Question

Find the sum of the integers between 10 and 30 including 10 and 30 which are not divisible by 3.​

Answer 1

Answer:

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum =

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2n

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2n

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2n (a+l)=

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2n (a+l)= 2

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2n (a+l)= 217

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2n (a+l)= 217

The odd integers between 2 and 100 which are divisible by 3 are 3, 9, 15, 21, ..., 99.Here, a=3 and d=6.a n =99⇒a+(n−1)d=99⇒3+(n−1)×6=99⇒6(n−1)=96⇒n−1=16⇒n=17Therefore,Required sum = 2n (a+l)= 217 (3+99)=867

Answer 2

Answer:

5

gudyoxkhchocljxhdhlxohd