Question

Among the natural numbers 1 to 49. Find the number for which the sum of all the preceding number and the sum of all the succeeding numbers are equal.

Answer 1

Hint: Here in this question we have to find The number for which the sum of all the preceding numbers and the sum of all the succeeding numbers are equal. First we have to write the sequence by using the arithmetic progression and further simplify by using the formula of sum of the first k natural number.

Complete step-by-step solution:
Given the natural number between 1 to 49
Let n be the number from 1 to 49 such that the sum of its preceding is equal to the sum of numbers succeeding.
Now According to arithmetic progression, an arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one and it is abbreviated as AP.
\[\therefore \] The AP formed will be
       \[1 + 2 + 3 + ... + \left( {n - 1} \right) = \left( {n + 1} \right) + \left( {n + 2} \right) + \left( {n + 3} \right) + ... + 49\]
As we know that the sum of the first k natural numbers is
\[{S_k} = 1 + 2 + ... + k = \dfrac{{k(k + 1)}}{2}\]. Then the above equation becomes
Suppose that \[{S_{49}} = \dfrac{1}{2} \times 49 \times 50 = 49 \times 25\]. Let this number be n. Then the sum of all preceding numbers is:
\[{S_{n - 1}} = \dfrac{{n\left( {n - 1} \right)}}{2}\]
The sum of all succeeding numbers are:
\[{S_{49}} - {S_n} = 25 \times 49 - \dfrac{{n\left( {n + 1} \right)}}{2}\]
These two sums are equal, thus
\[ \Rightarrow \,\,\,\dfrac{{n\left( {n - 1} \right)}}{2} = 25 \times 49 - \dfrac{{n\left( {n + 1} \right)}}{2}\]
Multiply both side by 2, then
\[ \Rightarrow \,\,\,n\left( {n - 1} \right) = 2 \times 25 \times 49 - n\left( {n + 1} \right)\]
On simplification, we get
\[ \Rightarrow \,\,\,{n^2} - n = 2450 - {n^2} - n\]
Add n on both side, then
\[ \Rightarrow \,\,\,{n^2} - n + n = 2450 - {n^2} - n + n\]
\[ \Rightarrow \,\,\,{n^2} = 2450 - {n^2}\]
Isolate to the LHS
\[ \Rightarrow \,\,\,2{n^2} = 2450\]
Divide both side by 2
\[ \Rightarrow \,\,\,{n^2} = 1225\]
Taking square root on both side
\[ \Rightarrow \,\,\,n = \pm \sqrt {1225} \]
\[ \Rightarrow \,\,\,n = \pm \sqrt {{{35}^2}} \]
\[ \Rightarrow \,\,\,n = \pm \,35\]
Consider only the positive number

Hence, 35 is the required number between 1 to 49.

Note: Generally we have different kinds of numbers namely, natural numbers, whole numbers, integers, rational numbers, irrational numbers and real numbers.In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total.