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Arithmetic-Geometric Progression       

Suppose a1, a2, a3, …. is an A.P. and b1, b2, b3, …… is a G.P. Then the sequence a1b1, a2b2, …, anbn is said to be an arithmetic-geometric progression. An arithmetic-geometric progression is of the form ab, (a+d)br, (a + 2d)br2, (a + 3d)br3, ……


Its sum Sn to n terms is given by
Sn = ab + (a+d)br + (a+2d)br2 +……+ (a+(n–2)d)brn–2 + (a+(n–1)d)brn–1.

Multiply both sides by r, so that rSn = abr+(a+d)br2+…+(a+(n–3)d)brn–2+(a+(n–2)d)brn–1+(a+(n–1)d)brn.


Subtracting we get
(1 – r)Sn = ab + dbr + dbr2 +…+ dbrn–2 + dbrn–1 – (a+(n–1)d)brn.

 = ab + dbr(1–rn–1)/(1–r)  (a+(n–1)d)brn

  Sn = ab/1–r + dbr(1–rn–1)/(1–r)2 – (a+(n–1)d)brn/1–r.


If –1 < r < 1, the sum of the infinite number of terms of the progression is

         limn Sn = ab/1–r + dbr/(1–r)2.

Example

Find the sum of series 1 . 2 + 2 . 22 + 3 . 23 +…+ 100 . 2100.

Solution

Let S = 1.2 + 2.22 + 3.23 +…+ 100.2100                           …… (1)
        ⇒ 2S = 1.22 + 2.23 +…+ 99.2100 + 100.2101                      …… (2)
        ⇒ –S = 1.2 + 1.22 + 1.23 +…+ 1.2100 – 100.2101
        ⇒ –S = 1.2 (2100–1/2–1) – 100.2101
        ⇒ S = –2101 + 2 + 100.2101 = 199.2101 + 2.

 

Example

Let r = 1/2, consider n (1/2)n for increasing value of n i.e.

Solution

n = 1 : 1. (1/2)1 = 1/2 = 0.5
        n = 2 : 2 × (1/2)2 = 1/2 = 0.5
        n = 3 : 3 × (1/2)3 = 0.375
        n = 10 : 10 (1/2)10 = 0.00976, and so on
        Thus we observe that as n → ∞
        n rn → 0 for |r| < 1.

 

Example

Evaluate 1 + 4/5 + 7/52 + 10/53 +…… to infinite terms.

Solution

 Let S = 1 + 4/5 + 7/52 + 10/53 + ………
        1/5 S = 1/5 + 4/52 + 7/53 ………
        Subtracting
        (1–1/5) S = 1 + 3/5 + 3/52 + 3/53 + ………
        4/5 S = 1/1–3/5                             (? It is infinite G.P.)
        ⇒ S =25/8


Example 11:

Let t1, t2, t3, ……, tm–1, tm, tm+1, be a sequence so that

        (i)     tm+1/tm = tm/tm–1 ……… constant                             (r)

                then tp = (t1)rp–1

  
        (ii)    tm+1/tm = tm/tm–1 = constant                                  (r)

                then tp = constant 1 + (constant 2) × rp–1

       
        (iii)    If the difference of difference of terms are in G.P. then

                tp = a + bp + crp–1, where r is the common ratio.

 Example 12:

 7. 14. 33. 88. 251. 738 …………

 

                                         1088_Arithmetic Geometric Progression.JPG 


Note:
324/108 = 108/36 = 36/12 = 3

  

     tp = a + bp + c 3p–1

        p=1   t1 = 7 = a + b + c

        p=2   t2 = 14 = a + 2b + 3c

        p=3   t3 = 33 = a + 3b + 9c

Solving, we get a = 3, b = 1, c = 3

     tp = 3 + p + 3. (3p–1)





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