Thursday, April 25, 2013

55-05 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-05

Blog-55

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

Examples on Summation.

D) Find the sum of first 100 natural numbers which are divisible by 6.


Solution:
1) For the natural number which are divisible by 6, a = 6, d = 6.
2) Here we can use the formula Sn = n * [ 2 a + (n -1) d ) ] / 2
          S100 = 100 * [ 2 (6) + (100 -1) (6) ) ] / 2
          S100 = 6 * 100 * [ 2 + 99 ] / 2
          S100 = 3 * 100 * [ 101 ]
          S100 = 100 * [ 303 ]
          S100 =  30300
3) Answer: The sum of the first 100 natural numbers which are divisible by 6, is S100 = 30300.


E) Find the sum of all odd natural numbers between 100 and 250.

Solution:
1) For odd natural number between 100 and 250, a = 101, d = 2.
2) First we will find the total number of terms between 100 & 250 which are odd.
     Here we use the formula tn =  a + (n-1) d 
     So,   tn =  101 + (n -1) * 2
        249   =  101 + (n -1) * 2
     (n - 1)   =  (249 - 101) / 2
     (n - 1)   =  (148) / 2
     (n - 1)   =  74
        n        =  75
3) Now we will use the formula Sn = n * [  a + tn ] / 2
          Sn = 75 * [ 101 + 249 ] / 2
          Sn = 75 * [ 350 ] / 2
          Sn = 75 * [ 175 ]
          Sn = 13125
4) Answer: The sum of odd natural numbers between 100 & 250, is Sn = 13125.


F) Find the sum of all 3-digit natural numbers which are divisible by 4.

Solution:

1) We know that the lowest 3 digit number is 100 & the largest one is 999.
2) So the numbers divisible by 4 between 100 & 999 are
     100, 104, 108, ----, 992, 996.
3) Here a = 100, d = 4 and last term = 996.
4) First we will find the total number of terms, so here,
                      t= a + ( n - 1 ) d, we have,
                  996 = 100 + ( n - 1 ) * 4
     4 *  ( n - 1 )  =  996 - 100 
     4 *  ( n - 1 )  =  896 
            ( n - 1 )  =  896 / 4
            ( n - 1 )  =  224
                    n    =  224 + 1
                    n    =  225
5) Now we will find Sn using the formula Sn = n * [  a + tn ] / 2.
                     Sn = 225 * [ 100 + 996 ] / 2
                     Sn = 225 * 4 * [ 25 + 249 ] / 2
                     Sn = 225 * 4 * [ 274 ] / 2
                     Sn = 225 * 4 * 137
                     Sn = 900 * 137
                     Sn = 123300
6) Answer: The sum of all 3-digit natural numbers which are divisible by 4 is Sn = 123300.


G) If t17 = 59 and t39 = 141 then find s55 .
Solution:

1) Let " a " be the first term & " d " be the common difference.
2) We know that, t= a + ( n - 1 ) d
                      t17 = a + ( 17 - 1 ) d = 59 and  t39 = a + ( 39 - 1 ) d = 141,
     So we have,
                               a + ( 16 ) d =    59
                               a + ( 38 ) d =  141
                   -----------------------------------------
                            2 a + ( 54 ) d = 200            --------------------- (1)

3) Here    Sn = n * [ 2 a + (n -1) d ) ] / 2
               S55 = 55 * [ 2 a + (54) d ) ] / 2
               S55 = 55 * [ 200 ] / 2
               S55 = 55 * [ 100 ]
               S55 = 5500
4) Answer: Here S55 = 5500.

Few more Problems on AP will be published in the next Blog.

Anil Satpute