The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P.
a = 1
d = 2
l = 49
l = a + (n − 1) d
49 = 1 + (n − 1)2
48 = 2(n − 1)
n − 1 = 24
n = 25
S_n = n/2 (a + l)
S₂₅ = 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S₃₀
= 30/2 [2(200) + (30 - 1) 50]

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S₃₀
= 30/2 [2(200) + (30 - 1) 50]

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.

Let the cost of 1^st prize be P.
Cost of 2^nd prize = P − 20
And cost of 3^rd prize = P − 40
It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
a = P
d = −20
Given that, S₇ = 700
7/2 [2a + (7 - 1)d] = 700

a + 3(−20) = 100
a − 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 − 1 = 1
S_n = n/2 [2a + (n - 1)d]
S₁₂ = 12/2 [2(1) + (12 - 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234
Therefore, 234 trees will be planted by the students.

perimeter of semi-circle = πr
_P1 = π(0.5) = π/2 cm
_P2 = π(1) = π cm
_P3 = π(1.5) = 3π/2 cm
_P1, P₂, P₃ are the lengths of the semi-circles

π/2, π, 3π/2, 2π, ....
P1 = π/2 cm
_P2 = π cm
d = P2- P1 = π - π/2 = π/2
First term = P1 = a = π/2 cm
S_n = n/2 [2a + (n - 1)d]
Therefor, Sum of the length of 13 consecutive circles
S₁₃ = 13/2 [2(π/2) + (13 - 1)π/2]
=  13/2 [π + 6π]
=13/2 (7π)  = 13/2 × 7 × 22/7
= 143 cm

It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a₂ − a₁ = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
S_n = 200
S_n = n/2 [2a + (n - 1)d]
S₁₂ = 12/2 [2(20) + (n - 1)(-1)]
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n − n²
n² − 41n + 400 = 0
n² − 16n − 25n + 400 = 0
n (n − 16) −25 (n − 16) = 0
(n − 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n = 25
a_n = a + (n − 1)d
a₁₆ = 20 + (16 − 1) (−1)
a₁₆ = 20 − 15
a₁₆ = 5
Similarly,
a₂₅ = 20 + (25 − 1) (−1)
a₂₅ = 20 − 24
= −4
Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

The distances of potatoes from the bucket are 5, 8, 11, 14…
Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are
10, 16, 22, 28, 34,……….
a = 10
d = 16 − 10 = 6
S₁₀ =?
S₁₀ = 12/2 [2(20) + (n - 1)(-1)]
= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.