NCERT Solutions Class 10 Maths Arithmetic Progressions Exercise 5.4

Core Mathematical Concepts in Exercise 5.4

Exercise 5.4 tests your logical reasoning by forcing you to look beyond simple direct applications of the formulas $a_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$. Instead, it introduces mathematical inequalities and real-world system modeling. For instance, when finding the first negative term of a decreasing AP, you must solve the inequality $a_n < 0$. This requires careful algebraic manipulation, especially since the common difference $d$ is negative, which reverses the inequality sign when dividing. Another hallmark of this exercise is setting up quadratic systems from contextual data, such as finding a specific house number in a row of houses where sum of preceding houses equals the sum of succeeding houses. Solving these requires translating raw word problems into clean, solvable algebraic models.

How to Find the First Negative Term of an AP

1. Identify Key Parameters — Extract the first term ($a$) and the common difference ($d$) directly from the given progression. Ensure $d$ is calculated correctly as $a_2 - a_1$.
2. Set Up the Inequality — To find a negative term, establish the condition that the $n$-th term must be less than zero: $a_n < 0$. Substitute the formula: $a + (n-1)d < 0$.
3. Solve for $n$ — Isolate the variable $n$. Crucial rule: When you divide or multiply both sides of an inequality by a negative number (since $d$ is negative for a decreasing sequence), you must reverse the inequality sign.
4. Determine the Integer Term Position — Since the position of a term $n$ must be a natural number ($1, 2, 3, \dots$), choose the smallest integer that is strictly greater than the fraction obtained in the previous step.

Avoid These Common Board Exam Mistakes

Watch out for the inequality sign flip! When solving $a + (n-1)d < 0$ where $d$ is negative, students often forget to reverse the inequality symbol when dividing by $d$. For example, if $-4n < -117$, dividing by $-4$ yields $n > 29.25$, not $n < 29.25$. Since $n$ must be a positive integer, we round up to the next integer, which is $30$. Never leave $n$ as a decimal or fraction in your final answer, as terms can only exist at whole numbered positions ($1st, 2nd, 3rd$ terms).

Practice Questions with Solutions

• Q: Find the first negative term of the AP: $121, 117, 113, \dots$ A: Step 1: Identify the components of the AP. Here, the first term $a = 121$ and the common difference $d = 117 - 121 = -4$. Step 2: Set up the inequality for the negative term. Let the $n$-th term $a_n < 0$. $a + (n-1)d < 0$ Substitute the values: $121 + (n-1)(-4) < 0$ $121 - 4n + 4 < 0$ $125 - 4n < 0$ Step 3: Solve the inequality. $-4n < -125$ Divide by $-4$ and reverse the inequality sign: $n > \frac{125}{4}$ $n > 31.25$ Step 4: Determine the integer value. The smallest integer greater than $31.25$ is $32$. Final answer: The first negative term of the AP is the 32nd term.
• Q: The sum of the third and seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. A: Step 1: Write down the equations based on the conditions. $a_3 + a_7 = 6 \implies (a + 2d) + (a + 6d) = 6 \implies 2a + 8d = 6 \implies a + 4d = 3 \implies a = 3 - 4d$ Step 2: Use the product condition. $a_3 \times a_7 = 8 \implies (a+2d)(a+6d) = 8$ Substitute $a = 3-4d$: $(3 - 4d + 2d)(3 - 4d + 6d) = 8$ $(3 - 2d)(3 + 2d) = 8$ Using identity $(x-y)(x+y) = x^2 - y^2$: $9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d^2 = \frac{1}{4} \implies d = \pm \frac{1}{2}$ Step 3: Solve for $a$ and find the sum ($S_{16}$) for both cases. Case 1: If $d = \frac{1}{2}$, then $a = 3 - 4(\frac{1}{2}) = 1$. $S_{16} = \frac{16}{2}[2(1) + (16-1)(\frac{1}{2})] = 8[2 + 7.5] = 8[9.5] = 76$. Case 2: If $d = -\frac{1}{2}$, then $a = 3 - 4(-\frac{1}{2}) = 5$. $S_{16} = \frac{16}{2}[2(5) + (16-1)(-\frac{1}{2})] = 8[10 - 7.5] = 8[2.5] = 20$. Final answer: The sum of the first sixteen terms can be either 76 or 20.
• Q: A row of houses is 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 $x$. A: Step 1: Define the sums mathematically. The houses are numbered $1, 2, 3, \dots, 49$. This is an AP where $a = 1, d = 1$. Sum of preceding houses = $S_{x-1}$ Sum of succeeding houses = Total Sum ($S_{49}$) - Sum up to $x$ ($S_x$) So, the condition is: $S_{x-1} = S_{49} - S_x$ Step 2: Apply the sum formula $S_n = \frac{n(n+1)}{2}$ since $a=1, d=1$. $\frac{(x-1)x}{2} = \frac{49(50)}{2} - \frac{x(x+1)}{2}$ Multiply the entire equation by 2 to clear denominators: $x(x-1) = 49(50) - x(x+1)$ $x^2 - x = 2450 - x^2 - x$ Step 3: Solve for $x$. $2x^2 = 2450$ $x^2 = 1225$ $x = \sqrt{1225} = 35$ (taking positive root as house numbers are positive). Final answer: The value of $x$ is 35.
• Q: A small terrace at a football ground comprises 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. Calculate the total volume of concrete required to build the terrace. A: Step 1: Understand the volume of concrete for each step. Volume = Length $\times$ Width (Tread) $\times$ Height (Rise). For 1st step: Volume $V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} = 6.25\text{ m}^3$. For 2nd step: Volume $V_2 = 50 \times \frac{1}{2} \times \frac{2}{4} = 12.5\text{ m}^3$. For 3rd step: Volume $V_3 = 50 \times \frac{1}{2} \times \frac{3}{4} = 18.75\text{ m}^3$. Step 2: Establish the arithmetic progression. The volumes form an AP: $6.25, 12.5, 18.75, \dots$ to $15$ terms. Here, $a = 6.25$, $d = 12.5 - 6.25 = 6.25$, and $n = 15$. Step 3: Calculate the total sum of 15 terms ($S_{15}$). $S_{15} = \frac{15}{2} [2a + (15-1)d]$ $S_{15} = \frac{15}{2} [2(6.25) + 14(6.25)]$ $S_{15} = \frac{15}{2} [12.5 + 87.5]$ $S_{15} = \frac{15}{2} [100]$ $S_{15} = 15 \times 50 = 750\text{ m}^3$. Final answer: The total volume of concrete required to build the terrace is 750 cubic meters.

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