Respuesta :
Answer:
Number of the houses as x =35
Step-by-step explanation:
The number of houses is 1, 2, 3, ..., 49
By observation, the numbers of houses are in an A.P.
Hence
First-term, a = 1
Common difference, d = 1
Let us assume that the number of xth house can be expressed as below:
We know that sum of n terms in an A.P. is given by the formula Sₙ = n/2 [2a + (n - 1) d]
Sum of number of houses preceding xth house = Sₓ ₋ ₁
Sₓ ₋ ₁ = (x - 1) / 2 [2a + ((x - 1) - 1)d]
= (x - 1) / 2 [2 × 1 + ( x - 2) × 1]
= (x - 1) / 2 [2 + x - 2]
= [x (x - 1)] / 2 ---------- (1)
By the given information we know that, sum of number of houses following xth house = S₄₉ - Sₓ
S₄₉ - Sₓ = 49 / 2 [2 × 1 + (49 - 1) × 1] - x / 2 [2 × 1 + (x - 1) × 1]
= 49 / 2 [2 + 48] - x / 2 (2 + x - 1)
= (49 / 2) × 50 - (x / 2) (x + 1)
= 1225 - [x (x + 1)] / 2 ---------- (2)
It is given that these sums are equal, that is equation (1) = equation(2)
x (x - 1) / 2 = 1225 - x (x + 1) / 2
x² / 2 - x / 2 = 1225 - x² / 2 - x / 2
On solving further we get,
x² = 1225
x = ± 35
As the number of houses cannot be a negative number, we consider the number of houses as x = 35
35 is the correct answer
Step-by-step explanation:
Given that the houses in a row are numbered consecutively from 1 to 49. We need to show that there exists a value of x such that the sum of numbers of houses preceding the house numbered x is equal to sum of the number of houses following x. Making it more simple for you; sum of (1, 2, 3, 4.... upto x - 1) is equal to sum of (x + 1), (x + 2), (x + 3).... upto 49.
→ Sum of (x - 1) terms = Sum of (49 - x) terms
"x" is the number of houses required. The sum of houses are in AP where the value of a is 1 and d is also 1; where 'a' is first term and 'd' is common difference. We know that sum of nth term of an AP is given by: T(n) = a + (n - 1)d and upto nth term by S(n) = n/2 (2a + (n - 1)d). And also by T(n) = S(n) - S(n - 1).
As per given problem:
→ (x - 1)/2 (Sum of first and last term) = (49 - x)/2 (Sum of first and last term)
→ (x - 1)(Sum of first and last term) = (49 - x) (Sum of first and last term)
Where first term and last term in LHS is 1 and (x - 1) & in RHS is (x + 1) and 49. Substitute them.
→ (x - 1) (1 + (x - 1)) = (49 - x) ((x + 1) + 49)
→ (x - 1) (1 + x - 1) = (49 - x) (x + 1 + 49)
→ (x - 1) (x) = (49 - x) (x + 50)
→ x² - x = 49x + 2450 - x² - 50x
On solving we get,
→ 2x² - 2450 = 0
→ 2x² = 2450
→ x² = 1225
→ x = ± 35
The number of houses is a positive integer. So negative value neglected. We left with x = 35. Therefore, the value of x is 35.
