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Arithmeticogeometric Progression Solved Examples

As the name suggests, an arithmeticogeometric progression is obtained when the corresponding terms of a geometric and arithmetic progression are multiplied by each other.

Mathematically, it can be expressed as

xn = bnhn, where

xn is the nth term of the arithmeticogeometric progression

bn is the nth term of the arithmetic progression

hn is the nth term of the geometric progression

Illustration:

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

Solution: Let us denote the given series by S.

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 (299–1/(2–1)) – 100.2101

⇒ S = –2100 + 2 + 100.2101 = 199.2100 + 2. 

Illustration:

Let r = 1/2, consider nrn for increasing value of n and then find its value as n tends to infinity.

Solution: We first consider n (1/2)n for increasing values of n.

       n = 1 we get 1. (1/2)1 = 1/2 = 0.5

       n = 2 gives 2 × (1/2)2 = 1/2 = 0.5

       n = 3 gives 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.

Illustration:

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

Solution: We denote the given sequence by S.

        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)                             (As it is infinite G.P.)

        ⇒ S =25/8 

Refer the following video for more on arithmeticogeometric progression



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. 

Illustration:

7. 14. 33. 88. 251. 738 …………

Solution: We can represent the given sequence in the form of a diagram as shown   



 

Remark: First we write down all the numbers in a straight line. Then we find out the difference between every pair of two terms. This is denoted in the second line. Again we find the difference between every pair of consecutive terms in the second line. Now we divide every successive term by the preceding term and see that the ratio in each case is 3.  

       324/108 = 108/36 = 36/12 = 3

Hence, using the formula listed above

∴      tp = a + bp + c 3p–1

        p=1 so t1 = 7 = a + b + c

        p=2 so t2 = 14 = a + 2b + 3c

        p=3 so t3 = 33 = a + 3b + 9c

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

⇒     tp = 3 + p + 3. (3p–1)

askIITians offers extensive study material which covers all the important topics of IIT JEE Mathematics. Topics like arithmetic progression, geometric progression and arithmeticogeometric progression have been covered along with numerous solved examples. It is important to have clear fundamentals in order to remain competitive in the JEE.   

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