Question 848987
Calculate the sum of all natural numbers from 100 to 200 both inclusive which are not divisible by 6
<pre>
First we'll find the sum of all natural numbers from 100 to 200, inclusive.
Then we'll find the sum of all the multiples of 6 from 102 to 198, inclusive.
Then we'll subtract them.

All natural numbers from 100 to 200, inclusive.

That's the arithmetic sequence 100,101,102,103,...,200
with a<sub>1</sub>=100, d=1, and a<sub>n</sub>=200


a<sub>n</sub> = a<sub>1</sub> + (n-1)d

To find the number of terms:

200 = 100 + (n-1)1
100 = n-1
101 = n

We use the sum formula

S<sub>n</sub> = {{{n/2}}}(a<sub>1</sub> + a<sub>n</sub>) 

S<sub>101</sub> = {{{101/2}}}(100 + 200) = {{{expr(101/2)(300)}}} = 15150

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The multiples of 6 from 102 to 198, inclusive, is the arithmetic
sequence 102,108,114,...,198, with a<sub>1</sub>=100, d=1, and a<sub>n</sub>=200

To find the number of terms:

a<sub>n</sub> = a<sub>1</sub> + (n-1)d

198 = 102 + (n-1)6
198 = 102+6n-6 
198 = 96+6n
102 = 6n
 17 = n

We use the sum formula

S<sub>n</sub> = {{{n/2}}}(a<sub>1</sub> + a<sub>n</sub>) 

S<sub>7</sub> = {{{17/2}}}(102 + 198) = {{{expr(17/2)(300)}}} = 2550

Subtracting,  15150-2550 = 12600.

Answer = 12600.

Edwin</pre>