Question

find the sum of first 40 positive integers divisible by 6.Also find the sum of first 20 positive integers divisible 5 or 7

Answer 1

positive integers divisible by 6 are
6,12,18,24...........
since difference is same,it is an AP
we need to find the sum of 40 integers
we can use formula
Sn= n/2(2a+(n-1)d)
here n=40,a=6&d=12-6=6
putting values in formula
Sn=n/2(2a+(n-1)d)
=40/2(2×6+(40-1)×6)
=20(12+39×6)
=20(12+234)
=20×246
=4920
therefore sum of the first 40 positive integers which are divisible by 6 is 4920

Answer 2

Answer:

let, sum of first 40+ve integers divisible by 6

6,12,18,24...............

since, a=6 & d=6 , n=40

to find :- S40=?

now ,

       Sn=n/2[2a + (n-1)d

       S40=40/2[2*6+39*6]

              =20*[12+234]

              =20*246

              =4920

let, sum of 20+ve integers divisible by 5 or 7

35,70,105,140...............

since, a=35 & d=35 , n=20

to find :- S20=?

now ,

             Sn=n/2[2a + (n-1)d

             S20=20/2[2*35+19*35]

                    =10[70+665]

                    =10*735

                    =7350          

Ans:-  sum of first 40+ve integers divisible by 6 is 4920.

          sum of 20+ve integers divisible by 5 or 7 is 7350.

I hope you can help me.

Thanks