18. Verify 576 is divisible by 2.3.6 and 9?

19. Write all the numbers between 100 and 200 which are divisible by 6?

20. Find the value of

3 1 1

4 3 4

21. The product of two numbers is 25. If one of the number is oz. then find the

22. The length of a rectangular field is 12 – mts and its area is 65*

Find its br

2

# 18. Verify 576 is divisible by 2.3.6 and 9?

### 2 thoughts on “18. Verify 576 is divisible by 2.3.6 and 9?<br />19. Write all the numbers between 100 and 200 which are divisible by 6?<br />20.”

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Answer:subject plz mark me briliest answer

Answer:18. 6

19. First no. between 100 and 200 that is divisible by 6 is 102

The last no. between 100 and 200 that is divisible by 6 is 198

Now the numbers between 100 and 200 that is divisible by 6:

102,102+6,102+6+6 ,….

So, it forma an AP

a = first term = 102

d = common difference = 6

a_n=198a

n

=198

Formula of nth term = a_n=a+(n-1)da

n

=a+(n−1)d

198=102+(n-1)6198=102+(n−1)6

198-102=(n-1)6198−102=(n−1)6

96=(n-1)696=(n−1)6

\frac{96}{6}=n-1

6

96

=n−1

16=n-116=n−1

17=n17=n

Sum of n terms = s_n=\frac{n}{2}(2a+(n-1)d)s

n

=

2

n

(2a+(n−1)d)

Substitute n =17

s_{17}=\frac{17}{2}(2(102)+(17-1)6)s

17

=

2

17

(2(102)+(17−1)6)

s_{17}=2550s

17

=2550

Hence the sum of integers between 100 and 200 that are divisible by 6 is 2550