GSEB Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.4

Question 1. Which term of AP 121, 117, 113, ........is its first negative term?
Answer: The given Arithmetic Progression (AP) is \( 121, 117, 113, \dots \).
Here, the first term \( a = 121 \).
The common difference \( d = 117 - 121 = -4 \).
Now, we will let the \( n^{th} \) term of the given AP be a negative term.
So, \( a_n < 0 \)
\( a + (n - 1)d < 0 \)
\( 121 + (n - 1)(-4) < 0 \)
\( 121 < (n - 1) \times 4 \)
\( n - 1 > \frac{121}{4} \)
\( n > \frac{121}{4} + 1 \)
\( n > 30.25 + 1 \)
\( n > 31.25 \)
The smallest whole number value for \( n \) is 32.
Hence, the 32nd term is the initial negative term of this AP.
In simple words: We find the first term and common difference. Then, we set the nth term to be less than zero and solve for n. The smallest whole number greater than our calculated n value will give us the first negative term's position.

Exam Tip: Remember to use the correct formula for the \( n^{th} \) term of an AP and always select the smallest integer greater than the calculated value of \( n \) for "first negative term" questions.

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, respectively.
Answer: Let 'a' represent the initial term and 'd' be the common difference of the provided AP.
As per the given information:
Third term + seventh term = 6
\( (a + 2d) + (a + 6d) = 6 \)
\( 2a + 8d = 6 \)
\( a + 4d = 3 \) .........(1) (dividing by 2)
And the product of the third term and seventh term = 8
\( (a + 2d)(a + 6d) = 8 \)
From equation (1), we obtain \( a = 3 - 4d \).
Substituting the value of 'a' into the second equation:
\( (3 - 4d + 2d)(3 - 4d + 6d) = 8 \)
\( (3 - 2d)(3 + 2d) = 8 \)
\( 9 - 4d^2 = 8 \)
\( -4d^2 = -1 \)
\( d^2 = \frac{1}{4} \)
\( d = \pm \frac{1}{2} \)

**Scenario I: When \( d = \frac{1}{2} \)**
Placing the value of 'd' into equation (1):
\( a + 4 \times \frac{1}{2} = 3 \)
\( a + 2 = 3 \)
\( a = 1 \)
For \( n = 16 \), the sum of the first sixteen terms of this AP is \( S_{16} \).
The formula for the sum of \( n \) terms is \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{16} = \frac{16}{2}[2 \times 1 + (16 - 1)\frac{1}{2}] \)
\( S_{16} = 8[2 + 15 \times \frac{1}{2}] \)
\( S_{16} = 8[2 + \frac{15}{2}] \)
\( S_{16} = 8[\frac{4+15}{2}] \)
\( S_{16} = 8[\frac{19}{2}] \)
\( S_{16} = 4 \times 19 = 76 \)

**Scenario II: When \( d = -\frac{1}{2} \)**
Substituting the value \( d = -\frac{1}{2} \) into equation (1):
\( a + 4(-\frac{1}{2}) = 3 \)
\( a - 2 = 3 \)
\( a = 5 \)
For \( n = 16 \), the sum of the initial sixteen terms of this AP is \( S_{16} \).
\( S_{16} = \frac{16}{2}[2 \times 5 + (16 - 1)(-\frac{1}{2})] \)
\( S_{16} = 8[10 + 15(-\frac{1}{2})] \)
\( S_{16} = 8[10 - \frac{15}{2}] \)
\( S_{16} = 8[\frac{20-15}{2}] \)
\( S_{16} = 8[\frac{5}{2}] \)
\( S_{16} = 4 \times 5 = 20 \)
Therefore, the sum of the first sixteen terms of the AP can be either 76 or 20.
In simple words: First, we use the given sum and product of the third and seventh terms to find the first term 'a' and common difference 'd'. We get two possible values for 'd'. Then, we use each 'd' value with the corresponding 'a' to calculate the sum of the first sixteen terms, resulting in two possible sums.

Exam Tip: For problems involving two conditions and finding AP elements, expect multiple possible solutions, especially when a quadratic equation for 'd' arises. Remember to calculate sums for all valid 'd' values.

Question 3. A ladder has rungs 25 cm apart. (see Fig.). 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 2.5 m apart, what is the length of the wood required for the rungs?
The ladder can be visualized as two slanted sides with horizontal rungs of decreasing length:

Answer: The distance between the top and bottom rungs is 2.5 m, which is \( 250 \) cm.
The rungs are \( 25 \) cm apart.
Number of rungs = \( \frac{250}{25} + 1 = 10 + 1 = 11 \).
It is stated that the length of the rungs gradually lessens from \( 45 \) cm at the bottom to \( 25 \) cm at the top.
Thus, the rung lengths create an Arithmetic Progression (AP) where:
The first term \( a = 45 \) cm (length of the bottom rung).
The last term \( l = 25 \) cm (length of the top rung).
The number of rungs \( n = 11 \).
The overall length of wood needed for the rungs is the sum of this AP:
\( S_n = \frac{n}{2}(a + l) \)
\( S_{11} = \frac{11}{2}(45 + 25) \)
\( S_{11} = \frac{11}{2}(70) \)
\( S_{11} = 11 \times 35 \)
\( S_{11} = 385 \) cm.
Converting to meters: \( 385 \) cm \( = 3.85 \) meters.
In simple words: First, count how many rungs there are by dividing the total height by the rung spacing and adding one. Then, since the rung lengths change evenly, we have an AP. Use the sum formula for an AP with the first, last, and total number of rungs to find the total wood needed.

