Q

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

Q : 4    The houses of a row are numbered consecutively from $\small 1$ to $\small 49$. Show that there is a value of  $\small x$ such that the sum of the numbers of the houses preceding the house numbered $\small x$ is equal to the  sum of the numbers of the houses following it. Find this value of $\small x$.                                                              [Hint : $\small S_{x-1}=S_4_9-S_x$

Views

It is given that the sum of the numbers of the houses preceding the house numbered $\small x$ is equal to the  sum of the numbers of the houses following it
And 1,2,3,.....,49 form an AP with a = 1 and d = 1
Now, we know that
$S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
Suppose their exist an n term such that ( n < 49)
Now, according to given conditions
Sum of first n - 1 terms of AP = Sum of terms following the nth term
Sum of first n - 1 terms of AP = Sum of whole AP - Sum of first m terms of AP
i.e.
$S_{n-1}=S_{49}-S_n$

$\frac{n-1}{2}\left \{ 2a+((n-1)-1)d \right \}=\frac{49}{2}\left \{ 2a+(49-1)d \right \}-\frac{n}{2}\left \{ 2a+(n-1)d \right \}$

$\frac{n-1}{2}\left \{ 2+(n-2) \right \}=\frac{49}{2}\left \{ 2+48 \right \}-\frac{n}{2}\left \{ 2+(n-1) \right \}$

$\frac{n-1}{2}\left \{ n\right \}=\frac{49}{2}\left \{ 50 \right \}-\frac{n}{2}\left \{ n+1 \right \}$

$\frac{n^2}{2}-\frac{n}{2}=1225-\frac{n^2}{2}-\frac{n}{2}$

$n^2 = 1225$
$n = \pm 35$

Given House number are not  negative so we reject n = -35

Therefore, sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35

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