Intuition The big pictureA geometric progression (GP) keeps multiplying by a fixed ratio r r r smaller than the last (in size), the terms shrink toward zero and the running total settles on a finite number. If terms stay the same size or grow, the total races off to infinity and there is no finite sum.
WHY it matters: infinite GPs appear everywhere — repeating decimals (0. 3 ‾ 0.\overline{3} 0. 3
A sequence a , a r , a r 2 , a r 3 , … a,\ ar,\ ar^2,\ ar^3,\dots a , a r , a r 2 , a r 3 , … first term a a a common ratio r = any term previous term r=\dfrac{\text{any term}}{\text{previous term}} r = previous term any term S ∞ = a + a r + a r 2 + ⋯ = ∑ n = 0 ∞ a r n . S_\infty = a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n. S ∞ = a + a r + a r 2 + ⋯ = ∑ n = 0 ∞ a r n .
The key word is if it exists . To make sense of an infinite sum, we watch the partial sums S N S_N S N N N N N → ∞ N\to\infty N → ∞
WHAT we want: a closed formula for S N = a + a r + ⋯ + a r N − 1 S_N = a + ar + \cdots + ar^{N-1} S N = a + a r + ⋯ + a r N − 1
HOW — the shift-and-subtract trick.
S N = a + a r + a r 2 + ⋯ + a r N − 1 S_N = a + ar + ar^2 + \cdots + ar^{N-1} S N = a + a r + a r 2 + ⋯ + a r N − 1
Multiply every term by r r r
r S N = a r + a r 2 + ⋯ + a r N − 1 + a r N rS_N = \quad\ ar + ar^2 + \cdots + ar^{N-1} + ar^{N} r S N = a r + a r 2 + ⋯ + a r N − 1 + a r N
Why this step? Multiplying by r r r
Subtract:
S N − r S N = a − a r N S_N - rS_N = a - ar^{N} S N − r S N = a − a r N
Why this step? Every middle term appears in both lines and cancels; only the very first (a a a a r N ar^N a r N
Factor and solve (valid when r ≠ 1 r\ne 1 r = 1
S N ( 1 − r ) = a ( 1 − r N ) ⟹ S N = a ( 1 − r N ) 1 − r S_N(1-r) = a(1-r^N) \implies \boxed{\,S_N = \dfrac{a(1-r^N)}{1-r}\,} S N ( 1 − r ) = a ( 1 − r N ) ⟹ S N = 1 − r a ( 1 − r N )
Now let N → ∞ N\to\infty N → ∞ only part that depends on N N N r N r^N r N
r N r^N r N
If ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 ⇒ r N → 0 \Rightarrow r^N \to 0 ⇒ r N → 0 ( 1 / 2 ) N (1/2)^N ( 1/2 ) N 0.5 , 0.25 , 0.125 , … 0.5,0.25,0.125,\dots 0.5 , 0.25 , 0.125 , …
If ∣ r ∣ > 1 |r|>1 ∣ r ∣ > 1 ∣ r N ∣ → ∞ |r^N|\to\infty ∣ r N ∣ → ∞
If r = 1 r=1 r = 1 a + a + a + ⋯ → ± ∞ a+a+a+\dots \to \pm\infty a + a + a + ⋯ → ± ∞
If r = − 1 r=-1 r = − 1 a , 0 , a , 0 , … a,0,a,0,\dots a , 0 , a , 0 , …
So the limit exists iff = = ∣ r ∣ < 1 = = ==|r|<1== == ∣ r ∣ < 1 == r N → 0 r^N\to 0 r N → 0
S ∞ = lim N → ∞ a ( 1 − r N ) 1 − r = a ( 1 − 0 ) 1 − r S_\infty = \lim_{N\to\infty}\frac{a(1-r^N)}{1-r} = \frac{a(1-0)}{1-r} S ∞ = lim N → ∞ 1 − r a ( 1 − r N ) = 1 − r a ( 1 − 0 )
Worked example Example 2 — repeating decimal0. 7 ‾ 0.\overline{7} 0. 7 0.7777 … = 0.7 + 0.07 + 0.007 + ⋯ 0.7777\ldots = 0.7 + 0.07 + 0.007 + \cdots 0.7777 … = 0.7 + 0.07 + 0.007 + ⋯
a = 0.7 a=0.7 a = 0.7 r = 0.1 r=0.1 r = 0.1 Why? Each block shifts one decimal place, i.e. × 0.1 \times 0.1 × 0.1 ∣ 0.1 ∣ < 1 |0.1|<1 ∣0.1∣ < 1 S ∞ = 0.7 1 − 0.1 = 0.7 0.9 = 7 9 . S_\infty = \dfrac{0.7}{1-0.1} = \dfrac{0.7}{0.9} = \dfrac{7}{9}. S ∞ = 1 − 0.1 0.7 = 0.9 0.7 = 9 7 . 7 / 9 7/9 7/9
a 1 − r \frac{a}{1-r} 1 − r a ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 Why it feels right: the formula "always works" algebraically and even gives a number.
