Question

What is the sum of all positive integers up to 1000, which are divisible by 5 and are not divisible by 2?
A. 10,050
B. 5050
C. 5000
D. 50,000

Answer 1

Hint: We will use the concept that we will take the difference of sum of all multiples of 5 upto 1000 and sum of multiples of 5 divisible by 2 which gives us the required value. Also, we know that a number which is multiples of 5 and divisible by 2 is divisible by 10. We will also use the formula of A.P as follows:
$\begin{align}
  & {{T}_{n}}=a+\left( n-1 \right)d \\
 & {{S}_{n}}=\dfrac{n}{2}\left( a+l \right) \\
\end{align}$
Where, a is the first term, l is the last term and d is the common difference of an A.P.

Complete step-by-step solution
We have been asked to find the sum of all positive integers up to 1000, which are divisible by 5 and are not divisible by 2.
= (sum of all numbers divisible by 5 upto 1000) – (sum of multiples of 5 divisible by 2)
= (sum of all multiples of 5 upto 1000) – (sum of all multiples of 10 upto 1000)
Now, the multiples of 5 upto 100 are as follows:
5, 10, 15, 20,……1000
It form an A.P with first term (a) = 5,
Common difference (d) = 5 and
The last term (l) = 1000
Let ${{n}^{th}}$ term be 1000.
We know the ${{n}^{th}}$ term of an A.P is given by,
$\begin{align}
  & {{T}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow a+\left( n-1 \right)d=1000 \\
 & \Rightarrow 5+\left( n-1 \right)5=1000 \\
 & \Rightarrow \left( 1+\left( n-1 \right) \right)5=1000 \\
 & \Rightarrow 1+n-1=\dfrac{1000}{5} \\
 & \Rightarrow n=200 \\
\end{align}$
Now, using the sum formula for an A.P i.e. ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$, we get
$\begin{align}
  & \Rightarrow {{S}_{200}}=\dfrac{200}{2}\left( 5+1000 \right) \\
 & =100\left( 1005 \right) \\
 & =100500 \\
\end{align}$
Similarly, the multiples of 10 upto 1000,
10, 20, 30, ……, 1000.
It forms an A.P with first term (a) = 10, last term (l) = 1000 and common difference (d) = 10.
Using ${{T}_{n}}=a+\left( n-1 \right)d$, we get,
$\begin{align}
  & a+\left( n-1 \right)d=1000 \\
 & \Rightarrow 10+\left( n-1 \right)10=1000 \\
 & \Rightarrow 10\left( 1+n-1 \right)=1000 \\
 & \Rightarrow 1+n-1=\dfrac{1000}{10} \\
 & \Rightarrow n=100 \\
\end{align}$
Using the sum formula, we get,
$\begin{align}
  & \Rightarrow {{S}_{100}}=\dfrac{100}{2}\left( 10+1000 \right) \\
 & =50\left( 1010 \right) \\
 & =50500 \\
\end{align}$
So, the required sum $=100500-50500$
$=50,000$
Therefore, the correct option is D.

Note: We can also solve it in less time if we try to find the numbers which are divisible by 5 but not divisible by 2. Then we get a series like 5, 15, 25, 35,….. which forms an A.P and by using the sum formula we get the required sum directly.