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

Here $a = 121$ and:

$$d = 117 - 121 = -4$$

For the $n$th term to be negative, $a_n < 0$:

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

$$\Rightarrow 125 - 4n < 0$$

$$\Rightarrow 4n > 125$$

$$\Rightarrow n > 31\dfrac{1}{4}$$

The smallest integer greater than $31\dfrac{1}{4}$ is 32.

Check:

$$a_{32} = 121 + 31(-4) = 121 - 124 = -3$$

∴ The 32nd term is the first negative term.

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

The sum of the third and seventh terms is 6, so:

$$(a + 2d) + (a + 6d) = 6$$

$$\Rightarrow 2a + 8d = 6$$

$$\Rightarrow a + 4d = 3 \quad \cdots (1)$$

The product of the third and seventh terms is 8, so:

$$(a + 2d)(a + 6d) = 8 \quad \cdots (2)$$

From equation (1):

$$a + 4d = 3$$

$$\Rightarrow a = 3 - 4d$$

Substituting $a = 3 - 4d$ in equation (2):

$$(a + 2d)(a + 6d) = 8$$

$$(3 - 4d + 2d)(3 - 4d + 6d) = 8$$

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

$$\Rightarrow 9 - 4d^2 = 8$$

$$\Rightarrow 4d^2 = 1$$

$$\Rightarrow d = \pm\dfrac{1}{2}$$

Case 1: $d = \dfrac{1}{2}$

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

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{16}$:

$$= \dfrac{16}{2}\left[2(1) + 15 \times \dfrac{1}{2}\right]$$

$$= 8 \times \dfrac{19}{2}$$

$$= 76$$

Case 2: $d = -\dfrac{1}{2}$

$$a = 3 - 4 \times \left(-\dfrac{1}{2}\right) = 5$$

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{16}$:

$$= \dfrac{16}{2}\left[2(5) + 15 \times \left(-\dfrac{1}{2}\right)\right]$$

$$= 8 \times \dfrac{5}{2}$$

$$= 20$$

∴ The sum of the first sixteen terms is 76 or 20.

Distance between the top and bottom rungs = $2\dfrac{1}{2}$ m = 250 cm

Number of rungs = $\dfrac{250}{25} + 1 = 11$

The lengths of the rungs form an AP with $a = 45$, $l = 25$ and $n = 11$.

Using $S_n = \dfrac{n}{2}(a + l)$, we find $S_{11}$:

$$= \dfrac{11}{2}(45 + 25)$$

$$= \dfrac{11 \times 70}{2}$$

$$= 385$$

∴ The length of wood required for the rungs is 385 cm (3.85 m).

The house numbers form an AP: $1, 2, 3, \ldots, 49$, with $a = 1$ and $d = 1$.

Sum of the houses before house $x$:

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

Sum of all 49 houses:

$$S_{49} = \dfrac{49 \times 50}{2} = 1225$$

Sum of the houses up to house $x$:

$$S_x = \dfrac{x(x+1)}{2}$$

The sum before house $x$ equals the sum after it, so:

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

$$\Rightarrow \dfrac{x(x-1)}{2} = 1225 - \dfrac{x(x+1)}{2}$$

$$\Rightarrow \dfrac{x(x-1)}{2} + \dfrac{x(x+1)}{2} = 1225$$

$$\Rightarrow \dfrac{x^2 - x + x^2 + x}{2} = 1225$$

$$\Rightarrow x^2 = 1225$$

$$\Rightarrow x = 35$$

Check:
Sum before house 35 = $\dfrac{34 \times 35}{2} = 595$
Sum after house 35 = $1225 - \dfrac{35 \times 36}{2} = 1225 - 630 = 595$

$$\Rightarrow 595 = 595$$

∴ Such a value exists, and $x = \mathbf{35}$.

Each step is solid concrete down to the ground, so the 2nd step is twice as high as the 1st, the 3rd three times as high, and so on.

Volume of 1st step = $\dfrac{1}{4} \times \dfrac{1}{2} \times 50 = \dfrac{25}{4}$ m³
Volume of 2nd step = $\dfrac{2}{4} \times \dfrac{1}{2} \times 50 = \dfrac{50}{4}$ m³
Volume of 3rd step = $\dfrac{3}{4} \times \dfrac{1}{2} \times 50 = \dfrac{75}{4}$ m³

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

Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$, we find $S_{15}$:

$$= \dfrac{15}{2}\left[2 \times \dfrac{25}{4} + 14 \times \dfrac{25}{4}\right]$$

$$= \dfrac{15}{2} \times 16 \times \dfrac{25}{4}$$

$$= \dfrac{15}{2} \times 100$$

$$= 750$$

∴ The total volume of concrete required is 750 m³.