Geometric progression (GP) — nth term, sum of n terms — derivations

1. Definition

Why $a\neq 0$ and $r\neq 0$ ? If $a=0$ every term is $0$ (a degenerate, boring sequence). If $r=0$ , after the first term everything is $0$ and dividing by a term becomes illegal.

2. The nth term — derived from scratch

Derivation (Feynman style, one multiplication at a time):

\begin{aligned} a_1 &= a\\ a_2 &= a_1 \cdot r = a r\\ a_3 &= a_2 \cdot r = a r^2\\ a_4 &= a_3 \cdot r = a r^3\\ &\;\;\vdots\\ a_n &= a\, r^{\,n-1} \end{aligned}

Why the exponent is $n-1$ not $n$ ? The first term uses zero multiplications ( $r^0=1$ ). Count the arrows, not the terms.

3. Sum of n terms — the "shift and subtract" trick

Derivation:

Let $S_n = a + ar + ar^2 + \cdots + ar^{\,n-1}. \tag{1}$

Multiply every term by $r$ : $rS_n = ar + ar^2 + ar^3 + \cdots + ar^{\,n}. \tag{2}$

Why multiply by $r$ ? It reproduces the same terms shifted one place, so they'll cancel.

Subtract (2) from (1):

S_n - rS_n = a - ar^{\,n}

Why does the middle vanish? Every interior term $ar^k$ appears in both lines, so it cancels. Only the very first term of (1) and the very last of (2) survive.

Factor:

S_n(1-r) = a(1-r^{\,n})

Divide by $(1-r)$ provided $r\neq 1$ :

\boxed{\,S_n = \dfrac{a(1-r^{\,n})}{1-r} = \dfrac{a(r^{\,n}-1)}{r-1}\,}

Why two forms? They're algebraically identical (multiply top & bottom by $-1$ ). Pick the one that keeps things positive to avoid sign slips.

Figure — Geometric progression (GP) — nth term, sum of n terms — derivations

5. Common mistakes (Steel-man + fix)

Recall Feynman: explain to a 12-year-old

Imagine a magic photocopier that always makes copies double the size (or half, or triple). Start with a sticker of size 3. Feed it in: 6, then 12, then 24 — each time it multiplies by the same number. That's a GP! To know the size after some copies, you just multiply that magic number by itself once for each copy you made — one fewer than the number of stickers, because the first sticker got copied zero times. And if you want the total area of ALL the stickers, there's a shortcut: instead of adding them one by one, you copy the whole pile, line it up shifted by one, and subtract — almost everything cancels and only the first and last bits are left. Neat!

Connections

Arithmetic Progression (AP) — GP multiplies where AP adds; $S_n$ derived by reversing there, by shifting here.
Sum of infinite GP — when $|r|<1$ , $r^n\to 0$ so $S_\infty = \dfrac{a}{1-r}$ .
Geometric Mean — middle term of a 3-term GP: $b=\sqrt{ac}$ .
Exponential functions — GP is the discrete version of $a\,r^{x}$ .
Compound interest — amounts form a GP with $r = 1+\frac{i}{100}$ .
Logarithms — used to solve $r^{n-1}=k$ for $n$ .

Flashcards

What defines a GP?

Ratio of consecutive terms is constant:

a_{n+1}/a_n = r

nth term of a GP with first term

a

, ratio

r

a_n = a\,r^{\,n-1}

Why is the exponent

n-1

and not

n

The first term is multiplied by

r

zero times (

r^0=1

); there are only

(n-1)

gaps to term

n

Sum of first

n

terms of a GP (

r\neq1

S_n = \dfrac{a(1-r^n)}{1-r} = \dfrac{a(r^n-1)}{r-1}

What trick derives the GP sum?

Multiply

S_n

r

to shift terms, then subtract — interior terms cancel (telescoping).

What is

S_n

when

r=1

S_n = na

(all terms equal

a

; the formula fails since

1-r=0

How do you find the common ratio

r

Divide a term by the previous one, not subtract.

Sum

8+4+2+1

using the formula?

a=8,r=\tfrac12,n=4

\dfrac{8(1-1/16)}{1/2}=15

When does the infinite GP sum exist and equal what?

When

|r|<1

; then

S_\infty=\dfrac{a}{1-r}

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, GP ka matlab hai ek aisi sequence jahan har agla term pichhle term ko ek fixed number se multiply karke banta hai. Yeh number hota hai common ratio $r$ . Jaise AP mein hum add karte hain (common difference $d$ ), GP mein hum multiply karte hain. Example: $3, 6, 12, 24$ — yahan har baar $\times 2$ ho raha hai, toh $r=2$ .

$n$ -th term ka formula $a_n = a\,r^{n-1}$ hai. Yaad rakho power $n$ nahi, $n-1$ hai — kyunki pehla term ko $r$ se zero baar multiply karte hain ( $r^0=1$ ). Pehle term se $n$ -th term tak jaane mein sirf $(n-1)$ chhalang (gaps) hoti hain, isliye $n-1$ .

Sum nikalne ka jugaad bahut pyara hai: pura sum $S_n$ ko $r$ se multiply karo — isse saare terms ek jagah aage khisak jaate hain. Ab dono ko subtract karo, toh beech ke saare terms cancel ho jaate hain, sirf shuru aur end bachte hain. Isse milta hai $S_n = \frac{a(1-r^n)}{1-r}$ . Bas dhyan rakho: agar $r=1$ ho toh denominator zero ban jaata hai, tab seedha $S_n = na$ use karo.

Yeh cheez real life mein important hai — compound interest, population growth, aur decay sab GP se chalte hain, kyunki inme baar-baar multiply hota hai. Exam mein trick yeh hai: $r$ hamesha divide karke nikalo (subtract mat karo, woh AP wali aadat hai), aur term-1 pe formula ko test karke confirm karo ki aapki power sahi hai.

Geometric progression (GP) — nth term, sum of n terms — derivations

1. Definition

2. The nth term — derived from scratch

3. Sum of n terms — the "shift and subtract" trick

4. Worked examples

5. Common mistakes (Steel-man + fix)

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Sequences & Series

Connections