Question

Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9

Answer 1

Answer:

i) 1683

ii)13367

Step-by-step explanation:

i) Numbers between 100 to 200 that are divisible by 9 ,form an A.P.

A.P.: 108 , 117 , .........., 198

here, a = 108 , an = 198 ,

d = a2-a1 = 117-108 = 9

We know that,

an = a +(n-1)×d

By applying the values we get,

198 = 108 + (n-1) × 9

198 = 108 + 9n - 9

198 = 99 + 9n

9n = 198 - 99

9n = 99

n = 11

Hence, number of terms between 100 to 200 that are divisible

by 9 = n = 11

Now,

Sn = n(a+an)/2

By applying values,we get

S11 = 11(108+198) / 2

S11 = ( 11 × 306 ) / 2

S11 = 11 × 153

S11 = 1683

Hence the sum of integers that are divisible by 9 between 100 and 200 is 1683.

ii) New A.P. of numbers between 100 to 200 =>

A.P.: 101 , 102, 103,......................, 200

Here, a = 101 , an = 200

d = a2- a1 = 102-101 = 1

We know that,

an = a + (n-1)× d

200 = 101 + (n-1) × 1

200 = 101 + n - 1

200 = 100 + n

n = 200-100 = 100

Hence numbers of terms between 100 and 200 = n = 100

Now,

Sn = n(a+an) / 2

By applying values,we get

S100 = 100(101+200) / 2

S100 = 50 × 301

S100 = 15050

Now, sum of integers between 100 to 200 that are not divisible

by 9 = sum of integers between 100 to 200 - sum of integers between 100 to 200 that are divisible by 9

= S100 - S11

= 15050 - 1683

= 13367

Hence sum of integers between 100 and 200 that are not divisible by 9 = 13367