NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4#

$a = 121$, $d = -4$.

For $a_n < 0$:

$$121 + (n-1)(-4) < 0$$

$$121 - 4n + 4 < 0$$

$$125 - 4n < 0$$

$$4n > 125$$

$$n > 31.25$$

The smallest integer value of $n$ 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.

Let the first term = $a$ and common difference = $d$.

$a_3 = a + 2d$ and $a_7 = a + 6d$.

$a_3 + a_7 = 2a + 8d = 6 \implies a + 4d = 3 \quad \cdots (1)$

$a_3 \times a_7 = 8 \quad \cdots (2)$

From (1): $a = 3 - 4d$. Substituting in (2):

$a_3 = (3-4d) + 2d = 3 - 2d$

$a_7 = (3-4d) + 6d = 3 + 2d$

$(3-2d)(3+2d) = 8$

$9 - 4d^2 = 8$

$4d^2 = 1 \implies d = \pm\dfrac{1}{2}$

Case 1: $d = \dfrac{1}{2}$ $a = 3 - 4 \times \dfrac{1}{2} = 1$

$$S_{16} = \frac{16}{2}\left[2(1) + 15\left(\frac{1}{2}\right)\right] = 8\left[2 + 7.5\right] = 8 \times 9.5 = 76$$

Case 2: $d = -\dfrac{1}{2}$ $a = 3 - 4 \times (-\dfrac{1}{2}) = 5$

$$S_{16} = \frac{16}{2}\left[2(5) + 15\left(-\frac{1}{2}\right)\right] = 8[10 - 7.5] = 8 \times 2.5 = 20$$

∴ Sum of first sixteen terms = 76 or 20.

Total distance between top and bottom rungs $= 2.5$ m $= 250$ cm.

Number of gaps between rungs $= \dfrac{250}{25} = 10$

Number of rungs $= 10 + 1 = 11$

The lengths of the rungs form an AP: $a = 45$ cm (bottom), $l = 25$ cm (top), $n = 11$.

$$\text{Total length} = S_{11} = \frac{11}{2}(45 + 25) = \frac{11 \times 70}{2} = \mathbf{385} \text{ cm}$$

Sum of house numbers from 1 to $x-1$:

$$S_{x-1} = \frac{(x-1)x}{2}$$

Sum of house numbers from $x+1$ to 49:

$$S_{\text{after}} = \frac{49 \times 50}{2} - \frac{x(x+1)}{2} = 1225 - \frac{x(x+1)}{2}$$

Setting them equal:

$$\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$$$$\frac{x(x-1) + x(x+1)}{2} = 1225$$$$\frac{x[(x-1)+(x+1)]}{2} = 1225$$$$\frac{x \cdot 2x}{2} = 1225$$$$x^2 = 1225 \implies x = 35$$

Verification: Sum before house 35 $= \dfrac{34 \times 35}{2} = 595$

Sum after house 35 $= 1225 - \dfrac{35 \times 36}{2} = 1225 - 630 = 595$ ✓

∴ $x = \mathbf{35}$ [Hence Proved]

The volume of concrete for each step forms an AP.

Volume of 1st step $= 50 \times \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{50}{8}$ m³

Volume of 2nd step $= 50 \times \dfrac{1}{2} \times \dfrac{2}{4} = \dfrac{100}{8}$ m³ (it is 2 layers high)

Volume of $n$th step $= 50 \times \dfrac{1}{2} \times \dfrac{n}{4} = \dfrac{50n}{8}$ m³

Actually, considering the solid nature — the $k$th step (from top) has height $k \times \dfrac{1}{4}$ m.

Volume of $k$th step $= 50 \times \dfrac{1}{2} \times \dfrac{k}{4} = \dfrac{50k}{8} = \dfrac{25k}{4}$ m³

These form an AP with $a = \dfrac{25}{4}$, $d = \dfrac{25}{4}$, $n = 15$.

$$\text{Total Volume} = \sum_{k=1}^{15} \frac{25k}{4} = \frac{25}{4} \times \frac{15 \times 16}{2} = \frac{25}{4} \times 120 = \frac{3000}{4} = \mathbf{750} \text{ m}^3$$