3.3.6
Ek AGP ka term aisa dikhta hai ( arithmetic part ) × ( geometric part ) 1 ⋅ 1 + 2 ⋅ x + 3 ⋅ x 2 + … 1 , 2 , 3 , … 1 , x , x 2 , … TRICK KYU KAAM KARTI HAI: ek pure GP ko sum karna aasaan hai kyunki ratio r shift hoti hai aur almost sab kuch cancel ho jaata hai. AGP mein multiply-and-shift almost cancel karta hai — lekin sliding arithmetic coefficient ek bacha hua pure GP chhod jaata hai, jise hum pehle se sum karna jaante hain. Toh hum mushkil series ko aasaan series mein reduce karke solve karte hain.
Definition Arithmetic-Geometric ProgressionEk aisi series jiska n AP aur ek GP ke n T n = AP part [ a + ( n − 1 ) d ] ⋅ GP part [ b r n − 1 ] b a , d S n = ∑ k = 1 n ( a + ( k − 1 ) d ) r k − 1 . == a == == d == == r ==
KAISE — derivation scratch se:
Har term ko r rS_n = \quad\;\; ar + (a+d)r^2 + \cdots + \big(a+(n-2)d\big)r^{n-1} + \big(a+(n-1)d\big)r^{n}. \tag{2}
Ye step kyun? r k k + 1 a + k d difference d
(1) mein se (2) subtract karo. r S n − r S n = a + [ d r + d r 2 + ⋯ + d r n − 1 ] − ( a + ( n − 1 ) d ) r n .
Ye step kyun? r 1 r n − 1 ( a + k d ) − ( a + ( k − 1 ) d ) = d a ( a + ( n − 1 ) d ) r n
Beech wala bracket ek pure GP hai jiska pehla term d r r n − 1 d r + d r 2 + ⋯ + d r n − 1 = 1 − r d r ( 1 − r n − 1 ) .
Toh
( 1 − r ) S n = a + 1 − r d r ( 1 − r n − 1 ) − ( a + ( n − 1 ) d ) r n .
Intuition Limit tab hi exist karta hai jab∣ r ∣ < 1 r n − 1 ∣ r ∣ < 1 r 0 ( a + ( n − 1 ) d )
KAISE: Boxed formula mein n → ∞ ∣ r ∣ < 1 r n → 0 r n − 1 → 0 ( n − 1 ) r n → 0
Worked example Example 1 — infinite sumS = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ ∣ x ∣ < 1 a = 1 , d = 1 , r = x S ∞ = 1 − x 1 + ( 1 − x ) 2 1 ⋅ x = ( 1 − x ) 2 ( 1 − x ) + x = ( 1 − x ) 2 1 . Ye step kyun? Hum directly infinite formula mein plug karte hain; do fractions ( 1 − x ) 2 ( 1 − x ) 2 1 = ∑ n ≥ 1 n x n − 1
Worked example Example 2 — famous numeric seriesS = n = 1 ∑ ∞ 2 n − 1 n = 1 + 2 2 + 4 3 + 8 4 + ⋯ a = 1 , d = 1 , r = 2 1 S = 1 − 2 1 1 + ( 1 − 2 1 ) 2 1 ⋅ 2 1 = 2 + 4 1 2 1 = 2 + 2 = 4. Ye step kyun? ∣ r ∣ = 2 1 < 1 1 , 2 , 2.75 , 3.25 , 3.625 , ⋯ → 4
Worked example Example 3 — finite sum method se (best exam habit)S 3 = 1 ⋅ 1 + 2 ⋅ 3 + 3 ⋅ 9 a = 1 , d = 1 , r = 3 , n = 3 1 + 6 + 27 = 34 S 3 = 1 + 2 ⋅ 3 + 3 ⋅ 9 3 S 3 = 1 ⋅ 3 + 2 ⋅ 9 + 3 ⋅ 27 S 3 − 3 S 3 = 1 + ( 3 ⋅ 9 − 2 ⋅ 9 ) + … S 3 = 1 − 3 1 + ( 1 − 3 ) 2 1 ⋅ 3 ( 1 − 3 2 ) − 1 − 3 ( 1 + 2 ⋅ 1 ) 3 3 . = − 2 1 + 4 3 ( 1 − 9 ) − − 2 3 ⋅ 27 = − 0.5 − 6 + 40.5 = 34. Ye step kyun? Ye confirm karta hai ki finite formula tab bhi kaam karta hai jab r > 1 infinite wale ko ∣ r ∣ < 1
Common mistake "Kisi bhi AGP ke liye1 − r a + ( 1 − r ) 2 d r Kyun sahi lagta hai: ye woh clean formula hai jo sabko yaad hai. Trap: ye infinite sum hai aur sirf tab valid hai jab ∣ r ∣ < 1 r = 3 Fix: finite n ∣ r ∣ ≥ 1 S n
Common mistake Leftover GP terms galat count karna.Kyun sahi lagta hai: tum expect karte ho bracket mein n Trap: subtraction ke baad pure-GP bracket mein n − 1 d r d r n − 1 a Fix: subtraction column-by-column likhkar physically dekho kaun se terms survive karte hain.
Common mistake Denominator mein( 1 − r ) 2 Kyun sahi lagta hai: GP sums mein sirf ek ( 1 − r ) Trap: arithmetic part ko ( 1 − r ) do baar divide hota hai — ek baar GP sum karne se, ek baar ( 1 − r ) Fix: yaad rakho: arithmetic tilt ⇒ squared denominator.
Recall Feynman: 12-saal ke bachche ko samjhaoSocho ek staircase hai jahan har step bhi aadhi hoti jaati hai . Step number (1, 2, 3, …) badhta rehta hai, lekin har step itni choti hoti jaati hai ki poori staircase ki total length finite hoti hai. Magic trick ye hai: staircase ki ek copy banao jo ek step shift ho, usse upar rakh do, aur almost sab kuch match hokar cancel ho jaata hai. Jo bachta hai wo ek simple shrinking pile hoti hai jise tum turant add kar sakte ho.
A over one-minus-r, plus tilt over the square. "
1 − r a + ( 1 − r ) 2 d r tilt , denominator squared ). Aur "|r|<1 or you can't win" — infinite AGP ko ∣ r ∣ < 1
Recall Pehle answers cover karo aur forecast karo!
AGP ka standard n ( a + ( n − 1 ) d ) r n − 1
AGP sum karne ki key manipulation? ::: Poori series ko r
Infinite AGP sum exist karne ki condition? ::: ∣ r ∣ < 1
Infinite AGP sum formula? ::: S ∞ = 1 − r a + ( 1 − r ) 2 d r
Arithmetic-geometric progression ko kya define karta hai? Har term = (AP term)×(GP term), yaani T n = ( a + ( n − 1 ) d ) r n − 1
Kaun si operation ek AGP ko known GP mein reduce karti hai? Ratio r d
Subtraction ke baad leftover GP mein kitne terms hote hain? n − 1 d r d r n − 1
AGP ka infinite sum aur uski condition? S ∞ = 1 − r a + ( 1 − r ) 2 d r ∣ r ∣ < 1
∑ n ≥ 1 n / 2 n − 1 4 a = 1 , d = 1 , r = 1/2
Arithmetic part ke liye denominator ( 1 − r ) 2 GP-sum ek ( 1 − r ) S n
∑ n ≥ 1 n x n − 1 ∣ x ∣ < 1 ( 1 − x ) 2 1
let n to infinity, r under 1
Arithmetic tilt correction
