## Tuesday, April 30, 2013

### 57-07 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-07

Blog-57

Dear Students,

Following two files are very important to improve your base for remembering the formulas. It is 100 % free for all the students of the world. Download these two files, take the Print Out of these files on two sides of the paper and practice these formulas for your benefit to get high scores in Mathematics Standard 10/ Grade 10.

Geometric Progression (GP) is a sequence: The terms in GP are obtained simply  multiplying by Non-Zero constant. This constant is known as "Common Ratio". In short, 2nd term divided by 1st term is same as 3rd term divided by 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 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 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 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 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 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 on 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 Simple Method of calculations will be published in the Next Blog.

Anil Satpute