3.3.3 · Maths › Sequences & Series
Ek geometric progression (GP) ek fixed ratio r se multiply hoti rehti hai. Agar har naya term pichle se chhota ho (size mein), toh terms zero ki taraf shrink karte hain aur running total ek finite number par settle ho jaata hai. Agar terms same size rehte hain ya badhte hain, toh total infinity ki taraf bhaag jaata hai aur koi finite sum hota hi nahi.
YE kyun matter karta hai: infinite GPs har jagah milte hain — repeating decimals (0. 3 ), fractal lengths, endless payments ki present value, aur "puri cheez sum karo" wale puzzles (aadha chalo, phir aadha aur...).
Ek sequence a , a r , a r 2 , a r 3 , … jisme pehla term a aur common ratio r = pichla term koi bhi term hota hai.
Iska infinite sum (agar exist karta hai) yeh hai:
S ∞ = a + a r + a r 2 + ⋯ = ∑ n = 0 ∞ a r n .
Key word hai agar exist karta hai . Ek infinite sum ko samajhne ke liye, hum partial sums S N (pehle N terms ka sum) dekhte hain aur poochte hain: kya yeh ek fixed number ki taraf jaate hain jab N → ∞ ?
KYA chahiye: S N = a + a r + ⋯ + a r N − 1 ke liye ek closed formula.
KAISE — shift-and-subtract trick.
S N = a + a r + a r 2 + ⋯ + a r N − 1
Har term ko r se multiply karo:
r S N = a r + a r 2 + ⋯ + a r N − 1 + a r N
Yeh step kyun? r se multiply karne par poori list ek jagah right shift ho jaati hai, isliye almost sab terms original ke saath align ho jaate hain — yeh cancel ho jaayenge.
Subtract karo:
S N − r S N = a − a r N
Yeh step kyun? Har beech wala term dono lines mein aata hai aur cancel ho jaata hai; sirf pehla (a ) aur bahar nikla hua aakhri (a r N ) bachta hai.
Factor karo aur solve karo (r = 1 ke liye valid):
S N ( 1 − r ) = a ( 1 − r N ) ⟹ S N = 1 − r a ( 1 − r N )
Ab N → ∞ karne do. Sirf r N hi N par depend karta hai.
r N kya karta hai?
Agar ∣ r ∣ < 1 : baar baar 1 se chhoti cheez se multiply karna ⇒ r N → 0 . (Socho ( 1/2 ) N : 0.5 , 0.25 , 0.125 , … )
Agar ∣ r ∣ > 1 : yeh blow up ho jaata hai, ∣ r N ∣ → ∞ .
Agar r = 1 : sum hai a + a + a + ⋯ → ± ∞ .
Agar r = − 1 : partial sums a , 0 , a , 0 , … flip karte hain — kabhi settle nahi hote.
Toh limit exist karta hai iff == ∣ r ∣ < 1 == , aur tab r N → 0 :
S ∞ = lim N → ∞ 1 − r a ( 1 − r N ) = 1 − r a ( 1 − 0 )
Worked example Example 1 — ek saaf geometric sum
2 + 1 + 2 1 + 4 1 + ⋯ find karo
a = 2 , r = 2 1 . Kyun? Har term pichle ka aadha hai, isliye r = 1/2 .
∣ r ∣ = 1/2 < 1 → converges. Pehle check kyun? Agar ∣ r ∣ ≥ 1 toh formula bekaaar hai.
S ∞ = 1 − r a = 1 − 2 1 2 = 1/2 2 = 4.
Worked example Example 3 — negative ratio
3 − 1 + 3 1 − 9 1 + ⋯
a = 3 , r = − 3 1 . Negative kyun? Signs alternate karte hain ⇒ r < 0 .
∣ r ∣ = 1/3 < 1 → converges.
S ∞ = 1 − ( − 3 1 ) 3 = 4/3 3 = 4 9 .
Worked example Example 4 — divergent, koi answer nahi
1 + 2 + 4 + 8 + ⋯
r = 2 , ∣ r ∣ = 2 ≥ 1 → diverges . Koi finite sum nahi hai.
