Arithmetic Progressions – Exercise 5.4 (Optional)

Advanced AP Problems

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

2. 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.

1. $a = 121, d = 117 - 121 = -4$.
Let $a_n < 0$.
$a + (n-1)d < 0$
$121 + (n-1)(-4) < 0$
$121 - 4n + 4 < 0$
$125 < 4n \Rightarrow n > \frac{125}{4} \Rightarrow n > 31.25$.
Since $n$ must be an integer, $n = 32$.
The 32nd term is the first negative term.

2. Let the AP be $a, a+d, a+2d, \dots$
Given: $a_3 + a_7 = 6 \Rightarrow (a+2d) + (a+6d) = 6 \Rightarrow 2a + 8d = 6 \Rightarrow a + 4d = 3 \Rightarrow a = 3 - 4d$.
Also given: $a_3 \times a_7 = 8 \Rightarrow (a+2d)(a+6d) = 8$.
Substitute $a = 3 - 4d$:
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
$9 - 4d^2 = 8 \Rightarrow 4d^2 = 1 \Rightarrow d^2 = \frac{1}{4} \Rightarrow d = \pm \frac{1}{2}$.

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

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

Answer: $S_{16} = 76$ or $20$.

Word Problems (Ladder, Houses, Terrace)

3. 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?

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

5. 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.

3. Total distance between top and bottom rungs = 2.5 m = 250 cm.
Distance between consecutive rungs = 25 cm.
Number of rungs $n = \frac{250}{25} + 1 = 10 + 1 = 11$.
Lengths of rungs form an AP with $a = 45$ (bottom) and $l = 25$ (top).
Total length of wood = Sum of lengths of 11 rungs.
$S_{11} = \frac{11}{2}(a + l) = \frac{11}{2}(45 + 25) = \frac{11}{2}(70) = 11 \times 35 = 385$ cm.
Answer: 3.85 m.

4. House numbers: $1, 2, 3, \dots, 49$. This is an AP with $a=1, d=1$.
Condition: Sum of numbers before $x$ = Sum of numbers after $x$.
$S_{x-1} = S_{49} - S_x$.
$\frac{x-1}{2}[1 + (x-1)] = \frac{49}{2}[1 + 49] - \frac{x}{2}[1 + x]$
$\frac{x(x-1)}{2} = \frac{49 \times 50}{2} - \frac{x(x+1)}{2}$
Multiply by 2:
$x^2 - x = 2450 - (x^2 + x)$
$x^2 - x = 2450 - x^2 - x$
$2x^2 = 2450 \Rightarrow x^2 = 1225 \Rightarrow x = \sqrt{1225} = 35$.
Answer: House number 35.

5. Number of steps $n = 15$. Length $L = 50$ m. Tread $B = \frac{1}{2}$ m.
Rise increases for each step: $H_1 = \frac{1}{4}, H_2 = \frac{2}{4}, H_3 = \frac{3}{4}, \dots$
Volume of step $k = L \times B \times H_k = 50 \times \frac{1}{2} \times (k \times \frac{1}{4}) = 25 \times \frac{k}{4} = \frac{25k}{4}$ m³.
Volumes form an AP: $\frac{25}{4}, \frac{50}{4}, \frac{75}{4}, \dots$
$a = \frac{25}{4}, d = \frac{25}{4}$.
Total Volume $S_{15} = \frac{15}{2}[2(\frac{25}{4}) + 14(\frac{25}{4})] = \frac{15}{2} \times \frac{25}{4} (2 + 14) = \frac{15}{2} \times \frac{25}{4} \times 16$.
$S_{15} = 15 \times 25 \times 2 = 750$ m³.
Answer: 750 m³.

Arithmetic Progressions – Exercise 5.4

Overview

Advanced AP Problems

Word Problems (Ladder, Houses, Terrace)