Question

Find the sum of natural numbers from 10 to 200 whose remainder is 3 when divisible by 4

Answer 1

Answer:

200

Step-by-step explanation:

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Answer 2

The sum would be 5040

Step-by-step explanation:

Since, the first number which gives 3 remainder when divided by 4 from 10 to 200 = 11,

Second number = 15,

Third number = 19,

........................., so on,

The last number = 199,

Thus, we get an AP that represents the number from 10 to 200 whose remainder is 3 when divisible by 4,

11, 15, 19, .......199

First term, a = 11,

Common difference, d = 15 - 11 = 4,

Let n be the total number of terms,

Last term = a + (n -1) d= 11 + (n-1)4

But, 11 + (n-1)4 = 199

(n-1)4 = 188

n - 1 = 47

n = 48

Hence, the sum of the above AP,

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

$S_{48}=\frac{48}{2}[22+(48-1)4]$

$=24(22+47\times 4)$

$=24(22+188)$

$=24(210)$

= 5040

#Learn more:

Find sum of the AP

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