Steel-man trap: plug in karne par 1 − 2 1 = − 1 milta hai, par yeh bakwaas hai — formula yahan invalid hai.
1 − r a use karna bina ∣ r ∣ < 1 check kiye
Kyun sahi lagta hai: formula algebraically "hamesha kaam karta hai" aur ek number bhi deta hai.
Kyun galat hai: derivation mein r N isliye drop kiya kyunki r N → 0 , jo ∣ r ∣ < 1 ke liye chahiye. ∣ r ∣ ≥ 1 ke liye, r N vanish nahi hota, isliye closed form jhooth hai (jaise 1 + 2 + 4 + ⋯ = − 1 ??).
Fix: 1 − r a use karne se pehle hamesha likho "∣ r ∣ < 1 , isliye yeh converge karta hai."
Common mistake Pehle term ka index confuse karna
Sahi lagta hai: a r se count shuru karte ho ya leading a bhool jaate ho.
Fix: a literally woh pehla number hai jo tumhe actually dikhta hai. 6 + 2 + 3 2 + ⋯ mein a = 6 hai, 2 nahi.
r ulta lena
Sahi lagta hai: pichla ÷ agla divide karna.
Fix: r = current term agla term (baad wala over pehle wala).
Recall Test yourself (answers chhupao)
Q: Kaun si ek condition guarantee karti hai ki infinite GP ka finite sum hoga?
A: ∣ r ∣ < 1 .
Q: Formula 1 − r a tak kyun collapse hota hai?
A: Kyunki r N → 0 hota hai S N = 1 − r a ( 1 − r N ) mein.
Q: ∑ n = 0 ∞ ( 1/3 ) n kya hai?
A: 1 − 1/3 1 = 2 3 .
Q: Kya 1 − 1 + 1 − 1 + ⋯ converge karta hai?
A: Nahi; r = − 1 , partial sums 1 , 0 , 1 , 0 oscillate karte hain.
Recall Feynman: ek 12-saal ke bacche ko samjhao
Socho ek chocolate bar hai. Tum aadha khaate ho, phir jo bacha uska aadha, phir uska aadha... Tum hamesha ke liye chhote chhote tukde khaate rehte ho. Total mein kitni chocolate khaoge? Poora bar — exactly 1. Bhale hi infinitely many bites hain, lekin woh itne tiny ho jaate hain ki sab kuch ek finite amount mein add ho jaata hai. Yeh tabhi kaam karta hai jab har bite pichle se chhota ho (ratio < 1 ). Agar har bite bada hota, toh kabhi ruko hi nahi aur koi total banta hi nahi.
"Shrink to sink." Terms zaroor shrink hone chahiye (∣ r ∣ < 1 ) tabhi sum ek value par sink karega. Aur formula: "a over (1 minus r)" — A-hOMinus-R , "a home minus r."
Geometric Progression — nth term
Sum of finite GP
Limits of Sequences (kyun r N → 0 )
Convergence of Series
Recurring Decimals as Fractions
Present Value & Perpetuities (finance application)
The infinite GP ∑ a r n converges iff ∣ r ∣ < 1
Sum of convergent infinite GP S ∞ = 1 − r a for ∣ r ∣ < 1
Closed form for partial sum S N of a GP S N = 1 − r a ( 1 − r N ) (for r = 1 )
Trick used to derive S N r se multiply karo, shift karo, aur subtract karo taaki beech ke terms cancel ho jayein
Why S ∞ = 1 − r a needs ∣ r ∣ < 1 Tabhi r N → 0 hota hai, jo r N wale term ko khatam karta hai
Value of 0. 3 as GP sum 1 − 0.1 0.3 = 3 1
Sum 2 + 1 + 2 1 + ⋯ 4
Does 1 + 2 + 4 + 8 + ⋯ have a finite sum? Nahi, r = 2 ≥ 1 , yeh diverge karta hai
What happens to partial sums when r = − 1 ? Woh oscillate karte hain aur kabhi settle nahi hote
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...