Geometric Progression (GP)

A sequence  a₁,  a₂,  a₃ , ………,a_n, … is called geometric progression, if each term is 

non-zero . 

By letting a₁= a, we obtain a geometric progression, a, ar, ar² , ar³ ,…., where a 

is called the first term and r is called the common ratio of the G.P.  

General Term of a GP 

 Let us consider a G.P. with first non-zero term ‘a’ and  common ratio ‘r’. 

The second term is obtained by multiplying a by r, thus a₂ = ar. 

 Similarly, third term is obtained by multiplying a₂  by r. Thus, 

              a₃ = a₂r = ar², …, and so on. 

Consider the pattern , a₁ = a = ar^1-1  ,  

                                         a₂ = ar = ar^2-1 ,  

                                          a₃ = ar² = ar^3-1, 

                                         ……………and so on. 

Thus, the n^th term of a GP is given by a_n = ar^n-1 . 

Finite and Infinite GP

 The GP a , ar , ar² , ……., ar^n-1 is called a finite GP and a GP a, ar ,ar²,……,ar^n-1,….. is called an infinite GP . 

Sum of First n terms of a GP

 Let the first term of a GP is ‘a’ and the common ratio be ‘r’ . 

Let S_n denote the sum of first n terms of GP .Then , 

           S_n = a + ar + ar² + …….+ar^n-1 

Case (1)

If r = 1 , then S_n = a +a +a +……+a (n terms ) = na 

Case(2)

If r ≠ 1 , then  

         S_n = a(1- rⁿ) /(1 -r)   , when r <1 

   or  

          S_n = a(r^n -1)/(r -1) , when r > 1 

Sum of an infinite GP

Let a , ar , ar² ,….., arⁿ ,….be an infinite series ,then  

         S = a /(1- r) 

Geometric Mean (GM)

The Geometric Mean of two positive numbers a and b is the number √ab . 

Let G1 , G2 ,G3, …….,Gn be the n numbers between positive numbers a and b such that 

                      a , G1 ,G2,……,Gn , b is a GP , then 

            Gn = arⁿ = a(b/a)ⁿ/(n+1) 

Relationship between AM and GM

Let A and G be AM and GM of two given positive real number a and b , respectively . 

Then ,  

       A = (a +b)/2 and  G = √ab 

Thus ,  

         A -G = ( √a -√b )² / 2  ≥ 0  

Therefore , A  ≥ G 

Sum to n terms of special series 

(1)1 + 2 + 3 +… + n (sum of first n natural numbers) 

                      S_n = n(n +1)/2 

(2) 1² +2² +3² + ……+n² (sum of squares of the first n natural numbers) 

                 S_n = n(n +1)(2n +1)/6 

(3) 1³ + 2³ +3³ +…..+n³ (sum of cubes of the first n natural numbers). 

              S_n = { n(n+1) }² / 4 

Note

If we have to choose three terms of a GP , then we should take a/r , a , ar . 

Solved examples

(1)Which term of the GP 16 , 8,4,2,… is 1/16 ? 

Sol.   Here , a = 16 and r = ½ 

Let 1/16 be the n^th term , then  

      T_n = ar^n-1 

       1/16 = 16 ×(1/2)^n-1 

        (1/2)^n-1 = 1/16×16 

                        = (1/2)⁸ 

Comparing , both sides ,we get ,  

                 n-1 = 8 or n =  9 

(2)Find sum of 8 terms of the series 1+3+9+27+….. 

Sol. This is a GP with a = 1 and r = 3/1 = 3 > 1 

Thus , S₈ = a(r^n -1)/(r -1) 

                = 1×(3^{8 -1})/(3-1) 

                 = (6561 -1)/2 

                 =6560 /2 = 3280 

(3)Find the sum of the series 1+1/2 +1/4 +….. up to infinite terms . 

Sol.      Here , a = 1 , r = ½ 

   Thus ,    S = a / (1- r) 

                     = 1/(1-1/2) 

                      = 1/1/2 

                       = 2 

(4)Find GM between the two numbers  3 and 27 . 

Sol.      Here , a = 3 , b = 27 

 Then ,     GM = √ab 

                       = √3×27 

                       = √81 

                        =9 

(5)Find the 4^th term from the last of the series 8,4,2,…..,1/128. 

Sol. This is in GP . 

In a GP , n^th term from the last term = l /r^n-1 

Here , l = 1/128 , r = ½ , n =4 

Therefore , 4^th term from the last = (1/128)/(1/2)^4-1 

                                                     = 8/128

                                                     = 1/16