Question

find the sum of all integer between 5 and 500 which are divisible by 7​

Answer 1

Step-by-step explanation:

the first integer divisible by 7 b/w 5 and 500

is 7

and last is 497

a=7 l=497

and 497 comes at 71

therefore n=71

Sn=n/2(a+l)

=71/2(7+497)

=71/2(504)

=71(252)

=17892

Answer 2

Answer:

The sum of all integers between 5 and 500, which are divisible by 7, is 17892.

Step-by-step explanation:

Let the first integer = 7

The last integer = 497

We assume the the all integer between 5 and 500, which are divisible by 7, are in Arithmetic Progression.

Then first term of A.P. (a) = 7

Last term of A.P. (l) = 497

Difference = 7  { because integers are divisible by 7 }

Total term (n) = ?

Using formula-

$T_{n} = a + (n - 1)d$

Where $T_{n}$ is nth term of A.P.

a is first term of A.P.

n is total term of A.P.

d is common difference

We have-  $T_{n} = 497$

a + (n-1)d = 497

7 + (n - 1)×7 = 497

(n - 1)×7 = 497 - 7

(n - 1)×7 = 490

$n-1 = \dfrac{490}{7}$

n - 1 = 70

n = 70 + 1

n = 71

i.e there are 71 integers between 5 and 500, which are divisible by 7.

Using formula-

$S_{n} = \dfrac{n}{2}[a+l]$

Then sum of 71 integers-

$S_{n} = \dfrac{71}{2}[7 + 497]$

              = $\dfrac{71}{2}*504$

             = 71 × 252

$S_{n} = 17892$

So, the sum of all integers between 5 and 500, which are divisible by 7, is 17892.