55-05 Basics of Arithmetic & Geometric Progression

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Examples of Summation.

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

Solution:
1) For the natural number which is divisible by 6, a = 6, d = 6.
2) Here we can use the formula S_n = 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 numbers 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 t_n =  a + (n-1) d 
     So,   t_n =  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 S_n = n * [  a + t_n ] / 2
          S_n = 75 * [ 101 + 249 ] / 2
          S_n = 75 * [ 350 ] / 2
          S_n = 75 * [ 175 ]
          S_n = 13125
4) Answer: The sum of odd natural numbers between 100 & 250, is S_n = 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_n = 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 S_n using the formula S_n = n * [  a + t_n ] / 2.
                     S_n = 225 * [ 100 + 996 ] / 2
                     S_n = 225 * 4 * [ 25 + 249 ] / 2
                     S_n = 225 * 4 * [ 274 ] / 2
                     S_n = 225 * 4 * 137
                     S_n = 900 * 137
                     S_n = 123300
6) Answer: The sum of all 3-digit natural numbers which are divisible by 4 is S_n = 123300.

G) If t₁₇ = 59 and t₃₉ = 141 then find s_{55 .}
Solution:

1) Let " a " be the first term & " d " be a common difference.
2) We know that, t_n = 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    S_n = 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.

In the next part, we will see a few examples and some essential formulae.

Click here for the next basics.

Anil Satpute