Question

find the sum of all multiples of 2 and 3 between 100 and 200

Answer 1

So AP is 102,108,114........198
then, d=6, a=102, an=198, n=?
So, we know that
an=a+(n-1)d
198=102+(n-1) 6
198-102=6n-6
96+6=6n
102=6n
So, 17 =n or
n=17
Now
Sn=n/2[a+an]
=17/2 [102+198]
=17/2[300]
=17 [150]
Sn= 2550

I hope it will help you
please mark as brainliest answer

Answer 2

multiples of 2 between 100 and 200 form and AP series
102,104,106....,198 with common difference 2
And the number of terms=(198-100)/2=49
sum of n term of an AP=(1st term+last term)n/2
                                     =(102+198)49/2
                                     =150*49=7350

Same for 3
102,105,108,....,198
Number of terms= (198-99)/3=99/3=33

Sum of the terms=(102+198)33/2
                           =150*33
                           =4950
Total sum=7350+49500=123000

now subtract the common multiple of both 2 and 3 =2550

SUM= 12300-2550=9750