Question

find the sum of all multiples of 2 or 3 between 100 and 200 (100 and 200 are not included)

Answer 1

for 2:
by using the formula s = a+(n-1)d
where s is the last term, a is the 1st term, d is the common difference
the first and last terms that are divisible by 2 between 100 to 200 without including 100 and 200 are 102 (51 times)  and 198 (99 times)
and the common difference is 2 since they are the multiples of 2
so, a=102, l /s=198, d=2
by substituiting them in formula we get
198=102+(n-1)2
n-1=198-102/2
n-1=96/2
n-1=48
n=48+1
n=49
so the number of terms that are divisible by 2 is 49 terms
S49 = 49/2 (102+198)
       = 49/2×300
       = 49×150
       = 7350
there fore the sum of the  49 terms is 7350

for 3:
by using the formula s = a+(n-1)d
where s is the last term, a is the 1st term, d is the common difference
the first and last terms that are divisible by 3 between 100 to 200 without including 100 and 200 are 102 (34 times) and 198 (66 times)
and the common difference is 3 since they are the multiples of 3
so, a=102, s=198, d=3
by substituiting them in formula we get
198=102+(n-1)3
n-1=198-102/3
n-1=96/3
n-1=32
n=32+1
n=33
so the number of terms that are divisible by 3 is 33 terms
Sn = n/2(a+l)
S49 = 33/2 (102+198)
       = 33/2×300
       = 33×150
       = 4950
there fore the sum of the 33 terms is 4950

Answer 2

F=102
D=2
XN=198
n=number of terms
n=$\frac{ x_{n}-f }{d} +1$
=198-102/2+1=49
SN=$\frac{n}{2} (F+ x_{n} )$
=$\frac{49}{2} (102+198)=7350$

f=102
d=3
xn=198
n=$\frac{ x_{n}-f }{d} +1$=33
 sn=$\frac{n}{2} (F+ x_{n} )$
   =33/2(102+198)=4950