Question

find the sum of all natural numbers between 1 and 100,which are divisible by 3.

Answer 1

Answer:

The sum of all natural numbers between 1 and 100, which are divisible by 3 are 1683.

Step-by-step explanation:

Given :  

Natural numbers between 1 and 100 which are divisible by 3 are 3, 6, 9, 12, ……….99

Here,  first term , a = 3, last term, l = 99, common Difference, d = 6 - 3 = 3

By using the formula ,an = a + (n - 1)d

99 = 3 + (n - 1)3

99 = 3 + 3n - 3

99 = 3n

n = 99/3

n = 33

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

S33 = 33/2 [3 + 99]

S33 = 33/2 × 102

S33 = 33 (51)  

S33 = 1683

Hence, the sum of all natural numbers between 1 and 100, which are divisible by 3 are 1683.

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Answer 2

Answer:

$Sum \: of \:all\: natural\\\: numbers\:1\:and \:100, \\which \:are\: divisible \: 3= 1683$

Step-by-step explanation:

3,6,9,12 ,...,99 are the numbers between 1 and 100 ,which are divisible by 3 are in A.P

first term (a) = 3,

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

last term (l) = 99

$\implies a+(n-1)d=99$

$\implies 3+(n-1)3=99$

/* Divide each term by 3,we get

$\implies 1+n-1=33$

$\implies n = 33$

$Sum \:of \: n \: terms (S_{n})=\frac{n}{2}(a+l)$

$Sum \: of \: 33\:terms\\ =S_{33}\\=\frac{33}{2}(3+99)\\=\frac{33}{2}\times 102\\=33\times 51\\=1683$

Therefore,

$Sum \: of \:all\: natural\\\: numbers\:1\:and \:100, \\which \:are\: divisible \: 3= 1683$

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