Geometric Progression - K Bhanumoorthy

 

 Geometric Progression. (GP)

 

If the terms in a sequence are such that, each term is a constant multiple of preceding term, then we say terms are said to be in a GP. The multiple is called as common ratio and denoted by “r”.

Eg :    4,8,16,32, 64       r is 2.

           8,4,2, 1, ½           r is ½.

           7, -14, 28, -56    r is -2.

 In general, we take terms in GP as a, ar, ar²,ar³ ar⁴- - -        r is “r”.

In this GP also we can find n ^th, term and also sum to n terms of GP.

To note:     n ^th , term of a GP  is       a. r^r-1,      - - - - - - - - -        I

Let S denote sum to n terms

      S = a + ar+ ar² + ar³+ar⁴+ - - - - - - - - - - - - + ar^n-1   . . . . . . .  II

       Multiply both sides by ‘ r’ ,

    rS = ar + ar²+ ar³+ ar⁴+  - - - - - - - - - - - - - - +arⁿ  . . . . . . . ..  III

 

subtracting equation III from II , we get ,

S – rS = a – a rⁿ   ;   S ( 1 – r ) = a ( 1 – rⁿ ); so ,  S = a ( 1 – r ⁿ ) /( 1 – r).

Here condition for r is r < 1.

Note 1:  . If r > 1, then S = a (r ⁿ – 1) / (r – 1).

Note 2:   a, b, c are 3 consecutive terms of a GP, then b is said to be geometric mean of   a and c.  Further b/ c = c / b,  ie  b ²  = ac.

 

Let us discuss some problems for practice.

 

1.     Find the 40^th term of GP ,2, 14,28,56 - - - - - - - --   

Ans: a = 2, r =7   r > 1, hence to use the formula   

 40 th term is   2. (7 )^39. Is the answer.

 

2.Find n, given that 1024 is the n ^th term of GP 4,8,16,32, - - - - - -

Ans : a = 4 and r = 2 ,using the formula ,we get  4x 2^n-1  = 1024,

 2² x 2^n-1 = 1024; 2 ^{2+n -1} = 1024; 2 ^{n +1} = 1024; 2 ^{n +1} = 2 ¹⁰ ,

  Since the base are the same powers are equal, n + 1 = 10,

 Hence n = 9. Is the answer.

 

3.find the sum to 18 terms of the GP  3,6,12, 24, - -- - - - --    - - - - - - -

Ans : a = 3, r = 2, r > 1, to use the formula ,

S = a (r ⁿ – 1) / (r – 1); 3. (2¹⁸ – 1) / (2 – 1)   = 3 (2 ¹⁸ – 1) is the answer.

 

4. If 5, x, y, 40 are in GP, find x and y.

    From the given data, x ² = 5y ; and  y ² = 40x  ; on substituting for y ,

Y = x ² / 5, we get x ⁴ = 25 x 40x, x ³ = 1000, hence x =10 and y = 20 

(For 5y = 100).

 

5. If the first, second and fifth terms of an AP are three consecutive terms of a GP, find the common ratio.

Ans: Let a, a+ d, a+2d, a+3d, a+4d be in AP, with ‘a’ as the first term, common difference as ‘d’.

As per given data in the question, a, a+ d, a+4d are three consecutive terms in GP.   - - - - - - - - I

 Ie using the principle of GP, we get (a+ d)² = a(a+4d),

                                  a²+2ad+d² = a² + 4d², we get d = 2a, by cancelling common terms.

 

Hence by substituting d = 2a in I, we get, a, a+2a, a+8a, ie  a, 3a,9a,

 

Hence the common ratio is 3.