Intuition The big pictureAn AGP term looks like ( arithmetic part ) × ( geometric part ) (\text{arithmetic part}) \times (\text{geometric part}) ( arithmetic part ) × ( geometric part ) 1 ⋅ 1 + 2 ⋅ x + 3 ⋅ x 2 + … 1\cdot 1 + 2\cdot x + 3\cdot x^2 + \dots 1 ⋅ 1 + 2 ⋅ x + 3 ⋅ x 2 + … 1 , 2 , 3 , … 1,2,3,\dots 1 , 2 , 3 , … 1 , x , x 2 , … 1,x,x^2,\dots 1 , x , x 2 , … WHY the trick works: a pure GP is easy to sum because multiplying by the ratio r r r shifts the series so almost everything cancels. In an AGP the multiply-and-shift almost cancels — but the sliding arithmetic coefficient leaves behind a leftover pure GP , which we already know how to sum. So we tame the hard series by reducing it to an easy one.
Definition Arithmetic-Geometric ProgressionA series whose n n n n n n AP and a GP :
T n = [ a + ( n − 1 ) d ] ⏟ AP part ⋅ [ b r n − 1 ] ⏟ GP part T_n = \underbrace{[a+(n-1)d]}_{\text{AP part}} \cdot \underbrace{[b\,r^{\,n-1}]}_{\text{GP part}} T n = AP part [ a + ( n − 1 ) d ] ⋅ GP part [ b r n − 1 ] b b b a , d a,d a , d S n = ∑ k = 1 n ( a + ( k − 1 ) d ) r k − 1 . S_n = \sum_{k=1}^{n} \big(a+(k-1)d\big) r^{\,k-1}. S n = ∑ k = 1 n ( a + ( k − 1 ) d ) r k − 1 . = = a = = ==a== == a == = = d = = ==d== == d == = = r = = ==r== == r ==
HOW — the derivation from scratch:
Multiply every term by r r 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}
Why this step? Multiplying by r r r k k k k + 1 k{+}1 k + 1 a + k d a+kd a + k d difference d d d
Subtract (2) from (1). Line up equal powers of r r r S n − r S n = a + [ d r + d r 2 + ⋯ + d r n − 1 ] − ( a + ( n − 1 ) d ) r n . S_n - rS_n = a + \big[dr + dr^2 + \cdots + dr^{n-1}\big] - \big(a+(n-1)d\big)r^{n}. S n − r S n = a + [ d r + d r 2 + ⋯ + d r n − 1 ] − ( a + ( n − 1 ) d ) r n .
Why this step? For r 1 r^1 r 1 r n − 1 r^{n-1} r n − 1 ( a + k d ) − ( a + ( k − 1 ) d ) = d (a+kd)-(a+(k-1)d)=d ( a + k d ) − ( a + ( k − 1 ) d ) = d a a a ( a + ( n − 1 ) d ) r n (a+(n-1)d)r^n ( a + ( n − 1 ) d ) r n
The middle bracket is a pure GP with first term d r dr d r r r r n − 1 n-1 n − 1 d r + d r 2 + ⋯ + d r n − 1 = d r ( 1 − r n − 1 ) 1 − r . dr + dr^2 + \cdots + dr^{n-1} = \frac{dr(1-r^{n-1})}{1-r}. d r + d r 2 + ⋯ + d r n − 1 = 1 − r d r ( 1 − r n − 1 ) .
So
( 1 − r ) S n = a + d r ( 1 − r n − 1 ) 1 − r − ( a + ( n − 1 ) d ) r n . (1-r)S_n = a + \frac{dr(1-r^{n-1})}{1-r} - \big(a+(n-1)d\big)r^{n}. ( 1 − r ) S n = a + 1 − r d r ( 1 − r n − 1 ) − ( a + ( n − 1 ) d ) r n .
