Question

find the sum of natural number between 1and 140 which are divisible ny 5​

Answer 1

$\huge\mathfrak\blue{Answer:}$

To Find:

Sum of natural numbers between 1 and 140 which are divisible by 5.

Solution:

Natural numbers divisible by 5 from 1 and 140 are

5, 10, 15, 20, 25..........

Total terms = 140/5 = 28

Common difference = 5

First term = 5

Sum of all the numbers= n/2(a+T)

= 28/2 ( 5 + 140 )

= 14 (145)

= 2030

Hence, the sum of natural numbers between 1 and 140 which are divisible by 5 is 2030.

Answer 2

$\huge{\orange{\fbox{\green{\fbox{\mathcal{\orange{An}\blue{sw}\green{er}}}}}}}$

$\large{\orange{\fbox{\green{\fbox{\mathcal{\orange{No}\green{te}}}}}}}$

• $\orange{Here,\;we \;can\; go \;with\;an \;AP \;with\; the\; given \;statement}$

• $\blue{A.P = 5 , 10 , ......... 140}$

• $\green{a_{n} = a+(n-1)d}$

• $\orange{Here , a = 5 \; d = 5}$

• $\blue{140=5+(n-1)5}$

• $\green{n = 28}$

$\large{\green{\fbox{\orange{Sum}}}}$

$\green{S_{n}=\dfrac{n}{2}(a+a_{n})}$

• $\orange{S_{n}=\dfrac{28}{2}(5+140)}$

• $\blue{S_{n}=\dfrac{14}{1}(145)}$

• $\green{S_{n}=2030}$