 The sum of the first three terms of a GP is $16$ and the sum of the next three terms is $128$. Find the first term, the common ratio and $S_n$.
  • 8 For a GP with $r\ne1$, show that $\dfrac{S_{2n}-S_n}{S_n}=r^n$. Interpret the numerator.
  • 9 Find the sum of the first $n$ terms of $5+55+555+\cdots$ .
Check: a GP's sum by adding its terms versus the formula
last term $ar^{n-1}$—
adding the terms—
formula $\tfrac{a(r^n-1)}{r-1}$—
1 $a=0.15$, $r=0.1$: $S_{20}=\dfrac{0.15\,(1-0.1^{20})}{0.9}=\dfrac16\left(1-10^{-20}\right)$.
2 $r=\sqrt3$: $S_n=\dfrac{\sqrt7\left(3^{n/2}-1\right)}{\sqrt3-1}=\dfrac{\sqrt7\,(\sqrt3+1)\left(3^{n/2}-1\right)}{2}$.
3 $r=-a$: $S_n=\dfrac{1-(-a)^n}{1+a}$.
4 $\displaystyle\sum2=22$ and $\displaystyle\sum_{k=1}^{11}3^k=\frac{3(3^{11}-1)}{2}=265\,719$. Total $265\,741$.
5 $2^n-1=1023$, $2^n=1024$, $n=10$.
6 $2\cdot3^{n-1}=486$, $3^{n-1}=243=3^5$, $n=6$. $S=\dfrac{lr-a}{r-1}=\dfrac{1458-2}{2}=728$.
7 $a(1+r+r^2)=16$ and $ar^3(1+r+r^2)=128$; dividing, $r^3=8$, $r=2$, and $7a=16$, $a=\tfrac{16}{7}$. $S_n=\tfrac{16}{7}(2^n-1)$.
8 $\dfrac{a(r^{2n}-1)-a(r^n-1)}{a(r^n-1)}=\dfrac{r^n(r^n-1)}{r^n-1}=r^n$. The numerator is the sum of terms $n+1$ to $2n$: each of them is $r^n$ times the matching term among the first $n$.
9 $5+55+555+\cdots=\tfrac59(9+99+999+\cdots)=\tfrac59\left[(10-1)+(100-1)+\cdots\right]=\tfrac59\left[\dfrac{10(10^n-1)}{9}-n\right]=\dfrac{50(10^n-1)}{81}-\dfrac{5n}{9}$. (Check $n=2$: $\tfrac{4950}{81}-\tfrac{10}{9}=60$ ✓.)
III
Infinite GPs.
  • 1 Find the sum to infinity of $6+1.2+0.24+\cdots$ .
  • 2 Find the sum to infinity of $5-\tfrac{20}{7}+\tfrac{80}{49}-\cdots$ .
  • 3 Evaluate $\displaystyle\sum_{k=1}^{\infty}\left(-\tfrac13\right)^k$.
  • 4 Write $0.\overline{36}$ as a fraction in lowest terms.
  • 5 Write $1.2\overline{3}$ as a fraction in lowest terms.
  • 6 For which values of $x$ does $1+2x+4x^2+8x^3+\cdots$ have a sum to infinity? Find the sum.
  • 7 The sum to infinity of a GP is $4$ and its first term is $3$. Find its first three terms.
  • 8 The sum to infinity of a GP is $15$ and the sum to infinity of the squares of its terms is $45$. Find the GP.
1 $r=0.2$: $\dfrac{6}{0.8}=7.5$.
2 $r=-\tfrac47$: $\dfrac{5}{1+4/7}=\dfrac{35}{11}$.
3 $a=r=-\tfrac13$: $\dfrac{-1/3}{4/3}=-\tfrac14$.
4 $\dfrac{0.36}{0.99}=\dfrac{36}{99}=\dfrac{4}{11}$.
5 $1.2+\dfrac{0.03}{0.9}=\dfrac65+\dfrac1{30}=\dfrac{37}{30}$.
6 $r=2x$; a sum exists when $-1\lt2x\lt1$, i.e. $-\tfrac12\lt x\lt\tfrac12$, and then it is $\dfrac{1}{1-2x}$.
7 $\dfrac{3}{1-r}=4$ gives $r=\tfrac14$: $3,\ \tfrac34,\ \tfrac3{16}$.
8 $\dfrac{a}{1-r}=15$ and the squares form a GP with ratio $r^2$: $\dfrac{a^2}{1-r^2}=45$. Dividing, $\dfrac{a}{1+r}=3$. So $15(1-r)=3(1+r)$, $r=\tfrac23$, $a=5$: the GP $5,\ \tfrac{10}{3},\ \tfrac{20}{9},\dots$ .
IV
Applications.
  • 1 ₹50 000 is invested at $6\%$ per year, compounded annually. Find the amount after $3$ years.
  • 2 A car bought for ₹8 00 000 loses $15\%$ of its value every year. Find its value after $4$ years.
  • 3 A colony of bacteria doubles every $30$ minutes. It starts with $500$ bacteria. How many are there after $4$ hours?
  • 4 A ball is dropped from $10\ \text{m}$ and always rebounds to $\tfrac45$ of the height it fell from. Find the total distance travelled before it comes to rest.
  • 5 A town of $2\,00\,000$ people grows by $2\%$ a year. Estimate its population after $3$ years.
  • 6 On day $1$ a message is sent to $3$ people. Each person who receives it forwards it the next day to $3$ people who have not had it. How many people have received the message by the end of day $6$?
  • 7 A patient takes a $100\ \text{mg}$ tablet every day; by the time of the next tablet, $40\%$ of the medicine in the body remains. Find the amount in the body just after the $5$th tablet, and the amount just after a tablet in the long run.
1 $50\,000\times1.06^3=50\,000\times1.191016=\text{₹}\,59\,550.80$.
2 Ratio $0.85$: $8\,00\,000\times0.85^4=8\,00\,000\times0.52200625=\text{₹}\,4\,17\,605$.
3 $4$ hours $=8$ doublings: $500\times2^8=1\,28\,000$ bacteria.
4 $10+2\cdot\dfrac{8}{1-4/5}=10+2\cdot40=90\ \text{m}$.
5 $2\,00\,000\times1.02^3=2\,12\,241.6\approx2\,12\,242$ people.
6 New receivers each day: $3,9,27,\dots,3^6$. Total $\dfrac{3(3^6-1)}{2}=1092$ people.
7 Just after the $n$th tablet: $100+40+16+\cdots=100\,\dfrac{1-0.4^n}{0.6}$ mg. For $n=5$: $\dfrac{100(1-0.01024)}{0.6}\approx164.96\ \text{mg}$. In the long run: $\dfrac{100}{1-0.4}\approx166.7\ \text{mg}$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for a term or the number of terms (1–2 marks), sums such as $0.7+0.77+\cdots$ or $5+55+\cdots$ (3 marks), recurring decimals and sums to infinity, and case studies on bouncing balls, interest and growth. JEE combines two conditions on $S_\infty$ and uses arithmetico-geometric sums.

