Tuesday, April 23, 2013

53-03 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-03

Blog-53
Dear Students,


Download following files for learning Formulas more effectively. Take the print out of these files and write your answers daily to improve your scores in 10th standard/grade.

[ Note: The following 2 files are available in the secured drive. While downloading, your email might be asked. Please provide it to download the files. I assure that your email-id will not be given to anybody ]

1) Algebra Formulas
2) Geometry Formulas


In the Previous Blog we had seen some important concepts of Arithmetic & Geometric Progression. 
1) Click Here to Read the Blog-60 (01 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-01)
2) Click Here to Read the Blog-61 (02 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-02)

E) Find tn for an AP where t5  = 19 & t15  = -21.

Solution:


1) Let " a " be the first term & " d " be the common difference.
2) So, t= a + ( n - 1 ) d
3) According to the problem, 
            t5   = a +  ( 4 )  d =  19
            t15 = a + (14 )  d = -21
                  (-)  (-)               (+)
  --------------------------------------------
                            - 10  d  =  40
                                  -  d  =  4
                                     d  =  - 4
4) We know that 
            t5   =  a +  ( 4 ) d =  19
                     a + 4 * (- 4) =  19
                             a  - 16 =  19
                                    a   =  19 + 16
                                    a   =  35
5) So,  t= a + ( n - 1 ) d
             t= 35 + ( n - 1 ) * (- 4)
             t= 35 - 4 n + 4
             t= 39 - 4 n
6) Answer: Here the nth term of an AP is  tn =  39 - 4 n

F) Find, how many 3 digit natural numbers are divisible by 3.

Solution:

1) We know that the lowest 3 digit number is 100 & the largest one is 999.
2) So the numbers divisible by 3 between 100 & 999 are
     102, 105, 108, ----, 996, 999.
3) Here a = 102, d = 3 and last term = 999.
4) So, from  t= a + ( n - 1 ) d, we have,
                  999 = 102 + ( n - 1 ) * 3
     3 *  ( n - 1 )  =  999 - 102 
     3 *  ( n - 1 )  =  897 
            ( n - 1 )  =  897 / 3
            ( n - 1 )  =  299
                    n    =  299 + 1
                    n    =  300
5) Answer: Here there are 300 three digit numbers which are divisible by 3.

Problems on Summation will be published in the next Blog.

Anil Satpute