Question

17) Find the sum of all the natural numbers between 100 and 300 which are divisible by 5.

Answer 1

$\textbf{- Answer -}$

Here, between 100 and 300, the first number which is divisible by 5 is 105 and the last number which is divisible by 5 is 295, and the common difference is 5.

Let, the nth term is 295

⇒ (first number) + (n - 1)(common difference) = 295

⇒ 105 + (n - 1)(5) = 295

⇒ 5n - 5 = 295 - 105

⇒ 5n = 190 + 5

⇒ 5n = 195

⇒ n = 39

So, between 100 and 300, there are 39 Natural numbers which are divisible by 5.

Hence, the required sum of all the 39 Natural numbers (divisible by 5)

= 39/2 (first number + last number)

= 39/2 (105 + 295)

= 39/2 × 400

= 39 × 200

= 7800

#$\textbf{MarkAsBrainliest}$

Answer 2

First number which is divisible by 5 between 100 and 300 is 105 and last number is 295


Common difference = 5
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First term + (n - 1)(d) = last term

105 + (n -1)(5) = 295

=> (n - 1)(5) = 295 - 105

=> (n - 1)(5) = 190

=> n - 1 = 190/5

=> n - 1 = 38

=> n = 38 + 1

=> n = 39

So, there are 39 numbers which are divisible by 5 between 100 and 300

××××××××××××××××××××××

Then,


S(n) = (n/2) × [ last term + first term]

=> S(39) = (39/2) × [100 + 300]

=> S(39) = (39 × 400)/2

=> S(39) = 39 × 200

=> S(39) = 7800



Sum of Numbers which are divisible by 5 between 100 and 300 is 7800




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I hope this will help you


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