Exam Tip: Pay close attention to unit conversions (cm to m) and correctly determine the number of items (rungs) in the sequence by adding 1 to the quotient of total distance by spacing.

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.
Answer: The house numbers in a row follow a sequence: 1, 2, 3, ..., up to 49.
These numbers create an Arithmetic Progression (AP).
Here, the first term \( a = 1 \), and the common difference \( d = 2 - 1 = 1 \).
The sum of \( n \) terms of an AP is given by \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
The sum of the numbers of houses preceding the house numbered \( x \) is \( S_{x-1} \).
\( S_{x-1} = \frac{x-1}{2}[2(1) + (x-1-1)(1)] \)
\( S_{x-1} = \frac{x-1}{2}[2 + x - 2] \)
\( S_{x-1} = \frac{x-1}{2}[x] \)
\( S_{x-1} = \frac{x(x-1)}{2} \)
The total sum of numbers of all houses is \( S_{49} \).
\( S_{49} = \frac{49}{2}[2(1) + (49 - 1)(1)] \)
\( S_{49} = \frac{49}{2}[2 + 48] \)
\( S_{49} = \frac{49}{2}[50] \)
\( S_{49} = 49 \times 25 \)
\( S_{49} = 1225 \)
The sum of the numbers of houses from 1 to \( x \) is \( S_x \).
\( S_x = \frac{x}{2}[2(1) + (x - 1)(1)] \)
\( S_x = \frac{x}{2}[2 + x - 1] \)
\( S_x = \frac{x}{2}[x + 1] \)
\( S_x = \frac{x(x+1)}{2} \)
According to the problem, the sum of numbers preceding house \( x \) equals the sum of numbers following house \( x \).
Sum of numbers following house \( x = S_{49} - S_x \).
So, \( S_{x-1} = S_{49} - S_x \)
\( \frac{x(x-1)}{2} = 49 \times 25 - \frac{x(x+1)}{2} \)
Multiply the entire equation by 2 to remove denominators:
\( x(x-1) = 49 \times 50 - x(x+1) \)
\( x^2 - x = 2450 - (x^2 + x) \)
\( x^2 - x = 2450 - x^2 - x \)
\( x^2 = 2450 - x^2 \)
\( 2x^2 = 2450 \)
\( x^2 = \frac{2450}{2} \)
\( x^2 = 1225 \)
\( x = \sqrt{1225} \)
\( x = 35 \)
Since \( x \) must be a positive integer and represent a house number, \( x = 35 \) is the correct value.
In simple words: We calculate the sum of numbers before house 'x' and the sum of all houses up to 'x'. The sum of houses after 'x' is found by subtracting the sum up to 'x' from the total sum. By setting the "before" sum equal to the "after" sum, we can solve for 'x'.

Exam Tip: Clearly define the sum notation (e.g., \( S_{x-1} \) for preceding houses and \( S_{49} - S_x \) for succeeding houses) and ensure careful algebraic manipulation, especially when dealing with negative signs.

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 figure). Calculate the total volume of concrete required to build the terrace.

Answer: We see that each step has a length of 50 m and a width (tread) of \( \frac{1}{2} \) m.
The height (rise) of each successive step increases by \( \frac{1}{4} \) m.
Height of the first step = \( \frac{1}{4} \) m.
The height of the second step is \( \frac{1}{4} \) m plus \( \frac{1}{4} \) m, which totals \( \frac{2}{4} \) m.
The third step's height is \( \frac{3}{4} \) m, and this pattern continues.
Let's consider the volumes of the concrete steps as \( V_1, V_2, V_3, \dots, V_{15} \).
The volume of a step is length \( \times \) breadth \( \times \) height.
Therefore, \( V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} \) m³.
\( V_2 = 50 \times \frac{1}{2} \times \frac{2}{4} \) m³.
\( V_3 = 50 \times \frac{1}{2} \times \frac{3}{4} \) m³.
... and so on.
The overall concrete volume needed is \( V = V_1 + V_2 + V_3 + \dots + V_{15} \).
\( V = (50 \times \frac{1}{2} \times \frac{1}{4}) + (50 \times \frac{1}{2} \times \frac{2}{4}) + (50 \times \frac{1}{2} \times \frac{3}{4}) + \dots + (50 \times \frac{1}{2} \times \frac{15}{4}) \)
We can factor out common terms:
\( V = 50 \times \frac{1}{2} \times \frac{1}{4} [1 + 2 + 3 + \dots + 15] \)
\( V = \frac{50}{8} [1 + 2 + 3 + \dots + 15] \)
\( V = \frac{25}{4} [1 + 2 + 3 + \dots + 15] \)
We use the formula for the sum of the first \( n \) natural numbers: \( S_n = \frac{n(n+1)}{2} \).
Here, \( n = 15 \).
Sum \( = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120 \).
Now, substitute this sum back into the volume equation:
\( V = \frac{25}{4} \times 120 \)
\( V = 25 \times 30 \)
\( V = 750 \) m³.
Thus, the total volume of concrete required is \( 750 \) m³.
In simple words: First, understand the dimensions of each step: length, breadth (tread), and height (rise). Notice that the height increases for each step, forming an arithmetic progression. Calculate the volume for each step and sum them up. Factor out common parts and use the formula for the sum of natural numbers to simplify the calculation.

Exam Tip: For volume calculations of stepped structures, recognize that while length and tread might be constant, the height changes. This often leads to a sum of an AP, where factoring out common dimensions simplifies the calculation significantly.