Question

find the sum of all the natural number between 300 and 500 which are divisible by 5

Answer 1

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

Answer 2

first number which is divisible by 5 after 300 is 305

so , a = 305

difference in each number is 5

so d = 5

last number which is divisible by 5 before 500 is 495

so , tn = 495

let first find how many such numbers are there ,

so tn = a + ( n - 1 ) d

495 = 305 + ( n - 1 ) × 5

( n - 1 ) × 5 = 495 - 305

( n - 1 ) × 5 = 190

n - 1 = 38

n = 39

so now sum is ,

Sn = n/2 [2a+(n-1)d]

Sn = 39 / 2 [2 × 305 + (39 - 1) × 5 ]

Sn = 39 / 2 [610 + 38 × 5 ]

Sn = 39 / 2 [610 + 190 ]

Sn = 39 / 2 [800]

Sn = 39 × 400

Sn = 15600

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for divisible by 11

first number which is divisible by 11 after 300 is 308

so , a = 308

difference in each number is 11

so d = 11

last number which is divisible by 11 before 500 is 495

so , tn = 495

let first find how many such numbers are there ,

so tn = a + ( n - 1 ) d

495 = 308 + ( n - 1 ) × 11

( n - 1 ) × 11 = 495 - 308

( n - 1 ) × 11 = 187

n - 1 = 17

n = 18

so now sum is ,

Sn = n/2 [2a+(n-1)d]

Sn = 18 / 2 [2 × 308 + (18 - 1) × 11 ]

Sn = 9 × [616 + 17 × 11 ]

Sn = 9 × [616 + 187 ]

Sn = 9 × 803

Sn = 7227