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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

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