Question

The sum of all natural numbers between 100 and 200 which are multiples of 3 is

Answer 1

so we have to find sum of
102+105+108+.........+198

SUM= 33/2(102+198)
= 33/2(300)
=33(150)
=4950

Answer 2

Answer:

The “sum” of all the multiples of 3 between the numbers “100 and 200” are 4950.  

Solution:

The first term after 100 which is being divisible by the number 3 (a) = 102

The last term before 200 which is being divisible by the number 3 ($t_n$ )= 198

Common difference (d) = 3

Let the no. of terms be denoted as n.

We know that the formula,

Number of terms n

$\begin{array} { c } { t _ { n } = a + ( n - 1 ) d } \\\\ { 198 = 102 + ( n - 1 ) 3 } \\\\ { 198 - 102 = 3 n - 3 } \\\\ { 96 = 3 n - 3 } \\\\ { 3 n = 96 + 3 = 99 } \\\\ { n = \frac { 99 } { 3 } = 33 } \end{array}$

Thus, sum of all 33 terms

$\begin{array} { c } { = \frac { n } { 2 } ( 2 a + ( n - 1 ) d ) } \\\\ { = \frac { 33 } { 2 } ( 2 \times 102 + ( 33 - 1 ) 3 ) } \\\\ { = \frac { 33 } { 2 } ( 204 + 96 ) } \\\\ { = 4950 } \end{array}$

Hence, the “sum” of all the multiples of 3 between the numbers “100 and 200” are 4950.