Question

The sum of all natural numbers between $ 1 $ and $ 100 $ which are multiples of $ 3 $ is
A. $ 1680 $
B. $ 1683 $
C. $ 1681 $
D. $ 1682 $

Answer 1

Hint: Here we are given a sequence indirectly and it is A.P. An arithmetic progression is defined as a mathematical sequence in which the difference between two consecutive terms is always a constant. Here, we need to apply the formula of the $ {n^{th}} $ term of an A.P to find the last term of our sequence. We are asked to calculate the sum of the given sequence. So, we need to apply the sum of n terms in A.P.
Formula to be used:
a) The formula to calculate the $ {n^{th}} $ term of the given arithmetic progression is as follows.
                               \[{a_n} = a + \left( {n - 1} \right)d\]
Where, $ a $ denotes the first term, $ d $ denotes the common difference, $ n $ is the number of terms, and $ {a_n} $ is the $ {n^{th}} $ term of the given arithmetic progression.
b) The formula to obtain the sum of n terms in A.P is as follows.
 $ {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $
Where, $ a $ denotes the first term, $ d $ denotes the common difference, $ n $ is the number of terms of A.P

Complete step by step answer:
The required sequence of the sum of all natural numbers between $ 1 $ and $ 100 $ which are multiples of $ 3 $ will be $ 3 + 6 + ..... + 99 $
Here, the first term is $ a = 3 $ and the common difference is $ d = 6 - 3 = 3 $
Also, the $ {n^{th}} $ _{term is} $ 99 $
Now, we shall apply the formula to calculate the $ {n^{th}} $ term of the given arithmetic progression
\[{a_n} = a + \left( {n - 1} \right)d\]
 $ \Rightarrow 99 = 3 + \left( {n - 1} \right)3 $
 $ \Rightarrow 99 = 3 + 3n - 3 $
 $ \Rightarrow 99 = 3n $
 $ \Rightarrow \dfrac{{99}}{3} = n $
 $ \therefore n = 33 $
Hence there are $ 33 $ terms in the sequence.
Now, we need to apply the formula of the sum of the n terms in A.P.
 $ {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $
 $ \Rightarrow {S_n} = \dfrac{{33}}{2}\left[ {2 \times 3 + \left( {33 - 1} \right)3} \right] $
             $ = \dfrac{{33}}{2}\left[ {6 + \left( {32} \right)3} \right] $
             $ = \dfrac{{33}}{2}\left[ {6 + 96} \right] $
             $ = \dfrac{{33}}{2} \times 102 $
             $ = 33 \times 51 $ $ = 1683 $
So, the correct answer is “Option B”.

Note: An arithmetic sequence or arithmetic progression is defined as a mathematical sequence in which the difference between two consecutive terms is always a constant. An arithmetic progression is abbreviated as A.P. We also need to know the three important terms, which are as follows.
A common difference $ \left( d \right) $ is a difference between the first two terms.
 $ {n^{th}} $ term\[({a_n})\]
And, Sum of the first $ n $ terms\[({S_n})\]