Question

Find the sum of all natural numbers not exceeding 1000 which are divisible by 4 but not by 8.
A. 62500
B. 62800
C. 64000
D. 65600

Answer 1

Hint: We will find the required sum by first separately calculating the sum of all natural numbers till 1000 divisible by 4 and that divisible by 8. The sum of all numbers divisible by 4 will also contain the sum of all numbers divisible by 8. Hence, the required sum will come out as the difference of the before mentioned two sums. We will find those sums by forming two separate APs and using the formula for the sum of an AP given by $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$. Then finding the difference, we will find the required sum.

Complete step by step answer:
Here, we have to find the sum of all natural numbers not exceeding 1000 which are divisible by 4 but not by 8.
For this, we will first separately find the sum of all natural numbers till 1000 divisible by 4 and that divisible by 8.
Now, the first natural number divisible by 4 is 4 itself. Till 1000, the last number divisible by 4 is 1000. Now every successive number divisible by 4 has a difference of 4 from the previous number. thus, if we form it as an AP, we can see that:
a=4
d=4
\[{{a}_{n}}=1000\]
Where, a is the first term, d is the common difference and ${{a}_{n}}$ is the $n^{th}$ term.
Now, we know that in an AP:
${{a}_{n}}=a+\left( n-1 \right)d$
Putting the required values in this formula, we get:
$\begin{align}
  & {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 1000=4+\left( n-1 \right)4 \\
 & \Rightarrow 996=\left( n-1 \right)4 \\
 & \Rightarrow n-1=\dfrac{996}{4}=249 \\
 & \Rightarrow n=250 \\
\end{align}$
Hence, there are a total of 250 terms.
Now, we know that the sum on ‘n’ terms of an AP is given as:
$S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Hence, the sum of all natural numbers not exceeding 1000 and divisible by 4 is given as:
$\begin{align}
  & {{S}_{1}}=\dfrac{250}{2}\left( 2\left( 4 \right)+\left( 250-1 \right)4 \right) \\
 & \Rightarrow {{S}_{1}}=\dfrac{250}{2}\left( 8+1000-4 \right) \\
 & \Rightarrow {{S}_{1}}=\dfrac{250}{2}\left( 1004 \right) \\
 & \Rightarrow {{S}_{1}}=125500 \\
\end{align}$
Thus, the sum of all terms divisible by 4 until 1000 is 125500.
Now, when we see the numbers divisible by 8, we know that the first natural number divisible by 8 is 8 and the last number not exceeding 1000 divisible by 8 is 1000. We also know that in there is a difference of 8 between every consequent term in the numbers divisible by 8.
Hence, if we form it as an AP, we will get:
a=8
d=8
${{a}_{n}}=1000$
Now, we know that in an AP:
${{a}_{n}}=a+\left( n-1 \right)d$
Thus, putting in the required values in this formula we get:
$\begin{align}
  & {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 1000=8+\left( n-1 \right)8 \\
 & \Rightarrow 992=\left( n-1 \right)8 \\
 & \Rightarrow n-1=\dfrac{992}{8}=124 \\
 & \Rightarrow n=125 \\
\end{align}$
Hence, there are a total of 125 terms.
Now, the sum of all the terms is given as:
$\begin{align}
  & {{S}_{2}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\
 & \Rightarrow {{S}_{2}}=\dfrac{125}{2}\left( 2\left( 8 \right)+\left( 125-1 \right)8 \right) \\
 & \Rightarrow {{S}_{2}}=\dfrac{125}{2}\left( 16+1000-8 \right) \\
 & \Rightarrow {{S}_{2}}=\dfrac{125}{2}\left( 1008 \right) \\
 & \Rightarrow {{S}_{2}}=63000 \\
\end{align}$
Thus, the sum of all terms divisible by 8 until 1000 is 63000.
Now, the sum of all natural numbers divisible by 4 till 1000 but not by 8 will be the difference of the sum of all numbers divisible by 4 and that by 8.
Hence, the required sum is given as:
\[\begin{align}
  & {{S}_{2}}-{{S}_{1}} \\
 & \Rightarrow 125500-63000 \\
 & \therefore 62500 \\
\end{align}\]
Thus, the required sum is 62500.

So, the correct answer is “Option A”.

Note: There is also another formula for the sum of an AP given as:
$S=\dfrac{n}{2}\left( a+l \right)$
Where, l is the last term of the series.
It is derived from the same formula. The derivation is given as:
$\begin{align}
  & S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\
 & \Rightarrow S=\dfrac{n}{2}\left( a+a+\left( n-1 \right)d \right) \\
\end{align}$
Now, we know that $a+\left( n-1 \right)d={{a}_{n}}$
Here, if the $n^{th}$ term is the last term, we can say that ${{a}_{n}}=l$.
Hence, we get:
$\begin{align}
  & S=\dfrac{n}{2}\left( a+a+\left( n-1 \right)d \right) \\
 & \Rightarrow S=\dfrac{n}{2}\left( a+l \right) \\
\end{align}$