Question

Find the sum of all 2digit natural number divisible by 5 full solution

Answer 1

Answer:

10+(n-1)*5=95,then n=18. Required sum=n/2(a+l)=18/2(10+95)=945.

Step-by-step explanation:

This answer is easy so I didn't give explanation

Answer 2

The two digit natural numbers which are divisible by 5 are in A.P with common difference 5,

So,

A.P = 5, 10, ... , 95

Here, a = 5, d = 5, l = 95

Find number of elements:

☛ L = a + ( n - 1)d

☛ 95 = 5 + 5n - 5

☛ 5n = 95

☛ n = 19

There are 19 two digit natural numbers divisible by 5.

We know,

$</p><p>\sf \quad S_{n} = \frac{n}{2} \{ a + l \}</p><p> \\ \\</p><p> \sf \rightarrow \quad S_{19} = \frac{19}{2} \{ 5 + 95 \} \\ \\</p><p> \sf \rightarrow \quad S_{19} = \frac{19}{\cancel{2}} \times \cancel{100} \\ \\</p><p> \sf \rightarrow \quad S_{19} = 19 \times 50 \\ \\</p><p> \sf \rightarrow \quad S_{19} = 950 \\ \\</p><p>$

Hence, The sum of first two digit natural numbers is 950