E1. Board · 1 mark The $6$th term of the GP $2,-6,18,\dots$ is (A) $486$   (B) $-486$   (C) $-1458$   (D) $1458$

Idea $r=-3$: $a_6=2(-3)^5=2(-243)=-486$.
Trap (D) is $2\cdot3^6$ — using $r^n$ instead of $r^{n-1}$ and dropping the sign; (A) drops only the sign.
Answer (B).

E2. Board · 2 marks Express $2.\overline{45}$ as a fraction $\tfrac pq$ using a geometric series.

Series $2.\overline{45}=2+0.45+0.0045+\cdots$; after the $2$, a GP with $a=0.45$, $r=0.01$. (1 mark)
Answer $2+\dfrac{0.45}{0.99}=2+\dfrac{5}{11}=\dfrac{27}{11}$. (1 mark)

E3. Board · 3 marks Find the sum of the first $n$ terms of $0.7+0.77+0.777+\cdots$ .

Rewrite $S=\tfrac79(0.9+0.99+0.999+\cdots)=\tfrac79\left[(1-0.1)+(1-0.01)+(1-0.001)+\cdots\right]$. (1 mark)
GP $=\tfrac79\left[n-(0.1+0.01+\cdots+0.1^n)\right]=\tfrac79\left[n-\dfrac{0.1(1-0.1^n)}{0.9}\right]$. (1 mark)
Answer $S=\tfrac79\left[n-\tfrac19\left(1-10^{-n}\right)\right]=\dfrac{7}{81}\left(9n-1+10^{-n}\right)$. Check $n=1$: $\tfrac{7}{81}(9-1+0.1)=0.7$ ✓. (1 mark)

E4. Board · Case study In a sports-science lab a ball is dropped from a height of $81\ \text{m}$ (from a tower) onto a hard floor. Each time it rebounds to $\tfrac23$ of the height it fell from. (i) How high does it rise after the third bounce? (1 mark) (ii) What total distance has it travelled when it hits the floor for the fourth time? (1 mark) (iii) What total distance does it travel before coming to rest? (2 marks)

(i) Rebound heights $54,36,24,\dots$: after the third bounce $81\left(\tfrac23\right)^3=24\ \text{m}$. (1 mark)
(ii) Down $81$, then up and down $54$, $36$, $24$: $81+2(54+36+24)=309\ \text{m}$. (1 mark)
(iii) set-up $81+2\,(54+36+24+\cdots)$ with an infinite GP, $a=54$, $r=\tfrac23$. (1 mark)
Answer $81+2\cdot\dfrac{54}{1/3}=81+324=405\ \text{m}$. (1 mark)

