Exercise 5.3 – Stepwise Solutions
Formulas Used:
Sₙ = n/2 [2a + (n − 1)d]
Sₙ = n/2 (a + l)
1. Find the sum of the following APs
(i) 2, 7, 12, … (10 terms)
a = 2, d = 5, n = 10
S₁₀ = 10/2 [2(2) + 9(5)]
= 5 [4 + 45]
= 5 × 49
= 245
(ii) −37, −33, −29, … (12 terms)
a = −37, d = 4, n = 12
S₁₂ = 12/2 [2(−37) + 11(4)]
= 6 [−74 + 44]
= 6 (−30)
= −180
(iii) 0.6, 1.7, 2.8, … (100 terms)
a = 0.6, d = 1.1, n = 100
S₁₀₀ = 100/2 [2(0.6) + 99(1.1)]
= 50 [1.2 + 108.9]
= 50 (110.1)
= 5505
(iv) 1/15, 1/12, 1/10, … (11 terms)
a = 1/15, d = 1/60, n = 11
S₁₁ = 11/2 [2(1/15) + 10(1/60)]
= 11/2 [2/15 + 1/6]
= 11/2 [4/30 + 5/30]
= 11/2 (9/30)
= 99/60
= 33/20
2. Find the sums
(i) 7 + 10½ + 14 + … + 84
a = 7, d = 3.5, l = 84
84 = 7 + (n − 1)3.5
77 = 3.5(n − 1)
n − 1 = 22
n = 23
S = 23/2 (7 + 84)
= 23/2 × 91
= 1046.5
(ii) 34 + 32 + 30 + … + 10
a = 34, d = −2, l = 10
10 = 34 + (n − 1)(−2)
−24 = −2(n − 1)
n − 1 = 12
n = 13
S = 13/2 (34 + 10)
= 13/2 × 44
= 286
(iii) −5 + (−8) + (−11) + … + (−230)
a = −5, d = −3, l = −230
−230 = −5 + (n − 1)(−3)
−225 = −3(n − 1)
n − 1 = 75
n = 76
S = 76/2 (−5 − 230)
= 38 (−235)
= −8930
4. Number of terms in AP: 9, 17, 25, … sum = 636
a = 9, d = 8
636 = n/2 [18 + 8(n − 1)]
636 = n/2 (8n + 10)
1272 = n(8n + 10)
8n² + 10n − 1272 = 0
n = 12
5.
a = 5, l = 45, S = 400
400 = n/2 (5 + 45)
400 = 25n
n = 16
d = (45 − 5)/15 = 40/15 = 8/3
6.
a = 17, l = 350, d = 9
350 = 17 + (n − 1)9
333 = 9(n − 1)
n − 1 = 37
n = 38
S = 38/2 (17 + 350)
= 19 × 367
= 6973
7.
d = 7, a₂₂ = 149
a + 21×7 = 149
a = 2
S₂₂ = 22/2 [4 + 21×7]
= 11 [4 + 147]
= 11 × 151
= 1661
8.
a₂ = 14, a₃ = 18
d = 4
a = 10
S₅₁ = 51/2 [20 + 50×4]
= 51/2 (220)
= 5610
9.
S₇ = 49
S₁₇ = 289
Solving gives:
a = 1, d = 2
Sₙ = n²
12.
First 40 multiples of 6:
S = 40/2 (6 + 240)
= 20 × 246
= 4920
13.
First 15 multiples of 8:
S = 15/2 (8 + 120)
= 15/2 × 128
= 960
14.
Odd numbers between 0 and 50:
1 to 49 (25 terms)
S = 25/2 (1 + 49)
= 25/2 × 50
= 625
Exercise 5.3 – Solutions (Questions 15–18)
15. Penalty for 30 Days Delay
Given:
First day penalty (a) = 200
Common difference (d) = 50
Number of days (n) = 30
Formula:
Sₙ = n/2 [2a + (n − 1)d]
S₃₀ = 30/2 [2(200) + 29(50)]
= 15 [400 + 1450]
= 15 × 1850
= 27,750
Total Penalty = ₹27,750
16. Seven Cash Prizes (Total ₹700)
Let first prize = a
Common difference (d) = −20
Number of prizes (n) = 7
700 = 7/2 [2a + 6(−20)]
700 = 7/2 (2a − 120)
700 = 7(a − 60)
a − 60 = 100
a = 160
Prizes:
160, 140, 120, 100, 80, 60, 40
17. Trees Planted
Classes: 1 to 12
Sections per class = 3
Total trees = 3(1 + 2 + 3 + … + 12)
= 3 × [12/2 (1 + 12)]
= 3 × 6 × 13
= 234
Total Trees = 234
18. Length of Spiral (13 Semicircles)
Radii form AP:
0.5, 1.0, 1.5, …
a = 0.5
d = 0.5
n = 13
a₁₃ = 0.5 + 12(0.5)
= 6.5
Sum of radii:
S = 13/2 (0.5 + 6.5)
= 13/2 × 7
= 45.5
Total length = π × 45.5
Using π = 22/7:
= 22/7 × 45.5
= 143 cm
Total Length = 143 cm
Exercise 5.3 – Solutions
19. 200 Logs Arrangement
Logs per row form an AP:
20, 19, 18, …
First term (a) = 20
Common difference (d) = −1
Total logs (Sₙ) = 200
Formula:
Sₙ = n/2 [2a + (n − 1)d]
200 = n/2 [2(20) + (n − 1)(−1)]
200 = n/2 [40 − (n − 1)]
200 = n/2 (41 − n)
400 = n(41 − n)
400 = 41n − n²
n² − 41n + 400 = 0
(n − 16)(n − 25) = 0
n = 16 or 25
Since number of rows cannot exceed 20 (first row has 20 logs),
n = 16
Top row logs:
a₁₆ = 20 + (16 − 1)(−1)
= 20 − 15
= 5
Number of rows = 16
Logs in top row = 5
20. Potato Race – Total Distance
Distances of potatoes from bucket form an AP:
5, 8, 11, 14, …
First term (a) = 5
Common difference (d) = 3
Number of potatoes (n) = 10
10th distance:
a₁₀ = 5 + 9(3)
= 5 + 27
= 32
Sum of distances (one way):
S = n/2 (a + l)
= 10/2 (5 + 32)
= 5 × 37
= 185
Total running distance (going and returning each time):
= 2 × 185
= 370 metres
Total distance = 370 m