Question

find the sum of all odd number divisible by 3 between 1 and 1000

Answer 1

3 +9 +15 + ... + 999 = 
999 =3 +(n-1)6 
999 =6n -3 
6n =1002 
n=167 
S =n/2(a1 +a167) 
=167/2(3 +999) 
=167(1002)/2 
=83667

Answer 2

Answer:

$Sum \: of \: all \: odd \: \\numbers \:divisible \: by \:3 \\between \: 1 \: and \:1000\\=166833$

Step-by-step explanation:

3,6,9,12,....,999 are the odd numbers divisible by 3 are in A.P

First term (a)= 3,

$common \: difference (d)=a_{2}-a_{1}=6-3=3$

$Last\:term (l)=999$

$\implies a+(n-1)d=999$

$\implies 3+(n-1)3=999$

/* divide each term by 3,we get

$\implies 1+n-1=333$

$\implies n = 333$

$Now,\\Sum \: n \: terms (S_{n})=\frac{n}{2}(a+l)\\=\frac{333}{2}(3+999)\\=\frac{333}{2}\times 1002\\=333\times 501\\=166833$

Therefore,

$Sum \: of \: all \: odd \: \\numbers \:divisible \: by \:3 \\between \: 1 \: and \: 1000\\=166833$

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