Question

Find the sum of all integers between 50 and 500 which are divisible by 9.

Answer 1

Answer:

Sn=(89/2)[55+495]=24475

Step-by-step explanation:

The first integer after 50, which is divisible by 5 is 55 and the last integer before 500, divisible by 5 is 495. Therefore, the corresponding AP series will be… 55,60,65,70,………..485,490,495.

Here, first term (a) = 55, Last term (an) =495, common difference (d)=5;

Thus, we know

an=a+(n−1)×d  ( where, n is the number of terms , which is unknown )

495=55+(n−1)×5  

n−1=(495–55)/5  

n=89  we also know sum of series formula for AP is  Sn=(n/2)[a+an]  

Sn=(89/2)[55+495]=24475

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Answer 2

((((((((******i have taken the number 7 can you please do the sum next by 9******))))))))))

Answer:

The sum of all integers between 50 and 500, which are divisible by 7 is 17696.

Step-by-step explanation:

Given :  

Integers between 50 and 500, which are divisible by 7  are 56, 63, 70,....... 497

Here,  

a = 56, d = 7, an(l) = 497

By using the formula ,an = a + (n - 1)d

497 = 56 + (n -1)7

497 - 56 =  7(n -1)

441 = 7(n -1)

n -1 = 441/7

n -1 = 63

n = 63 + 1  

n = 64

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

⇒ S64 = 64/2 (a + l)

⇒ S64 = 32 (56 + 497)

⇒ S64 = 32 (553)  

⇒ S64 = 17696

Hence, the sum of all integers between 50 and 500, which are divisible by 7 is 17696.

HOPE THIS ANSWER WILL HELP YOU….

((((((((******i have taken the number 7 can you please do the sum next by 9******))))))))))