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Q4:
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 $x$.
[Hint : $S_{x – 1} = S_{49} – S_x$]
Solution :
Given: A row of houses numbered consecutively from $1, 2, 3, \dots, 49$. Let $x$ be the number of a specific house in this row.
To Find: The value of $x$ such that the sum of the house numbers preceding $x$ is equal to the sum of the house numbers following $x$.
Step 1: Defining the Sums
The sequence of house numbers is an Arithmetic Progression (AP) where the first term $a = 1$ and the common difference $d = 1$.
The sum of the first $n$ terms of an AP is given by the formula: $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term.
The sum of all house numbers from $1$ to $49$ is: $S_{49} = \frac{49}{2}(1 + 49) = \frac{49}{2}(50) = 49 \times 25 = 1225$.
Step 2: Formulating the Equation
Let the sum of the houses preceding house $x$ be $S_{x-1}$. This is the sum of numbers from $1$ to $x-1$. $S_{x-1} = \frac{(x-1)}{2}(1 + x - 1) = \frac{(x-1)x}{2}$.
The sum of the houses following house $x$ is the total sum $S_{49}$ minus the sum of houses up to $x$. Sum following $x = S_{49} - S_x$. $S_x = \frac{x(x+1)}{2}$.
According to the problem, the sum of houses preceding $x$ equals the sum of houses following $x$: $S_{x-1} = S_{49} - S_x$.
Step 3: Algebraic Deduction
Substitute the expressions into the equation: $\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$.
Multiply the entire equation by $2$ to clear the denominators: $x(x-1) = 2450 - x(x+1)$.
Expand the terms: $x^2 - x = 2450 - (x^2 + x)$. $x^2 - x = 2450 - x^2 - x$.
Add $x$ to both sides: $x^2 = 2450 - x^2$.
Add $x^2$ to both sides: $2x^2 = 2450$.
Step 4: Solving for $x$
Divide by $2$: $x^2 = 1225$.
Take the square root of both sides: $x = \sqrt{1225}$. Since $x$ must be a positive integer representing a house number: $x = 35$.
Justification: Sum of houses preceding $35$ (i.e., $1$ to $34$): $S_{34} = \frac{34(35)}{2} = 17 \times 35 = 595$. Sum of houses following $35$ (i.e., $36$ to $49$): $S_{49} - S_{35} = 1225 - \frac{35(36)}{2} = 1225 - (35 \times 18) = 1225 - 630 = 595$. Since $595 = 595$, the condition is satisfied.
Final Answer: The value of $x$ is 35.
More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.4 (Optional)*
- Q1: Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find $n$ for $a_n < 0$]
- Q2: The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
- Q3: A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = $\frac{250}{25} + 1$]
- Q5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = $\frac{1}{4} \times \frac{1}{2} \times 50 m^3$]
CBSE Solutions for Class 10 Mathematics Arithmetic Progression
Chapters in CBSE - Class 10 Mathematics
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