Question

the housese of a row are number consecutively from 1 to 49.show that there is a value of x such that the sun of numbers of the houses preceding the house numbered x is equal to the sum of the number of find the value of x.

hint x -1 = 49 - x​

Answer 1

GIVEN :-

$\leadsto$Number of houses was 1,2,3,............49.

$\leadsto$By observation,the number of houses are in an A.P.

$\leadsto$Hence,

• First term, a = 1.

•Common difference, d = 1.

SOLUTION :-

$\leadsto$Let us assume that the number of xth house can be expressed as below:

$\leadsto$We know that, sum of n terms in A.P. = n/2[2a+(n-1)d]

$\leadsto$ Sum of number of houses preceding xth house , S = 1.

$\leadsto \frac{(x - 1)}{2}[2a + (x - 1 - 1)d ] \: \\ \leadsto \: \frac{x - 1}{2}[2(1) + (x - 2)(1) ] \\ \leadsto \: \frac{x - 1}{2}[2 + x - 2] \\ \leadsto \bold{ \frac{(x)(x - 1)}{2} }$

$\leadsto$ By the given we know that, Sum of number of houses following xth house= $S_{49}- s,$

$\leadsto \: \frac{49}{2} (2(1) + 49 - 1)(1)) - \frac{x}{2 } (2 + x - 1) \\\leadsto \frac{49}{2}(50) - \frac{x}{2} (x + 1) \\ \leadsto \bold{ 25(49) - \frac{x(x + 1)}{2} }$

$\leadsto$ It is given that these sums are equal to each other.

$\leadsto \: \frac{x(x - 1)}{2} = 25(49) - x( \frac{x + 1}{2}) \\ \leadsto \: \frac{x {}^{2} }{2} - \frac{x}{2}= 1225- \frac{x {}^{2} }{2} - \frac{x}{2} \\ \leadsto \: x {}^{2} = 1225 \\ => x = ±35$

$\leadsto$ As the number of houses cannot be a negative number we consider number of houses, x = 35

$\leadsto$ Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

Answer 2

Answer:

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