# Find the sum of all multiples of 3 b/w 100 and 200

2
by ishuvyshu045

2015-03-06T17:48:49+05:30

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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
:P : P :P :P
no yaar after answered i saw the question again n i am answering the similar question so it became late 2 edit
oh..no prob dear..!!..:)
:)
plz click at thanks if it helps u
2015-03-06T17:49:29+05:30
Multiples of 3 b/w 100 and 200 are :
102, 105. 108,........... 198
An = A + (n-1) d
198 = 102 + (n-1) 3
96= 3n-3
99= 3n
n = 33
Sn = n/2 (a+l)
S₃₃ = 33/2 (102+198)
S₃₃ = 33/2 ( 300) = 4950
silly.!!!!!!!????? howcome..?
wht i did..?
nthnq yaar
:)
kk...! so where r u frm?