NCERT Class 10 Mathematics — Arithmetic Progressions, Exercise 5.4 (Optional). All 5 questions solved.

The textbook marks this exercise as optional. A footnote on the page says these exercises are not from the examination point of view. They are still worth doing, because they test the one skill the earlier exercises take for granted: finding the AP inside the problem. Nobody hands you $a$ and $d$ here. You get the lengths of a ladder’s rungs, the numbers on a row of houses, the steps of a terrace, and you have to see that these form an arithmetic progression. After that, the formulae are the familiar ones:

$$a_n = a + (n-1)d, \qquad S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr] = \frac{n}{2}(a + l)$$

Key insight. Name the terms before you calculate anything. Say out loud what $a_1, a_2, a_3, \ldots$ stand for: the first rung, the second rung; the volume of the first step, the second step. Check that consecutive ones differ by a fixed amount, and count how many there are. Questions 3 and 5 turn on exactly this. Count the gaps instead of the rungs, or give every step the same height, and the formula faithfully returns a wrong answer.

Question 1

Which term of the AP: $121, 117, 113, \ldots$, is its first negative term?

[Hint: Find $n$ for $a_n < 0$]

Solution. The terms fall by $4$ each time, starting from $a = 121$, so eventually they must go below zero. The $n$th term is

$$a_n = 121 + (n-1)(-4) = 125 – 4n$$

The first negative term is the first $n$ for which $125 – 4n < 0$, that is,

$$n > \frac{125}{4} = 31\tfrac{1}{4}$$

The smallest whole number greater than $31\frac14$ is $32$. As a check, $a_{31} = 125 – 124 = 1$ is still positive, and $a_{32} = 125 – 128 = -3$ is the first below zero. Notice that no term equals zero, because $\frac{125}{4}$ is not a whole number. This is why the question needs an inequality rather than the equation $a_n = 0$.

Question 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.

Solution. Write both terms in terms of $a$ and $d$: $a_3 = a + 2d$ and $a_7 = a + 6d$.

The sum gives a linear equation:

$$(a + 2d) + (a + 6d) = 6 \;\Rightarrow\; a + 4d = 3$$

This is useful because $a + 4d$ is $a_5$, the term exactly midway between $a_3$ and $a_7$. So $a_3 = 3 – 2d$ and $a_7 = 3 + 2d$, and the product condition becomes a difference of squares:

$$(3 – 2d)(3 + 2d) = 8 \;\Rightarrow\; 9 – 4d^2 = 8 \;\Rightarrow\; d^2 = \frac{1}{4} \;\Rightarrow\; d = \pm\frac{1}{2}$$

Both signs are genuinely possible, so there are two APs.

When $d = \frac{1}{2}$: $a = 3 – 4d = 1$. The terms are $a_3 = 2$ and $a_7 = 4$, with sum $6$ and product $8$ as required.

$$S_{16} = \frac{16}{2}\Bigl[2(1) + 15 \cdot \frac{1}{2}\Bigr] = 8\Bigl(2 + \frac{15}{2}\Bigr) = 76$$

When $d = -\frac{1}{2}$: $a = 3 + 2 = 5$. Now $a_3 = 4$ and $a_7 = 2$, the same pair in the other order.

$$S_{16} = 8\Bigl[2(5) + 15 \cdot \Bigl(-\frac{1}{2}\Bigr)\Bigr] = 8\Bigl(10 – \frac{15}{2}\Bigr) = 20$$

Question 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?

[Hint: Number of rungs $= \dfrac{250}{25} + 1$]

Fig. 5.7 shows the ladder narrowing towards the top. The bottom rung is marked $45$ cm and the top rung $25$ cm. Neighbouring rungs are $25$ cm apart, and the top and bottom rungs are $2\frac12$ m apart.

Solution. “Decrease uniformly” means the rung lengths form an AP, so the only real work is counting the rungs. Put both distances in the same unit: $2\frac{1}{2}$ m $= 250$ cm. At $25$ cm apart, that span contains

$$\frac{250}{25} = 10 \text{ gaps}$$

and $10$ gaps sit between $11$ rungs, one more than the number of gaps, just as two rungs have one gap between them. So $n = 11$, with first term $45$ and last term $25$:

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

The common difference is never needed here, but for the record it is $\frac{25 – 45}{10} = -2$ cm, so the rungs measure $45, 43, 41, \ldots, 25$ cm.

