Question

find the sum of all odd numbers between 2 and 100 divisible by 3 ​

Answer 1

Answer:

a=3

nth term =100

d=3

Now;

nth term= a+d*(n-1)

= 100 = 3+ 3*(n-1)

= 100-3 = 3*(n-1)

= 97 = 3*(d-1)

= 97 = 3n-3

= 100 = 3n

=100/3 = n

33.3 = n (Approx. 33)

Therefore number of terms divisible by 3 are 33.

Hence number of odd numbers dividible by 3= 33+1/2

                                                                              = 17

Therefore,

Required sum =   2* n (1st term+last term)=  2 *17 (3+99)=867

Answer 2

Answer:

odd×odd is always odd including 1 and 3=3

is 3(1+3+5+7+9+11+13+15....33)

as 33 is the last no . multiplied by 3 upto 100

sum is use sum of aps formula for addition= {n/2×(2a+(n-1)d} therefore 259×3=867