Question

4. 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 nụmbered x is
equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: S.,= S.-SI​

Answer 1

Answer:

QUESTION:-

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 nụmbered x is

equal to the sum of the numbers of the houses following it. Find this value of x.

SOLUTION:-

☆We are given an AP ,namely

1 , 2 , 3 ,..., (x-1) ,x ,(x+1),...,49

Such that 1 + 2 + 3 + ... + (x-1) = (x+1) + (x+2) + ... 49

Thus , we hv Sx-1 = S49 - Sx--(i)

Using the formula,

$sn \: = \frac{n}{2} (a + l)$

in equation (i) ,we have

$\frac{(x - 1)}{2} \times {1 + (x - 1)} = \frac{49}{2} \times (1 + 49) - \frac{x}{2} \times (1 + x) \\ = > \: \frac{x(x - 1)}{2} + \frac{x(x + 1)}{2} = 1225 \\ = > 2 {x}^{2} = 2450 = > {x}^{2} = 1225 = > x = \sqrt{1225 } = 35 \\ hence \: x = 35$

Hence,number of houses following it =35.

Answer 2

${\huge {\bold {\boxed{\color{magenta}{answer}}}}}$

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First term, a = 1

Common difference, d=1

Let us say the number of xth houses can be represented as;

Sum of preceding the numbers of x = sum of following numbers of x

i.e. Sum of ( 1,2,3,….x-1) = sum of [(x+1), (x+2) ,….48,49]

That is 1 + 2 + 3 + …… + ( x-1) = ( x+1) + ( x+2) …… + 49

$\tt => (x-1)/2[1+x-1] = (49-x)/2[x+1+49]$

$\tt => (x-1)x=(49-x)(x+50)$

$\tt => x²-x=49x+2450-x²-50x$

$\tt => x²-x =2450-x²-x$

$\tt => 2x²=2450$

$\tt => x²=1225$

$\tt x=√1225$

$\tt x = 35$

Therefore, the value of x is 35

Hope it's Helpful.....:)