Table of Contents
Arithmetico-Geometric Series:
Series of the form a (b) + (a + d) (br) + (a + 2d) (br2) + (a + 3d) (br3) + ……… + (a + (n – 1) d) (brn-1) + ……… is called Arithmetico-Geometric Series (AG Series) which consists of two series i.e.
- Arithmetic Series → a + (a + d) + (a + 2d) + ………
- Geometric Series → b + br + br2 + ………
Sum to n-terms of Arithmetico-Geometric Series:
| Let Sn = a + (a + d) r + (a + 2d) r2 + ……… + (a + (n – 1) d) rn-1 …………. (i) Multiply (i) by r (common ratio):- rSn = ar + (a + d) r2 + (a + 2d) r3 + ……… + (a + (n – 1) d) rn …………. (ii) Subtract (ii) from (i):- Sn – rSn= a + dr + dr2 + dr3 + ……… + drn-1 – (a +  d) rn⇒ (1 – r) Sn = a + (dr) (rn-1 – 1)/(r – 1) – (a + d) rn⇒ Sn = a/(1 – r) + (dr) (1 – rn-1)/(1 – r)2 – (a + d) rn/(1 – r) |
Sum of Infinite Term of Arithmetico-Geometric Series:
| Let S∞ = a + (a + d) r + (a + 2d) r2 + …….. ∞ …………. (i) Multiplying (i) by r:- rS∞ = ar + (a + d) r2 + (a + 2d) r3 + …….. ∞ …………. (ii) Subtract (ii) from (i):- S∞ – rS∞= a + dr + dr2 + …….. ∞ ⇒ (1 – r) S∞= a + dr/(1 – r) ⇒ S∞= a/(1 – r) + dr/(1 – r)2 |
| Example- Show that 1 + 2x + 3x2 + …. to n terms (x < 1) is (1 – xn)/(1 – x)2 – nxn/(1 – x). Find also the sum when n → ∞. Solution- The given series is, 1 + 2x + 3x2 + …… to n terms The corresponding A.P. is 1, 2, 3, ……… It’s nth term is 1 + (n – 1) 1 = 1 + n – 1 =n The corresponding G.P. is 1, x, x2, ……… Its nth term is 1. xn-1 = xn-1 ∴ the nth term of the given series = (nth term of A.P.) (nth term of G.P.) = nxn-1 Let Sn = 1 + 2x + 3x2 + …… + (n – 1) xn-2 + nxn-1 …………. (i) Multiply (i) by x:- xSn = x + 2x2 + 3x3 + …… + (n – 1) xn-1 + nxn …………. (ii) Subtract (ii) from (i):- Sn – xSn = 1 + x + x2 + x3 + …… + xn-1 – nxn ⇒ (1 – x) Sn = Sum of n term of G.P. with a = 1, r = x – nxn ⇒ (1 – x) Sn = [1(1 – xn)/(1 – x)] – nxn ⇒ Sn = [(1 – xn)/(1 – x)2] – [nxn/(1 – x)] …………. (A) As n → ∞ and x < 1 ∴ xn → 0 ∴ Equation (A) becomes:- S∞ = [(1 – 0)/(1 – x)2] – [n (0)/(1 – x)] S∞ = 1/(1 – x)2 |
| Example- Sum to n terms of the series 1 + 2/3 + 3/32 + 4/33 + ……….. Solution-The given series is, 1 + 2/3 + 3/32 + 4/33 + ……….. Let Tn be its nth term. Tn = (nth term of A.P. 1, 2, 3, ………) (nth term of G.P. 1, 1/3, 1/32, ………..) ⇒ Tn = [1 + (n – 1)1] [1 (1/3)n-1] ⇒ Tn = n (1/3)n-1 Let Sn = 1 + 2/3 + 3/32 + 4/33 + ……….. + (n – 1) 1/3n-2 + n (1/3n-1) …………. (i) Multiply (i) by 1/3 (1/3) Sn = 1/3 + 2/32 + 3/33 + ……….. + (n – 1)/3n-1 + n/3n …………. (ii) Subtract (ii) from (i):- Sn – (1/3) Sn = 1 + 1/3 + 1/32 + 1/33 + ……….. + 1/3n-1 – n/3n ⇒ (1 – 1/3) Sn = (Sum of n term of G.P. with a = 1, r = 1/3) – n/3n ⇒ (2/3) Sn = 1[1 – (1/3n)]/[1 – (1/3)] – n/3n ⇒ (2/3) Sn = (3n – 1)/3n(2/3) – n/3n ⇒ Sn = (3n – 1)/3n(2/3)2 – n/(3n) (2/3) ⇒ Sn = 1/(3n) (2/3) [(3n – 1)/(2/3) – n] ⇒ Sn = 1/(2) (3n-1) [(3/2) (3n – 1) – n] ⇒ Sn = 1/(2) (3n-1) [(3n+1 – 3 – 2n)/2] |
| Example- If the sum to infinity of the series 3 + 5r + 7r2 + ….. is 44/9, find r. Solution- Let S∞ = 3 + 5r + 7r2 + ….. ∞ …………. (i) Multiply (i) by r:- rS∞ = 3r + 5r2 + 7r3 + ….. …………. (ii) Subtract (ii) from (i):- S∞ – rS∞ = 3 + 2r + 2r2 + 2r3 + ….. ∞ ⇒ (1 – r) S∞ = 3 + 2r/(1 – r) ⇒ S∞ = 3/(1 – r) + 2r/(1 – r)2 ⇒ S∞ = [3 (1 – r) + 2r]/(1 – r)2 ⇒ S∞ = (3 – r)/(1 – r)2 But S∞ = 44/9 ∴ (3 – r)/(1 – r)2 = 44/9 ⇒ 27 – 9r = 44 (1 – r)2 ⇒ 27 – 9r = 44 (1 + r2 – 2r) ⇒ 27 – 9r = 44 + 44r2 – 88r ⇒ -44r2 + 88r – 9r = 44 – 27 ⇒ -44r2 + 79r = 17 ⇒ 79r = 44r2 + 17 ⇒ 44r2 – 79r + 17 = 0 ⇒ 44r2 – 68r – 11r + 17 = 0 ⇒ 4r (11r – 17) – 1 (11r – 17) = 0 ⇒ (4r – 1) (11r – 17) = 0 Either 4r – 1 = 0 or 11r – 17 = 0 r = 1/4 or r = 17/11 |
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