Question

Find the sum of all natural number between 100 and 1000 which are multiples of 5

Answer 1

As the question is not clear whether 100 and 1000 would be included in the series or not, so I am assuming that 100 and 1000 won't be included.

So, the first number of the series which is greater than 100 and is divisible by 5 is 105 and the number last number of the series which is less than 1000 and divisible by 5 is 995

Number of terms in the series be N

∴ last term = first term + (N-1) common difference of the series

=> 995 = 105 + (N-1) 5

=> N = 179

Now, sum of the series

= N/2(2x first term + (N-1) common difference )

= 179/2 (2 x 105 + (179-1) 5)

= 98450 = Ans

Answer 2

$<b>$Multiples of 5 between 100 and 1000

So,

Smallest multiple of 5 between 100 and 1000 will be 105

Largest multiple of 5 between 100 and 1000 will be 995

$\rule{200}{2}$

This forms an ARITHMETIC PROGRESSION :-

The AP will be as follows

$\boxed{105,110,115....995}$

$\rule{200}{2}$

a = 105

d = 5

Now ,

We know that

$\boxed{a_{n} =a+(n-1)d}$

So ,

$105 + (n - 1)5 = 995 \\ \\ \\ (n - 1)5 = 995 - 105 = 890 \\ \\ \\n - 1 = \frac{890}{5} = 178 \\ \\ \\ n = 178 + 1 = 179$

There are 179 terms which are divisible by 5 between 100 and 1000.

$\rule{200}{2}$

We know that

$\boxed{s_{n} =\frac{n}{2}(a+l)}$

So,

$\boxed{s_{179} =\frac{179}{2}(105+995)}$

$= \frac{179}{2} \times 1100 \\ \\ \\ = 179 \times 550 \\ \\ \\ = 98450$

$\boxed{s_{179} =98450}$

$\rule{200}{2}$