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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