E5. JEE Main The sum to infinity of a GP is $3$ and the sum to infinity of the cubes of its terms is $\tfrac{27}{19}$. The common ratio is (A) $\tfrac13$   (B) $\tfrac23$   (C) $\tfrac32$   (D) $\tfrac12$

Idea $a=3(1-r)$, and the cubes form a GP with ratio $r^3$: $\dfrac{a^3}{1-r^3}=\dfrac{27}{19}$.
Solve $\dfrac{27(1-r)^3}{(1-r)(1+r+r^2)}=\dfrac{27}{19}$, so $19(1-r)^2=1+r+r^2$, i.e. $18r^2-39r+18=0$, $6r^2-13r+6=0$, $(3r-2)(2r-3)=0$.
Trap $r=\tfrac32$ (C) solves the quadratic but has $|r|\gt1$, so there is no sum to infinity — reject it.
Answer (B) $r=\tfrac23$ (and then $a=1$: check $\tfrac{1}{1-8/27}=\tfrac{27}{19}$ ✓).

E6. JEE Advanced Numerical answer. Find the value of $\displaystyle\sum_{n=1}^{\infty}\frac{2n-1}{2^n}$.

Idea This is arithmetico-geometric (🚀 box). Let $S=\tfrac12+\tfrac34+\tfrac58+\tfrac7{16}+\cdots$. Then $\tfrac12S=\tfrac14+\tfrac38+\tfrac5{16}+\cdots$.
Subtract $S-\tfrac12S=\tfrac12+\left(\tfrac24+\tfrac28+\tfrac2{16}+\cdots\right)=\tfrac12+\dfrac{1/2}{1-1/2}=\tfrac12+1=\tfrac32$.
Answer $S=3$.
⏱ Fast route $\sum\tfrac{2n}{2^n}-\sum\tfrac1{2^n}=2\cdot2-1=3$, using $\sum\tfrac{n}{2^n}=2$ from the 🚀 box.
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Do the rebound heights of a real bouncing ball form a geometric progression — and how far does the ball really travel before it stops?

Why it matters. Sports engineers rate balls by how well they bounce: cricket, tennis and table-tennis balls must pass rebound tests before they are approved for matches. If each rebound is a fixed fraction $r$ of the previous height, the heights form a GP and the total distance is a sum to infinity — a model you can test with a phone.

What to do.

  1. Tape a measuring tape (or marks every $10\ \text{cm}$) to a wall. Drop a ball from a measured height $h_0$ (about $1.5$–$2\ \text{m}$) next to it and film the bounces with your phone's slow-motion camera.
  2. Step through the video and read off the top of the first $4$–$5$ rebounds: $h_1,h_2,\dots$ . Repeat three times and average.
  3. Compute the ratios $\tfrac{h_1}{h_0},\tfrac{h_2}{h_1},\dots$ . Are they roughly constant? Take their mean as $r$.
  4. Predict the total distance $h_0+\dfrac{2h_0r}{1-r}$, and compare different balls (tennis, table tennis, rubber, cricket ball on grass).

Tools: A phone with slow motion, a measuring tape, two or three different balls, a spreadsheet

Hand in: Your table of heights and ratios, a graph of $h_n$ against $n$, your value of $r$ for each ball, the predicted total distance, and a paragraph on where the GP model fails (look at the last, tiny bounces).

Going further. If each bounce takes time proportional to $\sqrt{h_n}$, the bounce times form a GP with ratio $\sqrt r$. Does the ball stop bouncing in a finite time? Use your data to estimate it.

Instructor note: typical ratios on a hard floor are about $0.5$–$0.6$ for a tennis ball, $0.7$–$0.8$ for a table-tennis ball and $0.6$–$0.75$ for a basketball. Ratios usually drift down slightly for very small bounces (energy losses are not exactly proportional), which is a good discussion point. Since the bounce times form a GP with ratio $\sqrt r\lt1$, the total time is finite — the ball stops after a finite time despite "infinitely many" bounces.

🧪 Self-check quiz

0/5 correct

Q1. The common ratio of the GP $3,-6,12,-24,\dots$ is

Q2. A GP has first term $5$ and common ratio $2$. Its $6$th term is

Q3. The sum $1+3+9+27+81$ equals

Q4. The sum to infinity of $1-\tfrac12+\tfrac14-\tfrac18+\cdots$ is

Q5. Which of these infinite series has a sum?

← 8.1 Sequences and Series8.3 Arithmetic and Geometric Means →