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.

Answer 1

Hint: Here we have to find the missing number, so in this question we have to use the formula of the sum of n natural numbers. After that we have to imply the root expression into the given information. Also use mathematical expressions like multiplication, addition etc.

Complete step-by-step answer:
given that adding a certain number of consecutive natural number starting from one, by mistake he missed a number and gets a sum of $ 800 $
So, we know that the sum of $ n $ natural number $ = \dfrac{{n\left( {n + 1} \right)}}{2} $ .
In the question given that by mistake he missed a number and got a sum of $ 800 $ .
Therefore, let the number he missed by the mistake will be $ x $ . Where $ \left( {x \leqslant n} \right) $
So according to the given information in the question that gets a sum of $ 800 $
 $ \dfrac{{n\left( {n + 1} \right)}}{2} = 800 + x $
 $
\Rightarrow n\left( {n + 1} \right) = 2\left( {800 + x} \right) \\
  n\left( {n + 1} \right) = 1600 + 2x \;
 $
 $ \mathop n\nolimits^2 + n = 1600 + 2x $
According to the question it looks that $ \mathop n\nolimits^2 $ is near to $ 1600 $
 $
\Rightarrow \mathop n\nolimits^2 = 1600 \\
  n = \sqrt {1600} \\
  n = \sqrt {40 \times 40} \\
  n = 40 \;
 $
So,
 $
  \left( {n + 1} \right) = 40 + 1 \\
   = 41 \;
 $
Now putting $ n = 40 $ in the equation shown as
 $ n\left( {n + 1} \right) = 1600 + 2x $
So, we get
 $ n\left( {n + 1} \right) = 1600 + 2x $
 $ 40 \times 41 = 1600 + 2x $
 $ 1640 = 1600 + 2x $
 $ 2x = 1640 - 1600 $
 $ 2x = 40 $
 $ x = 20 $
So, the correct answer is “ $ 20 $ ”.

Note: : A natural number is an integer greater than zero. Natural numbers begin at one and increment to infinity. Natural numbers are numbers that we use to count. Natural Numbers are also called counting numbers. They are whole, non-negative numbers. Natural numbers denoted by $ N $ . Sum of all natural numbers is infinite.