Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.4
Question 1
Which term of the AP : 121, 117, 113, . . ., is its first negative term?
[Hint : Find n for an < 0]
Answer:
We have the A.P. having a = 121 and d = 117 – 121 = – 4
∴ an = a + (n – 1) d
= 121 + (n – 1) × (- 4)
= 121 – 4n + 4
= 125 – 4n
For the first negative term, we have
an < 0 =(125 – 4n) < 0 = 125 < 4n = 125/4 31 1⁄4
Thus, the first negative term is 32nd term.
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.
Answer:
We are given that the sum of the third and seventh terms of an arithmetic progression (AP) is 6 and their product is 8.
Step 1: The formula for the nth term of an AP is
T_n = a + (n – 1) * d
where
a is the first term
d is the common difference
T_n is the nth term.
Step 2: Using the given information
We know
- The sum of the third and seventh terms is 6: T_3 + T_7 = 6
- The product of the third and seventh terms is 8: T_3 * T_7 = 8
The third term is T_3 = a + 2d
The seventh term is T_7 = a + 6d
Step 3: Set up the equations
From the sum condition
T_3 + T_7 = (a + 2d) + (a + 6d) = 6
This simplifies to 2a + 8d = 6
So a + 4d = 3 (Equation 1)
From the product condition
T_3 * T_7 = (a + 2d)(a + 6d) = 8
Expanding this gives a^2 + 8ad + 12d^2 = 8 (Equation 2)
Step 4: Solve the system of equations
From Equation 1, solve for a
a = 3 – 4d
Substitute this into Equation 2
(3 – 4d)^2 + 8(3 – 4d)d + 12d^2 = 8
Simplifying this gives 9 – 4d^2 = 8
So d^2 = 1/4
Thus d = ± 1/2
Step 5: Find the value of a
Substitute d = ± 1/2 into a = 3 – 4d
If d = 1/2, then a = 1
If d = -1/2, then a = 5
Step 6: Find the sum of the first sixteen terms
The sum of the first n terms of an AP is
S_n = n/2 * (2a + (n – 1) * d)
For n = 16, calculate the sum for both cases
Case 1: a = 1 and d = 1/2
S_16 = 8 * (2 + 15 * 1/2) = 8 * 9.5 = 76
Case 2: a = 5 and d = -1/2
S_16 = 8 * (10 + 15 * -1/2) = 8 * 2.5 = 20
Step 7: Conclusion
The sum of the first sixteen terms of the AP is 20 when a = 5 and d = -1/2
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 21⁄2 m apart, what is the length of the wood required for the rungs?
[Hint : Number of rungs = 250/25 + 1]
Answer:
Step 1: Calculate the number of rungs
The rungs are 25 cm apart, and the distance between the top and bottom rungs is 250 cm.
Number of rungs = 250 / 25 = 10 rungs.
Step 2: Length of the rungs
The length of the rungs decreases from 45 cm at the bottom to 25 cm at the top. This forms an arithmetic progression, where:
The first term (a1) is 45 cm (bottom rung),
The last term (an) is 25 cm (top rung),
The common difference (d) is calculated as:
d = (25 – 45) / (10 – 1) = -20 / 9 = -2.22 cm.
Step 3: Total length of wood
To find the total length of wood, use the formula for the sum of an arithmetic progression:
Sn = n / 2 * (a1 + an)
Where:
n is the number of rungs (10),
a1 is the length of the first rung (45 cm),
an is the length of the last rung (25 cm).
Substitute the values:
S10 = 10 / 2 * (45 + 25) = 5 * 70 = 350 cm.
Step 4: Convert to meters
350 cm = 350 / 100 = 3.5 meters.
So, the total length of wood required for the rungs is 350 cm or 3.5 meters.
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 : Sx-1 = S49 – Sx]
Answer:
We are asked to find a house number xxx such that the sum of the house numbers before house xxx is equal to the sum of the house numbers after house xxx.
- The sum of house numbers before house xxx is the sum of the first x−1x-1x−1 numbers. This sum is given by the formula:
Sum before house xxx = (x – 1) * x / 2 - The sum of house numbers after house xxx is the total sum of all house numbers (from 1 to 49) minus the sum of the numbers from 1 to xxx. The total sum of the numbers from 1 to 49 is:
Total sum = 49 * 50 / 2 = 1225
So, the sum of the house numbers after house xxx is:
Sum after house xxx = 1225 – (x * (x + 1)) / 2 - Now, set the sum before house xxx equal to the sum after house xxx:
(x – 1) * x / 2 = 1225 – (x * (x + 1)) / 2 - Multiply both sides by 2 to get rid of the fractions:
(x – 1) * x = 2450 – x * (x + 1) - Expand both sides:
x^2 – x = 2450 – (x^2 + x) - Simplify the equation:
x^2 – x = 2450 – x^2 – x - Move all terms to one side:
2x^2 = 2450 - Solve for :
x^2 = 1225 - Take the square root of both sides:
x = 35
So, the value of xxx is 35. This means that the sum of the house numbers before house 35 is equal to the sum of the house numbers after house 35.
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 1⁄4 m and a tread of 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 = 1/4 × 1/2 × 50m3]
Answer:
To calculate the total volume of concrete required to build the terrace:
- Volume of Concrete for Each Step:
- The volume for each step increases by a constant amount.
- The formula for the volume of the n-th step is: Volume of the n-th step = (25n/4) m^3
- Volumes for Each Step:
- For the first step: Volume = (25 * 1) / 4 = 25/4 m^3
- For the second step: Volume = (25 * 2) / 4 = 50/4 = 25/2 m^3
- For the third step: Volume = (25 * 3) / 4 = 75/4 m^3
- This pattern continues for all 15 steps.
- Total Volume Using Arithmetic Series:
- The total volume of concrete for all 15 steps is the sum of an arithmetic series.
- The formula to find the sum of an arithmetic series is: Sum = (n / 2) * [2a + (n – 1) * d] where:
- a = 25/4 is the volume of the first step.
- d = 25/4 is the difference between the volumes of consecutive steps.
- n = 15 is the total number of steps.
- Substitute the Values:
- Sum = (15 / 2) * [2 * (25 / 4) + (15 – 1) * (25 / 4)]
- Sum = (15 / 2) * [(50 / 4) + (350 / 4)]
- Sum = (15 / 2) * (400 / 4)
- Sum = (15 / 2) * 100 = 750
So, the total volume of concrete required to build the terrace is 750 cubic meters.
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