Question

The sum of the numbers from 300 to 700 which are divisible by 5 ? (300 and 700 are inclusive)

Answer 1

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

Given:

• We have been given a question related to Arithmetic Progression

To Find:

• We have to find the sum of multiples of 5 between 300 and 700
• 300 and 700 are inclusive

Solution:

Let us form an AP of multiples of 5 between 300 and 700

The AP will be as follows

300 , 305 , 310 ................ 690 , 695 , 700

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On observing the series

$\boxed{\sf{First \: term ( a ) = 300}}$

$\boxed{\sf{Common \: Difference ( d ) = 5}}$

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$\underline{\large\mathfrak\orange{According \: to \: the \: Question}}$

$\implies \sf{Last \: Term = 700}$

$\implies \sf{a_n = 700}$

$\boxed{\sf{\pink{ a_n = a + ( n - 1 ) \: d}}}$

$\implies \sf{a + ( n - 1 ) \: d= 700}$

On Putting the Values:

$\\ \implies \sf{300 + ( n - 1 ) \: 5 = 700}$

$\implies \sf{5 \: ( n - 1 )= 400}$

$\implies \sf{n - 1 = \dfrac{400}{5}}$

$\implies \sf{n - 1 = 80}$

$\implies \boxed{\sf{n = 81}}$

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$\underline{\large\mathfrak\purple{Sum \: is \: as \: follows}}$

Let the required sum = S

$\implies \sf{ S = \dfrac{n}{2} \: ( First \: Term + Last \: term )}$

$\implies \sf{ S = \dfrac{n}{2} \: ( 300 + 700 )}$

Putting n = 81

$\implies \sf{ S = \dfrac{81}{2} \times 1000}$

$\implies \sf{ S = 81 \times 500}$

$\implies \sf{ S = 40,500 }$

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$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\large\boxed{\sf{\red{Required \: Sum = 40,500}}}$

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Answer 2

Answer:

40,500

Step-by-step explanation:

A number is said to be divisible with 5 when it's unit place is 5 or 0.

Therefore the no. between 300 to 700 which are divisible by 5 are:

300, 305, 310, 315, 320, 325, 330, 335, 340, 345, 350, 355, 360, 365, 370, 375, 380, 385, 390, 395, 400, 405, 410, 415, 420, 425, 430, 435, 440, 445, 450, 455, 460, 465, 470, 475, 480, 485, 490, 495, 500, 505, 510, 515, 520, 525, 530, 535, 540, 545, 550, 555, 560, 565, 570, 575, 580, 585, 590, 595, 600, 605, 610, 615, 620, 625, 630, 635, 640, 645, 650, 655, 660, 665, 670, 675, 680, 685, 690, 695, 700.

Sum of these numbers are:

300 + 305 + 310 + 315 + 320 + 325 + 330 + 335 + 340 + 345 + 350 + 355 + 360 + 365 + 370 + 375 + 380 + 385 + 390 + 395 + 400 + 405 + 410 + 415 + 420 + 425 + 430 + 435 + 440 + 445 + 450 + 455 + 460 + 465 + 470 + 475 + 480 + 485 + 490 + 495 + 500 + 505 + 510 + 515 + 520 + 525 + 530 + 535 + 540 + 545 + 550 + 555 + 560 + 565 + 570 + 575 + 580 + 585 + 590 + 595 + 600 + 605 + 610 + 615 + 620 + 625 + 630 + 635 + 640 + 645 + 650 + 655 + 660 + 665 + 670 + 675 + 680 + 685 + 690 + 695 + 700 = 40,500

                              OR Simple way to do is:

First number to be divisible by 5 in between 300 to 700 is 300

Last number to be divisible by 5 in between 300 to 700 is 700

So in an AP (Arithmetic Progression)

$a^{n}$ = a + (n-1)d

Here, a = 300

$a^{n}$ = 700

d (common difference between each factors of 5) = 5

Therefore, 700 = 300 + (n-1)5

=> 700 - 300 = (n-1)5

=> 400/5 = n-1

=> 80 = n-1

=> n = 80 + 1

=> n = 81

Let  the total sum be S:

S = $\frac{n}{2}$ (a + $a^{n}$ )

S = $\frac{81}{2} (300 + 700)$

S = $\frac{81}{2}$ × 1000

S = 81 × 500

S = 40,500