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 preteding the house numbered x is
equal to the sum of the numbers of the houses following it. Find this value of r.
(Hint: S(x-1)=s49-sx]​

Answer 1

Given :

1, 2, 3, ..... 49 consecutive numbers.

Sum of preceding numbers of x = Sum of following numbers of x. i.e.

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

Find :

The value of x.

Solution :

Now,

$\Rightarrow\:\sf{S_1\: =\: 1\: + \:2 \:+\: 3 \:+ .... +\: (x \:-\: 1)}$

$\Rightarrow\:\sf{S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)[2 \: \times \: 1 \: + \: (x - 1 - 1)1]}$

$\Rightarrow\:\sf{S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)[2 \: + \: (x - 2)1]}$

$\Rightarrow\:\sf{S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)(2 \: + \: x - 2)}$

$\Rightarrow\:\sf{S_1 \:= \:\bigg(\dfrac{x - 1}{2} \bigg)}$

$\Rightarrow\:\sf{S_1 \:= \:x\bigg(\dfrac{x - 1}{2} \bigg)}$

Now,

$\Rightarrow\:\sf{S_2 \:= \:\bigg(\dfrac{49 - x}{2} \bigg)[2 (x + 1 )\: + \: (49 - x - 1)1]}$

$\Rightarrow\:\sf{S_2 \:= \:\bigg(\dfrac{49 - x}{2} \bigg)(2x \:+ \: 2 \: - \: x \:+ \:48)}$

$\Rightarrow\:\sf{S_2 \:= \:\bigg(\dfrac{49 - x}{2} \bigg)(x\:+\:50)}$

According to question,

$\sf{S_1\:=\:S_2}$

So,

$\Rightarrow\:\sf{x\bigg(\dfrac{x - 1}{2} \bigg)\:=\:\bigg(\dfrac{49 - x}{2} \bigg)(x\:+\:50)}$

$\Rightarrow\:\sf{x(x\:-\:1)\:=\:(49\:-\:x)(x\:+\:50)}$

$\Rightarrow\:\sf{x^2\:-\:x\:=\:49x\:+\:2450\:-\:x^2\:-\:50x}$

$\Rightarrow\:\sf{x^2\:-\:x\:=\:-\:x^2\:-\:x\:+\:2450}$

$\Rightarrow\:\sf{2x^2\:=\:2450}$

$\Rightarrow\:\sf{x^2\:=\:1225}$

$\Rightarrow\:\sf{x\:=\:35}$

Therefore, Number is 35.

Sum of 1, 2, 3, .... (35 - 1) = Sum of [(35 + 1), (35 + 2) .... 48, 49)]

Sum of 1, 2, 3, .... 34 = Sum of (36, 37, .... 48, 49)

Answer 2

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 preteding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of r. (Hint: S(x-1)=s49-sx]

Formula and concept used :---

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

______________________________

Solution :----

it has been given that , 1 to 49 their are consecutive numbers , and x is a number in between such that sum of preceding numbers of x is Equal to sum of following numbers of x...

Or, we can say that now,

→ sum of ( 1,2,3,4,5,.... upto x-1) = sum of {(x+1), (x+2) ,.... upto 49}

→ Sum of First (x-1) terms = Sum of (49-x) terms .

Putting values in Sum of AP formula (Told above) and comparing both now, we get,

→ (x-1)/2 [ First term + last term] = (49-x)/2 [ First term + last term ]

2 will be cancel from both sides ,

→ (x-1) [ 1 + (x -1) ] = (49-x) [ (x+1) + 49]

→ (x-1) * x = (49-x)(x+50)

→ x² - x = 49x + 2450 - x² - 50x

Taking all to LHS side now, (sign will change)

→ x² + x² -x + 50x - 49x - 2450 = 0

→ 2x² -50x + 50x - 2450 = 0

→ 2x² - 2450 = 0

→ 2x² = 2450

Dividing both sides by 2 ,

→ x² = 1225

→ x² = 35 * 35

Square - root both sides now,

→ x = ±35 .

(As negative value not possible).

→ x = 35 .

So, Value of x is 35 , Than we can say that ,

( 1+2+3+4________34) = (36+37+38+39__________49)