Question

From natural numbers 1 to 288, find the number for which the sum of numbers smaller than that number equals the sum of numbers greater than that number.

Answer 1

Hint: Here, we will use the concept of arithmetic progressions to solve this question. We will find two sequences using the condition given in question. Then, we will find the sum of these sequences and equate them, we will reach our required answer.

Formula Used:
In an Arithmetic Progression, sum of \[n\] terms is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].

Complete step-by-step answer:
We will first find the number for which the sum of numbers smaller than that number equals the sum of numbers greater than that number.
Now, let the number be \[n\].
Hence, the sum of numbers smaller than \[n\] is,
\[1 + 2 + 3 + ......n - 1\]
Here, the first term, \[a = 1\], common difference, \[d = 1\] and the number of terms \[ = n - 1\].
Substituting the values in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] to find the sum of \[n - 1\] terms, we get
\[{S_{n - 1}} = \dfrac{{n - 1}}{2}\left[ {2\left( 1 \right) + \left( {n - 1 - 1} \right)\left( 1 \right)} \right]\]
\[ \Rightarrow {S_{n - 1}} = \dfrac{{n - 1}}{2}\left[ {2 + n - 2} \right]\]
Simplifying the expression, we get
\[ \Rightarrow {S_{n - 1}} = \dfrac{{n\left( {n - 1} \right)}}{2}\]………………………….(1)
Also, the sum of numbers greater than \[n\] are:
\[n + 1 + n + 2 + .......288\]
Here, the first term, \[a = n + 1\], the common difference, \[d = 1\] and number of terms\[ = 288 - n\].
Substituting the values in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] to find the sum of these \[288 - n\]terms, we get
\[{S_{288 - n}} = \dfrac{{288 - n}}{2}\left[ {2\left( {n + 1} \right) + \left( {288 - n - 1} \right)\left( 1 \right)} \right]\]
Simplifying the expression, we get
\[ \Rightarrow {S_{288 - n}} = \dfrac{{288 - n}}{2}\left[ {2n + 2 + 287 - n} \right]\]
Solving further, we get
\[ \Rightarrow {S_{288 - n}} = \dfrac{{288 - n}}{2}\left[ {n + 289} \right]\]………………………….(2)
Now, as per the question, the sum of numbers smaller than \[n\] equals the sum of numbers greater than \[n\].
Hence, from the equations (1) and (2), we get
\[\dfrac{{n\left( {n - 1} \right)}}{2} = \dfrac{{288 - n}}{2}\left[ {n + 289} \right]\]
Multiplying both sides by 2 and solving the brackets, we get
\[ \Rightarrow {n^2} - n = 288n - {n^2} + 288 \times 289 - 289n\]
Cancelling out the same terms, we get
\[ \Rightarrow 2{n^2} = 288 \times 289\]
Dividing both sides by 2, we get
\[ \Rightarrow {n^2} = 144 \times 289\]
We know that 144 is the square of 12 and 289 is the square of 17. Therefore, the above equation can also be written as:
\[ \Rightarrow {n^2} = {\left( {12} \right)^2} \times {\left( {17} \right)^2}\]
Taking square root on both sides, we get
\[ \Rightarrow n = \pm 12 \times 17 = \pm 204\]
Now, we will reject the negative term as \[n\] cannot be negative as it is a natural number.
Hence, the required term is 204.

Note: Here, the two series formed are in AP. Arithmetic Progressions or AP are the sequences in which the difference between the consecutive terms is constant. Other than arithmetic progression, there are two more progressions and they are geometric progression and harmonic progression. Geometric progression is the sequence or the series in which two consecutive terms differ by the same common ratio. Harmonic progression is a sequence or series which is the reciprocal of arithmetic progression.