HOW TO FIND THE SUM OF TERMS WHICH ARE MULTIPLES BETWEEN TWO NUMBERS

Example 1 :

If the sum of the first n terms of an AP is 4n - n²  what is the first term (that is S)? what is the sum of first two terms?what is the second term? similarly find the 3rd,the 10th and the nth terms.

Solution :

S_n  =  4n - n²

Sum of first two terms  =  4

To find the second term, we have to subtract the sum of two terms from first term

  =  4 - 3  =  1

Second term  =  1

a₃  =  a + 2d

d = second term - first term  =  1 - 3  =  -2 

 a₃ =  3 + 2(-2)  =  3 - 4 = -1

 a₁₀  =  a + 9 d

=  3 + 9(-2)

= 3 - 18

= -15

Example 2 :

Find the sum of first 40 positive integers divisible by 6.

Solution :

The positive integers which are divisible by 6 are 

6, 12, 18,...........

6 + 12 + 18 + ...........

a = 6, d = 12 - 6  =  6, n = 40

S_n  =  (n/2) [ 2 a + (n - 1) d]

S₄₀  =  (40/2) [2(6) + (40-1) (6)]

=  20 [12 + 39(6)]

=  20 [12 + 234]

=  20 [246]

=  4920

Example 3 :

Find the sum of first 15 multiples of 8.

Solution :

first 15 multiples of 8 are

8, 16, 24,................

8 + 16 + 24 + ...........

a = 8         d = 16 - 8       n = 15

                   = 8

 S_n  =   (n/2) [ 2 a + (n - 1) d]

 S₁₅ = (15/2) [2(8) + (15-1) (8)]

  =  (15/2) [16 + 14(8)]

  =  (15/2) [16 + 112]

  =  (15/2) [128]

  =  15 [64]

  =  960

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