Question

Ankush was given a problem of adding a certain number of consecutive natural numbers starting from $1$, by mistake he missed a number and a sum of $800$ was obtained. Find the number he missed.

• A. $15$
• B. $20$
• C. $25$
• D. $10$

Answer 1

Answer: (B)

$20$

We know that the sum of $n$ natural numbers is $\frac{n(n-1)}{2}$;

It is given that Ankush have missed a number, so, let the missing number be $x$ where $(x\le n)$ and if he obtained a sum a sum of $800$ then,

$\frac{n(n-1)}{2}=800+x\Rightarrow n(n-1)=2(800+x)\Rightarrow n(n-1)=1600+2x$

It shows that $(n-1{)}^2$ is approximate to $1600$ which implies that:

${(n-1)}^2=1600\Rightarrow n-1=\sqrt{1600}\Rightarrow n-1=40\Rightarrow n=40+1\Rightarrow n=41$

Now, put $n=41$ in the equation $n(n-1)=1600+2x$ as shown below:

$n(n-1)=1600+2x\Rightarrow 41(41-1)=1600+2x\Rightarrow 41\times 40=1600+2x\Rightarrow 1640=1600+2x$

$\Rightarrow 2x=1640-1600\Rightarrow 2x=40\Rightarrow x=\frac{40}{2}\Rightarrow x=20$

Hence, the number he missed is $20$.