Intuition Why a limit exists only when∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 r n − 1 r^{n-1} r n − 1 ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 r r r 0 0 0 ( a + ( n − 1 ) d ) (a+(n-1)d) ( a + ( n − 1 ) d )
HOW: In the boxed formula let n → ∞ n\to\infty n → ∞ ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 r n → 0 r^{n}\to 0 r n → 0 r n − 1 → 0 r^{n-1}\to 0 r n − 1 → 0 ( n − 1 ) r n → 0 (n-1)r^n\to 0 ( n − 1 ) r n → 0
Worked example Example 1 — infinite sumFind S = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ S = 1 + 2x + 3x^2 + 4x^3 + \cdots S = 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 a = 1 , d = 1 , r = x a=1,\ d=1,\ r=x a = 1 , d = 1 , r = x S ∞ = 1 1 − x + 1 ⋅ x ( 1 − x ) 2 = ( 1 − x ) + x ( 1 − x ) 2 = 1 ( 1 − x ) 2 . S_\infty = \frac{1}{1-x} + \frac{1\cdot x}{(1-x)^2} = \frac{(1-x)+x}{(1-x)^2}=\frac{1}{(1-x)^2}. S ∞ = 1 − x 1 + ( 1 − x ) 2 1 ⋅ x = ( 1 − x ) 2 ( 1 − x ) + x = ( 1 − x ) 2 1 . Why this step? We plug straight into the infinite formula; the two fractions combine over ( 1 − x ) 2 (1-x)^2 ( 1 − x ) 2 1 ( 1 − x ) 2 = ∑ n ≥ 1 n x n − 1 \dfrac{1}{(1-x)^2}=\sum_{n\ge1} n x^{n-1} ( 1 − x ) 2 1 = ∑ n ≥ 1 n x n − 1
Worked example Example 2 — famous numeric seriesEvaluate S = ∑ n = 1 ∞ n 2 n − 1 = 1 + 2 2 + 3 4 + 4 8 + ⋯ \displaystyle S = \sum_{n=1}^{\infty} \frac{n}{2^{n-1}} = 1 + \frac{2}{2} + \frac{3}{4} + \frac{4}{8}+\cdots S = n = 1 ∑ ∞ 2 n − 1 n = 1 + 2 2 + 4 3 + 8 4 + ⋯ a = 1 , d = 1 , r = 1 2 a=1,\ d=1,\ r=\tfrac12 a = 1 , d = 1 , r = 2 1 S = 1 1 − 1 2 + 1 ⋅ 1 2 ( 1 − 1 2 ) 2 = 2 + 1 2 1 4 = 2 + 2 = 4. S = \frac{1}{1-\frac12} + \frac{1\cdot\frac12}{(1-\frac12)^2} = 2 + \frac{\frac12}{\frac14} = 2 + 2 = 4. S = 1 − 2 1 1 + ( 1 − 2 1 ) 2 1 ⋅ 2 1 = 2 + 4 1 2 1 = 2 + 2 = 4. Why this step? ∣ r ∣ = 1 2 < 1 |r|=\tfrac12<1 ∣ r ∣ = 2 1 < 1 1 , 2 , 2.75 , 3.25 , 3.625 , ⋯ → 4 1,2,2.75,3.25,3.625,\dots \to 4 1 , 2 , 2.75 , 3.25 , 3.625 , ⋯ → 4
Worked example Example 3 — finite sum by the method itself (best exam habit)Sum S 3 = 1 ⋅ 1 + 2 ⋅ 3 + 3 ⋅ 9 S_3 = 1\cdot 1 + 2\cdot 3 + 3\cdot 9 S 3 = 1 ⋅ 1 + 2 ⋅ 3 + 3 ⋅ 9 a = 1 , d = 1 , r = 3 , n = 3 a=1,d=1,r=3,n=3 a = 1 , d = 1 , r = 3 , n = 3 1 + 6 + 27 = 34 1+6+27=34 1 + 6 + 27 = 34 S 3 = 1 + 2 ⋅ 3 + 3 ⋅ 9 S_3=1+2\cdot3+3\cdot9 S 3 = 1 + 2 ⋅ 3 + 3 ⋅ 9 3 S 3 = 1 ⋅ 3 + 2 ⋅ 9 + 3 ⋅ 27 \;3S_3=1\cdot3+2\cdot9+3\cdot27 3 S 3 = 1 ⋅ 3 + 2 ⋅ 9 + 3 ⋅ 27 S 3 − 3 S 3 = 1 + ( 3 ⋅ 9 − 2 ⋅ 9 ) + … S_3-3S_3 = 1 + (3\cdot9-2\cdot9)+\dots S 3 − 3 S 3 = 1 + ( 3 ⋅ 9 − 2 ⋅ 9 ) + … S 3 = 1 1 − 3 + 1 ⋅ 3 ( 1 − 3 2 ) ( 1 − 3 ) 2 − ( 1 + 2 ⋅ 1 ) 3 3 1 − 3 . S_3=\frac{1}{1-3}+\frac{1\cdot3(1-3^{2})}{(1-3)^2}-\frac{(1+2\cdot1)\,3^{3}}{1-3}. S 3 = 1 − 3 1 + ( 1 − 3 ) 2 1 ⋅ 3 ( 1 − 3 2 ) − 1 − 3 ( 1 + 2 ⋅ 1 ) 3 3 . = 1 − 2 + 3 ( 1 − 9 ) 4 − 3 ⋅ 27 − 2 = − 0.5 − 6 + 40.5 = 34. =\frac{1}{-2}+\frac{3(1-9)}{4}-\frac{3\cdot27}{-2}=-0.5 -6 +40.5 = 34. = − 2 1 + 4 3 ( 1 − 9 ) − − 2 3 ⋅ 27 = − 0.5 − 6 + 40.5 = 34. Why this step? Confirms the finite formula also works when r > 1 r>1 r > 1 infinite one needs ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1
a 1 − r + d r ( 1 − r ) 2 \frac{a}{1-r}+\frac{dr}{(1-r)^2} 1 − r a + ( 1 − r ) 2 d r Why it feels right: it's the clean formula everyone memorizes. The trap: it is the infinite sum and is valid only if ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 . With r = 3 r=3 r = 3 Fix: for finite n n n ∣ r ∣ ≥ 1 |r|\ge1 ∣ r ∣ ≥ 1 S n S_n S n
Common mistake Mis-counting the leftover GP terms.Why it feels right: you expect n n n The trap: after subtraction the pure-GP bracket has n − 1 n-1 n − 1 d r dr d r d r n − 1 dr^{n-1} d r n − 1 a a a Fix: write the subtraction column-by-column and physically see which terms survive.
