57-07 Basics of Arithmetic & Geometric Progression

Click here for the previous basics.

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 2^nd term divided by the 1^st 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, ... t_n-1, t_n are the terms in GP then t_2/t_{1 =} t_3/t_{2 =} t_4/t_{3 = ..... =} r.
Find the n^th 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 n^th term of the GP 2, 6, 18...

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

3) Answer: The nth term of the GP is t_n = 2 x (3) ^{(n – 1)}

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

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

Solution:
1) Here a = 1 and r = 2.
2) So      t_n = a x (r) ^{(n – 1)}
                t_n = 1 x (2) ^{(n – 1)}
                t_n = (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: 7^th and 13^th 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 1^st n terms of the GP.

Solution:
1) Let 1^st term be "a" and the Common Ratio be "r".
2) We know that 
                          s_n  =  a + a r + a r² + a r³ + ------ + a r ^{(n – 1)}
                        r s_n  =         a r + a r² + a r³ + ------ + a r ^{(n – 1)} + a r ^{(n)
                              ( - )                 ( - )      ( - )      ( - )                        ( - )              ( - )
            ------------------------------------------------------------------------------------------------------------}
             ( 1 - r )  s_n  =  a  + -------------------------------------------+ - a r ⁽ⁿ⁾ 
             ( 1 - r )  s_n  =  a  - a r ⁽ⁿ⁾ 
                          s_n  =  [a (1  -  rⁿ ) ] / ( 1 - r )
Note: 
1) If r < 1 then s_n  =  [a (1  -  rⁿ ) ] / ( 1 - r )
2) If r > 1 then s_n =  [a ( rⁿ  - 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 
       s_n  =  a + a r + a r² + a r³ + ------ + a r ^{(n – 1)      put r = 1 we get,}
       s_n  =  a + a + a + ----- + n-times
       s_n  =  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, s_n =  [a ( rⁿ  - 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.

Click here for the next basics.

Anil Satpute