Why it's wrong: the derivation dropped r N r^N r N only because r N → 0 r^N\to0 r N → 0 ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 ∣ r ∣ ≥ 1 |r|\ge1 ∣ r ∣ ≥ 1 r N r^N r N 1 + 2 + 4 + ⋯ = − 1 1+2+4+\cdots=-1 1 + 2 + 4 + ⋯ = − 1 Fix: always state "∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 before using a 1 − r \frac{a}{1-r} 1 − r a
Common mistake Confusing first term indexFeels right: you start counting from a r ar a r a a a Fix: a a a 6 + 2 + 2 3 + ⋯ 6+2+\tfrac23+\cdots 6 + 2 + 3 2 + ⋯ a = 6 a=6 a = 6 2 2 2
r r r Feels right: dividing previous ÷ next.
Fix: r = next term current term r=\dfrac{\text{next term}}{\text{current term}} r = current term next term
Recall Test yourself (hide answers)
Q: What single condition guarantees an infinite GP has a finite sum?
A: ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 Q: Why does the formula collapse to a 1 − r \frac{a}{1-r} 1 − r a A: Because r N → 0 r^N\to 0 r N → 0 S N = a ( 1 − r N ) 1 − r S_N=\frac{a(1-r^N)}{1-r} S N = 1 − r a ( 1 − r N ) Q: What is ∑ n = 0 ∞ ( 1 / 3 ) n \sum_{n=0}^\infty (1/3)^n ∑ n = 0 ∞ ( 1/3 ) n A: 1 1 − 1 / 3 = 3 2 \frac{1}{1-1/3}=\frac32 1 − 1/3 1 = 2 3 Q: Does 1 − 1 + 1 − 1 + ⋯ 1-1+1-1+\cdots 1 − 1 + 1 − 1 + ⋯ A: No; r = − 1 r=-1 r = − 1 1 , 0 , 1 , 0 1,0,1,0 1 , 0 , 1 , 0
Recall Feynman: explain to a 12-year-oldImagine a chocolate bar. You eat half, then half of what's left, then half of that... You keep eating smaller and smaller pieces forever. How much chocolate do you eat in total? The whole bar — exactly 1. Even though there are infinitely many bites, they get so tiny that everything adds up to a finite amount. That only works because each bite is smaller than the last (ratio < 1 <1 < 1 bigger , you'd never stop and there'd be no total.
"Shrink to sink." Terms must shrink (∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 "a over (1 minus r)" — A-hOMinus-R , "a home minus r."
Geometric Progression — nth term Sum of finite GP Limits of Sequences (why r N → 0 r^N\to0 r N → 0 Convergence of Series Recurring Decimals as Fractions Present Value & Perpetuities (finance application)
The infinite GP ∑ a r n \sum ar^n ∑ a r n Sum of convergent infinite GP S ∞ = a 1 − r S_\infty=\dfrac{a}{1-r} S ∞ = 1 − r a for
∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 Closed form for partial sum S N S_N S N S N = a ( 1 − r N ) 1 − r S_N=\dfrac{a(1-r^N)}{1-r} S N = 1 − r a ( 1 − r N ) (for
r ≠ 1 r\ne1 r = 1 )
Trick used to derive S N S_N S N Multiply by
r r r , shift, and subtract so middle terms cancel
Why S ∞ = a 1 − r S_\infty=\frac{a}{1-r} S ∞ = 1 − r a ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 Only then
r N → 0 r^N\to0 r N → 0 , killing the
r N r^N r N term
Value of 0. 3 ‾ 0.\overline{3} 0. 3 0.3 1 − 0.1 = 1 3 \frac{0.3}{1-0.1}=\frac13 1 − 0.1 0.3 = 3 1 Sum 2 + 1 + 1 2 + ⋯ 2+1+\frac12+\cdots 2 + 1 + 2 1 + ⋯ Does 1 + 2 + 4 + 8 + ⋯ 1+2+4+8+\cdots 1 + 2 + 4 + 8 + ⋯ No,
r = 2 ≥ 1 r=2\ge1 r = 2 ≥ 1 , it diverges
What happens to partial sums when r = − 1 r=-1 r = − 1 They oscillate and never settle
S_N times 1-r equals a times 1-r^N
S_N equals a times 1-r^N over 1-r
S_infinity equals a over 1-r
Repeating decimals like 0.777...
Intuition Hinglish mein samjhoDekho, infinite GP ka matlab hai ek aisi list jisme har agla term pichhle term ko ek fixed number r r r a , a r , a r 2 , … a, ar, ar^2, \dots a , a r , a r 2 , … ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1
Formula nikaalne ka jugaad simple hai. Pehle N N N S N S_N S N r r r S N = a ( 1 − r N ) 1 − r S_N=\frac{a(1-r^N)}{1-r} S N = 1 − r a ( 1 − r N ) N N N ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 r N → 0 r^N \to 0 r N → 0 0.5 , 0.25 , 0.125 , … 0.5,0.25,0.125,\dots 0.5 , 0.25 , 0.125 , … S ∞ = a 1 − r S_\infty=\frac{a}{1-r} S ∞ = 1 − r a
Sabse important baat: ye formula use karne se pehle hamesha check karo ki ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 r = 2 r=2 r = 2 1 1 − 2 = − 1 \frac{1}{1-2}=-1 1 − 2 1 = − 1
Real life me ye kaam ka hai: 0. 3 ‾ 0.\overline{3} 0. 3 = 1 / 3 =1/3 = 1/3 a 1 − r \frac{a}{1-r} 1 − r a
