Question

Sum of all natural nos. between 100 and 200 which are divisible by 4

Answer 1

so sum of all natural numbers between 100 to 200 which are divisible by 4 are 3900

Answer 2

$\huge{\mathfrak{\green{S}}}{\mathfrak{o}}{\mathfrak{\orange{l}}}{\mathfrak{\red{u}}}{\mathfrak{\pink{t}}}{\mathfrak{\blue{i}}}{\mathfrak{\purple{o}}}{\mathfrak{\gray{n}}}$

3900

$\large{\blue{\mathrm{To \: Find :-}}}$

Number of terms(n)

Sum of all natural numbers which are divisible by 4.

$\large{\green{\mathrm{Given:-}}}$

A.P :- 100, 104 , 108,......... 200.

First term(a) = 100

Common Difference(d) = 4

Last term(L) = 200

$\large{\blue{\mathrm{Solution:-}}}$

Finding number of terms

Using Formula

$\huge{\boxed{\boxed{\red{A_{n} = a +(n - 1)d}}}}$

_______________[Put Values]

⇒200 = 100 + (n - 1)4

⇒200 = 100 + 4n - 4

⇒200 = 96 + 4n

⇒200 - 96 = 4n

⇒104 = 4n

⇒104/4 = n

⇒26 = n

$\large{\boxed{\orange{Number \: of \: terms(n) = 26}}}$

Finding sum

Using Formula

$\huge{\boxed{\boxed{\red{S_{n} = \frac{n}{2}[2a + (n - 1)d]}}}}$

_______________[Put Values]

$\sf{S_{n} = \frac{26}{2}[2 * 100 + (26 - 1){\times}4}]$

$\sf{S_{26} = 13(200 + 100)}$

$\sf{S_{26} = 13(300)}$

$\bf{S_{26} = 3900}$

${\red{S_{26}}} = {\blue{3900}}$

Same question answered by me :-

https://brainly.in/question/1630129