Question

(b) Find the sum of all the multiples of 2 and 3 but 6 between 100 and 200?​

Answer 1

Answer:

there fore the sum of the 33 terms is 4950

Step-by-step explanation:

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 substituting 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)

S 49 = 33/2 (102+198)

      = 33/2×300

      = 33×150

      = 4950

there fore the sum of the 33 terms is 4950