Question

The sum of all natural numbers between 500 and 1000 which are divisible by 13

Answer 1

hi!
here is your answer !

The numbers between 500 & 1000
divisible by 13:

507 , 520, ........... 988

here

t2 - t1 = t3 - t2 = 13

This forms an A.P

a = 507 , d = 13 , tn = 988

We know that,

tn = a +(n - 1) d.......................[formula]
988 = 507 + (n - 1) 13
988 = 507 +13n - 13
988 = 494 + 13n
988 - 494 = 13n
494 = 13n
13n = 494
n = 494/13
n = 38

We know that,

$sn = \frac{n}{2}(2a + (n - 1)d)$


$s38 = \frac{38}{2}(2a+ (38 - 1)d)$


$s38 = 19(2a + 37d)$


$s38 = 19(2 \times 507 + 37 \times 13)$


$s38 = 19(1014 + 481)$


$s38 = 19(1495)$


$s38 = 28405$

Therefore,
The sum of all numbers between 500
& 1000 divisible by 13 is 28,405

hope this helps ! ! !

Answer 2

Answer:

S38 = 28405

Step-by-step explanation:

natural number between 500 and 1000 = 501,502,503...........999.

divisible of 13 in between 500 and 1000 = 39×13 =507

76×13 =988

507,520,533.......988.

a=507 , d=a2 - a1 , l=988=An

=520-507

=13

An=a+(n-1)d

988=507+(n-1)d

988-507=(n-1)13

481=(n-1)13

481/13=n-1

37=n-1

37 + 1 = n

38=n

Sn = n/2 [2a + (n-1)d]

S38 = 38/2 [2×507+(38-1)13]

=19[1014+37×13]

=19[1014 + 481]

S38 =19[1495]

S38 = 28405

The sum of all natural numbers between 500 and 1000 which are divisible by 13 is 28405

HOPE IT WILL HELP YOU !!