Which term of the AP: 121, 117, 113... is its first negative term?

Using an 31.25. The 32nd term is the first negative term.

Find the house number x such that sum of house numbers preceding x equals sum of house numbers following it.

Using S_{x-1} = S_{49} - S_x, 2x^2 = 2450 gives x = 35. House number is 35.

Class 10 Maths Chapter 5 Exercise 5.4 NCERT Solutions

Question 1 Inequalities & First Negative Term

Which term of the AP: $121, 117, 113, \dots$, is its first negative term?
[Hint: Find $n$ such that $a_n < 0$]

Step 1: Identify AP parameters

First term $a = 121$.
Common difference $d = 117 - 121 = -4$.

Step 2: Set up inequality for $a_n < 0$

$a_n = a + (n - 1)d < 0$
$121 + (n - 1)(-4) < 0$
$121 - 4n + 4 < 0 \implies 125 - 4n < 0$
$4n > 125 \implies n > \dfrac{125}{4} \implies n > 31.25$

Step 3: Determine least integer value

Since the term index $n$ must be a natural number, the smallest integer greater than $31.25$ is $n = 32$.
Verification: $a_{32} = 121 + 31(-4) = 121 - 124 = -3 < 0$.

The 32nd term is the first negative term of the AP (value = -3).

Question 2 Symmetric Sum and Product System

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.

Step 1: Use sum condition to relate $a$ and $d$

$a_3 + a_7 = (a + 2d) + (a + 6d) = 2a + 8d = 6$
Dividing by $2$: $a + 4d = 3 \implies a = 3 - 4d$.

Step 2: Use product condition to solve for $d$

$a_3 \times a_7 = (a + 2d)(a + 6d) = 8$
Substitute $a = 3 - 4d$:
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
$9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d^2 = \dfrac{1}{4} \implies d = \pm\dfrac{1}{2}$.

Step 3: Compute $S_{16}$ for both cases

Case 1: $d = \dfrac{1}{2}$
$a = 3 - 4\left(\dfrac{1}{2}\right) = 3 - 2 = 1$.
$S_{16} = \dfrac{16}{2}\left[2(1) + 15\left(\dfrac{1}{2}\right)\right] = 8[2 + 7.5] = 8(9.5) = \mathbf{76}$.

Case 2: $d = -\dfrac{1}{2}$
$a = 3 - 4\left(-\dfrac{1}{2}\right) = 3 + 2 = 5$.
$S_{16} = \dfrac{16}{2}\left[2(5) + 15\left(-\dfrac{1}{2}\right)\right] = 8[10 - 7.5] = 8(2.5) = \mathbf{20}$.

The sum of the first 16 terms is 76 or 20.

Question 3 Uniform Decrease & Ladder Rungs

A ladder has rungs $25\text{ cm}$ apart. The rungs decrease uniformly in length from $45\text{ cm}$ at the bottom to $25\text{ cm}$ at the top. If the top and the bottom rungs are $2\dfrac{1}{2}\text{ m}$ apart, what is the length of the wood required for the rungs?

Step 1: Find total number of rungs

Total distance between top and bottom rungs $= 2\dfrac{1}{2}\text{ m} = 2.5 \times 100\text{ cm} = 250\text{ cm}$.
Distance between consecutive rungs $= 25\text{ cm}$.
Total number of rungs $n = \dfrac{\text{Total Distance}}{\text{Spacing}} + 1 = \dfrac{250}{25} + 1 = 10 + 1 = 11$.

Step 2: Sum the arithmetic progression of lengths

The lengths decrease uniformly, forming an AP with:
First rung (bottom) $a = 45\text{ cm}$, Last rung (top) $l = 25\text{ cm}$, Number of rungs $n = 11$.
Total length of wood $= S_{11} = \dfrac{n}{2}(a + l) = \dfrac{11}{2}(45 + 25) = \dfrac{11}{2}(70) = 11 \times 35 = 385\text{ cm}$.
In metres: $\dfrac{385}{100} = 3.85\text{ m}$.

Total length of wood required is 385 cm (or 3.85 metres).

Question 4 Row of Houses & Equilibrium Point

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: $S_{x-1} = S_{49} - S_x$]

Step 1: Express preceding and following sums

The house numbers are $1, 2, 3, \dots, 49$, which forms an AP with $a = 1, d = 1$.
Sum of first $n$ natural numbers is $S_n = \dfrac{n(n + 1)}{2}$.
Sum of house numbers preceding house $x$ ($1$ to $x - 1$):
$S_{x-1} = \dfrac{(x - 1)(x - 1 + 1)}{2} = \dfrac{x(x - 1)}{2}$

Sum of house numbers following house $x$ ($(x + 1)$ to $49$):
$S_{49} - S_x = \dfrac{49(50)}{2} - \dfrac{x(x + 1)}{2} = 1225 - \dfrac{x(x + 1)}{2}$

Step 2: Equate and solve for $x$

Given $S_{x-1} = S_{49} - S_x$:
$\dfrac{x(x - 1)}{2} = 1225 - \dfrac{x(x + 1)}{2}$
$\dfrac{x(x - 1)}{2} + \dfrac{x(x + 1)}{2} = 1225$
$\dfrac{x^2 - x + x^2 + x}{2} = 1225 \implies \dfrac{2x^2}{2} = 1225 \implies x^2 = 1225$
$x = \sqrt{1225} = 35$ (since house number is positive).

The required value of x is 35 (House number 35).

Question 5 Solid Concrete Terrace Volume

A small terrace at a football ground comprises of $15$ steps each of which is $50\text{ m}$ long and built of solid concrete. Each step has a rise of $\dfrac{1}{4}\text{ m}$ and a tread of $\dfrac{1}{2}\text{ m}$. Calculate the total volume of concrete required to build the terrace.

Step 1: Analyze step dimensions and volume formula

For each step:
Length $L = 50\text{ m}$
Tread (width) $B = \dfrac{1}{2}\text{ m}$
The height increases with each step as it is solid from ground level:
Step 1 Height $= \dfrac{1}{4}\text{ m}$
Step 2 Height $= 2 \times \dfrac{1}{4}\text{ m}$
$\dots$
Step $k$ Height $= k \times \dfrac{1}{4}\text{ m}$

Step 2: Establish the volume series

Volume of $k^{\text{th}}$ step $= \text{Length} \times \text{Tread} \times \text{Height}_k = 50 \times \dfrac{1}{2} \times \left(k \times \dfrac{1}{4}\right) = \dfrac{25}{4}k\text{ m}^3$.
The volumes of the 15 steps form an AP:
$\dfrac{25}{4}, \dfrac{50}{4}, \dfrac{75}{4}, \dots$ for $n = 15$ steps.
Here $a = \dfrac{25}{4}\text{ m}^3$, common difference $d = \dfrac{25}{4}\text{ m}^3$.

Step 3: Calculate total concrete volume

Total volume $S_{15} = \dfrac{n}{2}[2a + (n - 1)d]$
$S_{15} = \dfrac{15}{2}\left[2\left(\dfrac{25}{4}\right) + 14\left(\dfrac{25}{4}\right)\right]$
$S_{15} = \dfrac{15}{2} \times \dfrac{25}{4} [2 + 14] = \dfrac{15 \times 25}{8} \times 16 = 15 \times 25 \times 2 = 750\text{ m}^3$.

Total volume of concrete required to build the terrace is 750 m³.

Arithmetic Progressions - Exercise 5.4

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