Question 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: $S_{x-1} = S_{49} – S_x$]

Solution. The house numbers $1, 2, 3, \ldots, 49$ form an AP with $a = d = 1$, so the sum of the first $m$ of them is $S_m = \frac{m(m+1)}{2}$.

The houses before house $x$ are numbered $1$ to $x – 1$, and their sum is $S_{x-1}$. The houses after it are numbered $x + 1$ to $49$, and their sum is all $49$ minus the first $x$, that is, $S_{49} – S_x$. House $x$ itself belongs to neither side. The condition is therefore

$$\frac{(x-1)x}{2} = \frac{49 \times 50}{2} – \frac{x(x+1)}{2}$$

Multiply by $2$ and collect the $x$ terms:

$$x^2 – x + x^2 + x = 2450 \;\Rightarrow\; 2x^2 = 2450 \;\Rightarrow\; x^2 = 1225 \;\Rightarrow\; x = 35$$

taking the positive root, since a house number is positive. Because $35$ is a whole number between $1$ and $49$, such a house really exists, and that is the “show that” part of the question. As a check, the houses before it add up to $\frac{34 \times 35}{2} = 595$. The fourteen houses after it add up to $\frac{14}{2}(36 + 49) = 595$.

Question 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.

[Hint: Volume of concrete required to build the first step $= \frac{1}{4} \times \frac{1}{2} \times 50 \text{ m}^3$]

Fig. 5.8 shows the terrace as a solid flight of steps rising from the ground, the rise of $\frac14$ m and tread of $\frac12$ m marked on the lowest step, and the $50$ m length running along the terrace. Seen from the side, each step is a solid block standing on the ground:

Solution. The hint treats each step as a separate cuboid of solid concrete, $50$ m long and $\frac{1}{2}$ m deep (its tread), standing on the ground. What changes from step to step is the height. The first step is one rise high, $\frac{1}{4}$ m. The second stands one rise higher, $\frac{2}{4}$ m, and in general the $k$th step is $\frac{k}{4}$ m high. Its volume is

$$V_k = \frac{k}{4} \times \frac{1}{2} \times 50 = \frac{25k}{4} \text{ m}^3$$

So the volumes $\frac{25}{4}, \frac{50}{4}, \frac{75}{4}, \ldots$ form an AP with $a = d = \frac{25}{4}$ and $n = 15$:

$$S_{15} = \frac{15}{2}\Bigl[2 \cdot \frac{25}{4} + 14 \cdot \frac{25}{4}\Bigr] = \frac{15}{2} \times 16 \times \frac{25}{4} = \frac{15}{2} \times 100 = 750 \text{ m}^3$$

Equivalently, $\frac{25}{4}(1 + 2 + \cdots + 15) = \frac{25}{4} \times 120 = 750$.

Common mistakes

• Counting gaps instead of rungs in question 3. $250 \div 25 = 10$ is the number of spaces between rungs. There is always one more rung than gaps, so $n = 11$. Taking $n = 10$ gives $350$ cm. The hint’s “$+1$” is there for exactly this reason.
• Mixing units in question 3. The rungs are $25$ cm apart but the span is $2\frac12$ m. Dividing $2.5$ by $25$ produces nonsense; convert to $250$ cm first.
• Keeping only $d = +\frac{1}{2}$ in question 2. Both roots of $d^2 = \frac14$ satisfy the conditions, because the two APs use the same pair of terms, $2$ and $4$, in opposite orders. Reporting only $76$ loses half the answer.
• Answering with the value instead of the position in question 1. The question asks which term, so the answer is the $32$nd term, not $-3$. Stopping at $n = 31$, the last positive term, is the other slip.
• Including house $x$ in one of the sums in question 4. The houses “following” house $x$ start at $x + 1$, so their sum is $S_{49} – S_x$, not $S_{49} – S_{x-1}$. Using the wrong one gives an equation with no whole-number solution.
• Giving every step the same height in question 5. Each step stands on the ground, so the $k$th is $\frac{k}{4}$ m tall. Multiplying the first step’s volume by $15$ gives $93.75 \text{ m}^3$, far too little.

Practise next

• Exercise 5.3 — the core sum exercise, where $a$, $d$ and $n$ are given directly. It is worth revising if any step here felt unfamiliar.
• Chapter 4, Exercise 4.2 — quadratic equations by factorisation, which settle question 2 above and several parts of Exercise 5.3.
• Chapter 7, Exercise 7.1 — the next chapter, where the distance formula replaces sequences.