Common mistake Forgetting( 1 − r ) 2 (1-r)^2 ( 1 − r ) 2 Why it feels right: GP sums have a single ( 1 − r ) (1-r) ( 1 − r ) The trap: the arithmetic part gets divided by ( 1 − r ) (1-r) ( 1 − r ) twice — once from summing the GP, once from the ( 1 − r ) (1-r) ( 1 − r ) Fix: remember: arithmetic tilt ⇒ squared denominator.
Recall Feynman: explain to a 12-year-oldImagine a staircase where each step is also getting shorter by half . The step number (1, 2, 3, …) keeps counting up, but each step is so much tinier that the whole staircase has a total, finite length. The magic trick: build a copy of the staircase shifted one step over, lay it on top, and almost everything matches up and cancels. What's left is a simple shrinking pile you can add up instantly.
A over one-minus-r, plus tilt over the square. "
a 1 − r \dfrac{a}{1-r} 1 − r a + d r ( 1 − r ) 2 \dfrac{dr}{(1-r)^2} ( 1 − r ) 2 d r tilt , denominator squared ). And "|r|<1 or you can't win" — infinite AGP needs ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1
Recall Cover the answers and forecast first!
Standard n n n ( a + ( n − 1 ) d ) r n − 1 \big(a+(n-1)d\big)r^{\,n-1} ( a + ( n − 1 ) d ) r n − 1
Key manipulation to sum an AGP? ::: Multiply the whole series by r r r
Condition for the infinite AGP sum to exist? ::: ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1
Infinite AGP sum formula? ::: S ∞ = a 1 − r + d r ( 1 − r ) 2 S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2} S ∞ = 1 − r a + ( 1 − r ) 2 d r
What defines an arithmetic-geometric progression? Each term = (AP term)×(GP term), i.e.
T n = ( a + ( n − 1 ) d ) r n − 1 T_n=(a+(n-1)d)r^{n-1} T n = ( a + ( n − 1 ) d ) r n − 1 .
What operation reduces an AGP to a known GP? Multiply by ratio
r r r and subtract, so consecutive AP coefficients differ by just
d d d , leaving a pure GP.
How many terms are in the leftover GP after subtraction? n − 1 n-1 n − 1 terms (from
d r dr d r to
d r n − 1 dr^{n-1} d r n − 1 ).
Infinite sum of AGP and its condition? S ∞ = a 1 − r + d r ( 1 − r ) 2 S_\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2} S ∞ = 1 − r a + ( 1 − r ) 2 d r , valid for
∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 .
Evaluate ∑ n ≥ 1 n / 2 n − 1 \sum_{n\ge1} n/2^{n-1} ∑ n ≥ 1 n / 2 n − 1 4 4 4 (with
a = 1 , d = 1 , r = 1 / 2 a=1,d=1,r=1/2 a = 1 , d = 1 , r = 1/2 ).
Why is the denominator ( 1 − r ) 2 (1-r)^2 ( 1 − r ) 2 The GP-sum gives one
( 1 − r ) (1-r) ( 1 − r ) and factoring
S n S_n S n gives another, so it's squared.
∑ n ≥ 1 n x n − 1 \sum_{n\ge1} n x^{n-1} ∑ n ≥ 1 n x n − 1 ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 1 ( 1 − x ) 2 \frac{1}{(1-x)^2} ( 1 − x ) 2 1 .
let n to infinity, r under 1
Arithmetic tilt correction
Intuition Hinglish mein samjhoDekho, AGP ka matlab hai har term do cheezon ka product: ek arithmetic part (jaise 1, 2, 3, … jo seedha count karta hai) aur ek geometric part (jaise 1 , r , r 2 , … 1, r, r^2, … 1 , r , r 2 , … r r r pure GP bach jaati hai, jise hum already jaante hain.
Main formula (infinite ke liye) yaad rakho: S ∞ = a 1 − r + d r ( 1 − r ) 2 S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2} S ∞ = 1 − r a + ( 1 − r ) 2 d r square hota hai, yeh mat bhoolna. Sabse important baat: yeh infinite formula sirf tab valid hai jab ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 ho. Agar r r r
Exam tip: agar sirf n n n 1 + 2 2 + 3 4 + 4 8 + ⋯ = 4 1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\dots = 4 1 + 2 2 + 4 3 + 8 4 + ⋯ = 4 a = 1 , d = 1 , r = 1 2 a=1,d=1,r=\tfrac12 a = 1 , d = 1 , r = 2 1 1 , 2 , 2.75 , 3.25 , … 1,2,2.75,3.25,\dots 1 , 2 , 2.75 , 3.25 , …
