Question

class 10 find the sum of all natural number less than 100, which are divisible by 5​

Answer 1

Answer:

1200

Step-by-step explanation:

The smallest natural numbr divisible by 4=4

the largest number inside 100 =96

thus a=4

d=8-4=4

n=?

an=96

an=a+(n-1)d

96=4+(n-1)4

96-4=4n-4

92+4=4n

96=4n

n=96/4

n=24

you can directly find n by directly multiplying 96/4

now,

sn=n/2(2a+(n-1)d)

or

sn=n/2(a+l)

so when we do it,we get

sn=24/2(4+96)

sn=1200

hope it helps!!

please mark as brainliest if it helped!!!

Answer 2

Given ,

The AP is 5 , 10 , 15 , .... 95

Here ,

First term , a = 5

Common difference , d = 5

Last term , l = 95

We know that , the nth term of an AP is given by

$\green{ \large \sf \fbox{ \fbox{A_{n} = a + (n - 1)d }}}$

Substitute the values , we obtain

$\hookrightarrow \sf 95 = 5 + (n - 1)5 \\ \\\hookrightarrow \sf 90 = (n - 1)5 \\ \\\hookrightarrow \sf 18 = n - 1 \\ \\\hookrightarrow \sf n = 19$

Thus , number of term of AP is 19

Thus , number of term of AP is 19Now , the sum of first n terms of AP is given by

$\large \pink{ \sf\fbox{ \fbox{S_{n} = \frac{n}{2} (a + l) }}}$

Substitute the values , we obtain

$\hookrightarrow \sf S_{19} = \frac{19}{2} (5 + 95) \\ \\\hookrightarrow \sf S_{19} = \frac{19}{ \cancel{2}} \times \cancel{100} \\ \\\hookrightarrow \sf S_{19} = 19 \times 50 \\ \\ \hookrightarrow \sf S_{19} = 950$

$\large \red{\sf \underline{ \underline{ \: \: Or \: \: \: }}}$

$\hookrightarrow \sf 5 + 10 + 15 + .... + 95 \\ \\\hookrightarrow \sf 5(1 + 2 + 3 + .... + 19) \\ \\\hookrightarrow \sf 5( \frac{19(19 + 1)}{2} ) \: \: \: \bigg\{ sum \: of \: first \: n \: natural \: no. = \frac{n(n + 1)}{2} \bigg\} \\ \\\hookrightarrow \sf 5 \times 19(10) \\ \\\hookrightarrow \sf 950$

Hence , the sum of all natural number less than 100 , which are divisible by 5 is 950