Question

Sum of natural number from 200 to 600 which is neither divisible by 8 nor 12

Answer 1

Both 200 and 600 are multiples of 8. So right away we remove 200 and 600 in the beginning itself.

Now for the sum of numbers from 201 to 599.

a = 201, d = 1 and the number of terms from 201 to 599 = 399

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

= (399/2)[2*201 + (399–1)*1]

= (399/2)[402 + 398]

= 399*400 = 159600 …(1)

Next the AP which is a multiple of 8 is 208, 216 …..592

a = 208, d = 8

For the number of terms in this series: a1 =208, d = 8, an = 592

an = a1 + (n-1)d, so

592 = 208 +(n-1)*8,

592 = 208 + 8n - 8,

8n = 592- 208 +8 = 392, so n = 392/8 = 49

There are thus 49 terms from 208 to 592 and their sum is

S(8) = (49/2)[2*208 +(49–1)*8]

= (49/2)[416 + 48*8] = (49/2){ 416+ 384] = (49/2)*800 = 19600 …(2)

Next for the numbers that are multiples of 12 from a1 = 204 to an = 588, d = 12

an = a1 + (n-1)d, so

588 = 204 +(n-1)*12,

588 = 204 + 12n - 12

12n = 588 -204 + 12 = 396, or

n = 396/12 = 33.

Sum of numbers from 204 to 588 is as follows

S(12) = (33/2)[2*204 + (33–1)*12] = (33/2)[408 + 32*12]

= (33/2)[408 + 384) = (33/2)*792 = 13068 …(3)

So from the total sum from 201 to 599 deduct the sum of multiples of 8 and add the sum of multiples of 12. Therefore we have (1) - (2) + (3), or

159600 - 19600 + 13068 = 153068

Answer = 153068.