Wednesday, April 24, 2013

54-04 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-04

Blog-54
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)
3) Click Here to Read the Blog-62 (03 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-03)


Examples on Summation.

A) Find the sum of the first n terms of an AP 7, 3, -1, -5, ---.

Solution:
1) Here, a = 7, d = 3 - 7 = - 4 , so d = - 4 
[ Note: Let us check whether given sequence is an AP or not Simply by calculating d for the next pair. i.e. d = - 1 - 3 = - 4. which is same as previous d so it is an AP ]
2) Now we can use the formula 

        Sn = n * [ 2 a + (n - 1) d ) ] / 2
        Sn = n * [ 2 * (7)+ (n - 1) * (- 4 ) ] / 2
        Sn = n * [ 14 + (- 4 n + 4) ] / 2
        Sn = n * [ 14 - 4 n + 4) ] / 2
        Sn = n * [ 18 - 4 n ) ] / 2
        Sn = n * [ 18/2 - 4 n /2 ]
        Sn = n * [ 9 - 2 n ]
3) Answer: So the sum of first n terms of an AP is  Sn = n * [ 9 - 2 n ] 

B) Find the sum of all the terms of an AP  2, 5, 8, 11, --- , 299.

Solution:
1) Here, a = 2, d = 5 - 2 = 3, so d = 3 and last term l = tn = 299.
2) To find n, we need to use the formula, 
             t= a + ( n - 1 ) d
          299  = 2 + ( n - 1 ) * ( 3 )
          297  = ( n - 1 ) * ( 3 )
          297 / 3 = ( n - 1 )
          99  = ( n - 1 )
          99 + 1 = n
          n = 100
3) Now we can use the formula,       Sn = n * ( a + l ) / 2
          Sn = n * ( a + l ) / 2
          Sn = 100 * ( 2 + 299 ) / 2
          Sn = 100 * ( 301 ) / 2
          Sn = 50 * ( 301 )
          Sn = 15050
4) Answer: The sum of the terms of an AP  2, 5, 8, 11, --- , 299, is Sn = 15050.

Note: The Above problem in different way:

C) Find the sum of first 100 terms of an AP, 2, 5, 8, 11, ---.

Solution:
1) Here, a = 2, d = 5 - 2 = 3, so d = 3 and n = 100.

2) Here we can use the formula Sn = n * [ 2 a + (n-1) d ) ] / 2
          Sn = n * [ 2 a + (n-1) d ) ] / 2
          Sn = 100 * [ 2 (2) + (100-1) (3) ) ] / 2
          Sn = 50 * [ 4 + (99) (3) ) ]
          Sn = 50 * [ 4 + 297 ]
          Sn = 50 * [ 301 ]
          Sn = 15050
3) Answer: The sum of the first 100 terms of an AP  2, 5, 8, 11, ---, is Sn = 15050.

D) Find the sum of first n natural numbers.

Solution:
1) For the natural numbers, a = 1, d = 1.
2) Here we can use the formula Sn = n * [ 2 a + (n-1) d ) ] / 2
          Sn = n * [ 2 (1) + (n-1) (1) ) ] / 2
          Sn = n * [ 2 + n - 1 ] / 2
          Sn = n * [ n + 1 ] / 2
3) Answer: The sum of the first n natural numbers, is Sn = n * [ n + 1 ] / 2.

Some examples of this session will be discussed in the next Blog.

Anil Satpute