Tuesday, April 30, 2013

57-07 Basics of Arithmetic & Geometric Progression

Geometric Progression (GP) is a sequence: The terms in GP are obtained simply by multiplying by a Non-Zero constant. This constant is known as the "Common Ratio". In short, the 2nd term divided by the 1st term is the same as the 3rd term divided by the 2nd term and so on for all the terms of the sequence called as Geometric Progression (GP).

If t1, t2, t3, ... tn-1, tn are the terms in GP then t2/t1 = t3/t2 = t4/t3 = ..... = r.
Find the nth term of GP.

Solution:
1) Let " a " be the 1st term and " r " be the common ratio.
2) By the definition of GP,
Now we will see some examples of GP.

A) Find the nth term of the GP 2, 6, 18...

Solution:
1) Here a = 2 and r = 6/2 = 3.
2) We know that 
               tn  =  a x (r)( n – 1 )
               tn  =  2 x (3)( n – 1 )

3) Answer: The nth term of the GP is tn = 2 x (3) (n – 1)

Note: Above formula can be used to find any term in GP.

B) Find 7th and 13th term of the GP if a = 1 and r = 2

Solution:
1) Here a = 1 and r = 2.
2) So      tn = a x (r) (n – 1)
                tn = 1 x (2) (n – 1)
                tn = (2) (n – 1)
 3)            t7 = (2) (7 – 1)
                 t7 = (2) (6)
                 t7 = 64

 4)           t13 = (2) (13 – 1)
                t13 = (2) (12)
                t13 = 4096
5) Answer: 7th and 13th term of the GP are t7 = 64 and t13 = 4096.

Remember: For any GP, t1 = a, t2 = a r, t3 = a (r) (2), t4 = a (r) (2)  so on.

A) Find the sum of 1st n terms of the GP.

Solution:
1) Let 1st term be "a" and the Common Ratio be "r".
2) We know that 
                          sn  =  a + a r + a ra r+ ------ + a r (n – 1)
                        r sn  =         a r + a ra r+ ------ + a r (n – 1) + a r (n)
                              ( - )                 ( - )      ( - )      ( - )                        ( - )              ( - )
            ------------------------------------------------------------------------------------------------------------
             ( 1 - r )  sn  =  a  + -------------------------------------------+ - a r (n) 
             ( 1 - r )  sn  =  a  - a r (n) 
                          sn  =  [a (1  -  r) ] / ( 1 - r )
Note: 
1) If r < 1 then sn  =  [a (1  -  r) ] / ( 1 - r )
2) If r > 1 then sn =  [a ( r - 1 ) ] / ( r  - 1 )
3) Very Important:
     If r = 1 then we can't use the above formula as we get " 0 " in the denominator. So in this case, we have 
       sn  =  a + a r + a ra r+ ------ + a r (n – 1)      put r = 1 we get,
       sn  =  a + a + a + ----- + n-times
       sn  =  n x a

Examples of Summation:

A) Find s10  of the GP 2, 4, 8, 16 -----

Solution:
1) Here a = 2 and r = 4 / 2 = 2
2) Since r = 2 > 1, we can use the formula, sn =  [a ( r - 1 ) ] / ( r  - 1 )
3) So,           s10 =  [2 ( 210  - 1 ) ] / ( 2  - 1 )
                     s10 =  [2 ((25)2  - 1 ) ] 
                     s10 =  [2 ((32)2  - 1 ) ] 
                     s10 =  [2 (1024  - 1 ) ] 
                     s10 =  [2 (1023) ] 
                     s10 =  [2046
4) Answer: Here s10 =  [2046

Note: Please understand the simple method of finding the square of any two-digit number:
The Simple method to find the square will be published in some other Blog.

Some more Problems with the Simple Method of calculations will be published in the Next Blog.

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