Question

find sum of all natural no. s between 500 and 1000 which are divisible by 13
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Answer 1

Step-by-step explanation:

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)sn=

2

n

(2a+(n−1)d)

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

2

38

(2a+(38−1)d)

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

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

s38 = 19(1014 + 481)s38=19(1014+481)

s38 = 19(1495)s38=19(1495)

s38 = 28405s38=28405

Therefore,

The sum of all numbers between 500

& 1000 divisible by 13 is 28,405

Answer 2

Answer:

First number after 500 and divisible by 13 = 507 and number just less than 1000 and divisible by 13 = 998

Hence sequence

507,520,.........,988(nterms)

∴988=507+(n−1)13

∴(n−1)=481/17=37

n=38

So, there are 38 numbers between 500 & 1000 which are divisible by 13.