Question

find the sum of integers between 100 and 200 divisible by 9 or not divisible by 9

Answer 1

Answer:

Step-by-step explanation:

here series will be 108,117,.......................................198

So a=108

    l=198

   d=117-108=9

n= ?

nth term= a + (n-1)d

198= 108+(n-1)9

198= 108+9n-9

198-99= 9n

99/9= n

n= 11

Now, sum of n terms

Sum of first 11 terms=$\frac{n}{2}(a+l)$

=11/2(108+198)

=11/2(306)

=11×153

=1683

Answer 2

Given: integers between 100 and 200.

To find: Sum of integers between 100 and 200 not divisible by 9.

The sum of the integers between 100 and 200 which is not divisible by 9

= (sum of total numbers between 100 and 200) – ( sum of total numbers between 100 and 200 which is divisible by 9).

Let the required sum be S

S = S1 - S2

Where S1 is the sum of AP 101, 102, 103, - - - , 199

And S2 is the sum of AP 108, 117, 126, - - - - , 198

For S1  

First term, a = 101

Common difference, d = 199

Let n be no of terms

Then,

an = a + (n - 1) d

199 = 101 + (n - 1 )1

98 = (n - 1)

n = 99

now, Sum of this AP

 [ as last term is given]

= 99(150 )

= 14850

For S1

First term, a = 108

Common difference, d = 9

Last term, an = 198

Let n be no of terms

Then,

an = a + (n - 1) d

198 = 108 + (n - 1 )9

10 = (n - 1 )

n = 11

now, Sum of this AP

= 11(153)

= 1683

Therefore

S = S1 - S2

= 14850 - 1683

= 13167