Question

The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of numbers of houses preceeding the house numbered x is equal to sum of the numbers of houses following x. Find value of x.
Please elaborate all steps.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, the houses are numbered consecutively from 1 to 49.

Now, we have to find a value of x such that sum of numbers of houses preceeding the house numbered x is equal to sum of the numbers of houses following x.

So, it means, we have to find the value of x such that

$\rm \: 1+2+3 +\cdots+(x - 1) = (x + 1) + (x + 2) + \cdots + 49$

Now, Consider

$\rm \: 1+2+3 +\cdots+(x - 1)$

We know,

$\boxed{ \rm{ \:1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{(x - 1)(x - 1 + 1)}{2}$

$\rm \: = \: \dfrac{x(x - 1)}{2}$

So,

$\boxed{ \rm{ \:1 + 2 + 3 + \cdots + (x - 1) = \: \dfrac{x(x - 1)}{2}}} - - - (1)$

Now, Consider

$\rm \: (x + 1) + (x + 2) + \cdots + 49$

can be rewritten as

$\rm \: = (1 + 2 + 3 + \cdots + 49) - (1 + 2 + 3 + \cdots + x)$

$\rm \: = \: \dfrac{49(49 + 1)}{2} - \dfrac{x(x + 1)}{2}$

$\rm \: = \: \dfrac{49 \times 50 - x(x + 1)}{2}$

$\rm \: = \: \dfrac{49 \times 50 - {x}^{2} - x}{2}$

So,

$\boxed{ \rm{ \:(x + 1) + (x + 2) + \cdots + 49 = \dfrac{49 \times 50 - {x}^{2} - x}{2}}} - - - (2)$

So, Consider again

$\rm \: 1+2+3 +\cdots+(x - 1) = (x + 1) + (x + 2) + \cdots + 49$

$\rm \: \dfrac{x(x - 1)}{2} = \dfrac{49 \times 50 -{x}^{2} - x }{2}$

$\rm \: \dfrac{ {x}^{2} - x}{2} = \dfrac{49 \times 50 -{x}^{2} - x }{2}$

$\rm \: {x}^{2} - x = 49 \times 50 - {x}^{2} - x$

$\rm \: {x}^{2} = 49\times 50 - {x}^{2}$

$\rm \: 2{x}^{2} = 49 \times 50$

$\rm \: {x}^{2} = 49 \times 25$

$\rm\implies \:x = 7 \times 5 = 35$

So, there exist a house number x = 35, such that sum of numbers of houses preceeding the house numbered 35 is equal to sum of the numbers of houses following 35

$\rule{190pt}{2pt}$

Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Answer 2

$\small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

$\rule{300pt}{0.1pt}$

Solution :-

Sum of the numbers of houses preceding the house numbered

$\rm X=1+2+3+\ldots+(X-1)$

$\[ \begin{array}{l} \displaystyle\rm =\frac{(X-1)}{2}[2 \times 1+(X-1-1)] \\ \\\displaystyle\rm =\frac{X(X-1)}{2} \end{array} \]$

Sum of the numbers of houses following the house numbered

$\rm X=(X+1)+(X+2)+(X+3)+\ldots +49$

Using the formula, we get

$\text{Sum of n terms, \( \rm S_{n}=\dfrac{n}{2}(a+l) \)}$

(Here, a is the first term and l is the last term of an AP.)

$\[ \begin{array}{l} \displaystyle\rm=\frac{(49-X)}{2}[49+(X+1)] \\ \\ \displaystyle\rm=\frac{(49-X)(50+X)}{2} \end{array} \]$

Sum of the numbers of houses following the house numbered X= Sum of the numbers of houses preceding the house numbered X

$\[ \begin{array}{l} \displaystyle\rm \Rightarrow \frac{(49-X)(50+X)}{2}=\frac{X(X-1)}{2} \\\\ \displaystyle\rm \Rightarrow(49-X)(50+X)=X(X-1) \\\\ \displaystyle\rm \Rightarrow 2450+49 X-50 X-X^{2}=X^{2}-X \\\\ \displaystyle\rm \Rightarrow 2 X^{2}=2,450 \\ \\ \displaystyle\rm\Rightarrow X^{2}=1,225 \\ \\ \fcolorbox{red}{indigo}{\color{cyan} \displaystyle\rm★ \: \: ⇒ X=35 \: \: ★} \end{array} \]$

Hence, at X=35, the sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.