Exercise 5.4 (Optional) – Stepwise Solutions
1. Which term of the AP 121, 117, 113, … is its first negative term?
a = 121, d = −4
an = a + (n − 1)d
an = 121 + (n − 1)(−4)
an = 121 − 4(n − 1)
an = 125 − 4n
For first negative term:
125 − 4n < 0
125 < 4n
n > 31.25
Smallest integer n = 32
Answer: 32nd term
2. Sum of 3rd and 7th terms is 6 and their product is 8
a3 = a + 2d
a7 = a + 6d
a3 + a7 = 6
2a + 8d = 6
a + 4d = 3 …(1)
(a + 2d)(a + 6d) = 8 …(2)
From (1): a = 3 − 4d
Substitute in (2):
(3 − 4d + 2d)(3 − 4d + 6d) = 8
(3 − 2d)(3 + 2d) = 8
9 − 4d² = 8
4d² = 1
d = ±1/2
If d = 1/2 → a = 1
AP: 1, 1.5, 2, …
S16 = 16/2 [2(1) + 15(1/2)]
= 8[2 + 7.5]
= 8 × 9.5
S16 = 76
3. Ladder Problem
Distance between top and bottom rungs = 2.5 m = 250 cm
Distance between rungs = 25 cm
Number of rungs = 250/25 + 1 = 10 + 1 = 11
Lengths form AP:
a = 45 cm, l = 25 cm, n = 11
S = n/2 (a + l)
S = 11/2 (45 + 25)
= 11/2 × 70
= 11 × 35
Total wood required = 385 cm
4. Houses numbered 1 to 49
Total sum = 49/2 × 50 = 1225
Let house number = x
Sum before x = x(x − 1)/2
Sum after x = 1225 − x(x + 1)/2
Set equal:
x(x − 1)/2 = 1225 − x(x + 1)/2
Simplifying:
x² = 1225
x = 35
Answer: x = 35
5. Volume of Concrete
Number of steps = 15
Each step length = 50 m
Rise = 1/4 m
Tread = 1/2 m
Volume of first step = (1/4 × 1/2 × 50)
= 50/8 = 6.25 m³
Volumes form AP:
a = 6.25
d = 6.25
n = 15
S = n/2 [2a + (n − 1)d]
S = 15/2 [12.5 + 87.5]
= 15/2 × 100
= 15 × 50
Total volume = 750 m³
Final Answers Summary
1) 32nd term
2) S16 = 76
3) 385 cm
4) x = 35
5) 750 m³