Example 1: Using an Arithmetic Series to Model and Solve a Problem
There are 25 trees at equal distance of 5m in a line with a water tank, the distance of the water tank from the nearest tree being 10m. A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next. Find the total distance covered by the gardener in order to water all the trees.

Solution:
Distance covered by the gardener to water the first tree and return to the water tank
=10m+10m=20m.
Distance covered by the gardener to water the second tree and return to the water tank
=15m+15m=30m.
Distance covered by the gardener to water the third tree and return to the water tank =20m+20m=40m and soon.
∴ Total distance covered by the gardener to water all the trees =20m+30m+40m+⋯ up to 25 terms
This series is an arithmetic series.
Here, a=20, d=30-20=10 and n=25.
Using the formula, S_n=½n[2a+(n-1)d] we get
S₂₅=½⋅25⋅[2⋅20+(25-1)⋅10]
=25⋅½⋅(40+240)
=25⋅140
=3500.
Hence, the total distance covered by the gardener to water all the trees 3500m.

Question 1. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance [in metres) run by a competitor is 2×5+2×(5+3)]
Solution:
Total distance travel to pick the first potato =2×5=10
Total distance travel to pick the second potato =2×(5+3)=16
Similarly, the series of distances travelled to pick the potatoes are 10, 16, 22, 28, ….
Here, a=10, d=16-10=6 and n=10.
The sum of n terms of an AP is given by S_n=½n[2a+(n-1)d].
S₁₀=½⋅10⋅[2(10)+(10-1)⋅6]
=5⋅[20+54]
=5⋅74=370 m
Hence, total distance travelled by the competitor is 370 m.

Q2. In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3m apart in a straight line. There are 10 potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and he continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution:
Distance covered by the competitor to pick and drop the first potato =2⋅5m=10m Distance covered by the competitor to pick and drop the second potato =2⋅(5+3)m=2⋅8m=16m.
Distance covered by the competitor to pick and drop the third potato
=2⋅(5+3+3)m=2⋅11m=22m and so on.
∴ Total distance covered by the competitor =10m+16m+22m+⋯ up to 10 terms
This is an arithmetic series.
Here, a=10, d=16-10=6 and n=10.
Using the formula, S_n=½n[2a+(n-1)d], we get
S₁₀=½⋅10⋅[2⋅10+(10-1)⋅6]
=5⋅(20+54)
=5⋅74
=370.
Hence, the total distance the competitor has to run is 370m.

Q3. Students created a trapezoid from the cans they had collected for the food bank. There were 10 rows in the trapezoid. The bottom row had 100 cans. Each consecutive row had 5 fewer cans than the previous row. How many cans were in the trapezoid?
Solution:
The numbers of cans in the rows form an arithmetic sequence with first 3 terms 100, 95, 90, …
The total number of cans is the sum of the first 10 terms of the arithmetic series:

100+95+90+⋯
Use:
S_n=½n(2a₁+d(n-1))
Substitute: n=10, a₁=100, d=-5
S₁₀=½⋅10⋅(2⋅100-5⋅(10-1))
S₁₀=5⋅(200-45)=5⋅155
S₁₀=775
There were 775 cans in the trapezoid.

Q4. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Each section of each class will plant tree =3× Class, therefore
Total number of tree planted by class I =3×1=3
Total number of tree planted by class II =3×2=6
Total number of tree planted by class III =3×3=9
Similarly, the series of trees planted by classes are as follows: 3, 6, 9, …, 36.
Here, a=3, d=6-3=3 and n=12.
The sum of n terms of an AP is given by S_n=½n[2a+(n-1)d].

S₁₂=½⋅12⋅[2(3)+(12-1)(3)]
=6⋅(6+33)
=6⋅39=234
Hence, the total number of tree planted by the students is 234.

Q5. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two section, find how many trees were planted by student. Which value is shown in the question?
Solution:
Number of trees planted by the students of each section of class 1 =2.
There are two sections of class 1.
∴ Number of trees planted by the students of class 1 =2⋅2=4.
Number of trees planted by the students of each section of class 2 =4.
There are two sections of class 2.
∴ Number of trees planted by the students of class 2 =2⋅4=8.
Similarly,
Number of trees planted by the students of class 3 =2⋅6=12.
So, the number of trees planted by the students of different classes are 4, 8, 12, … .
∴ Total number of trees planted by the students =4+8+12+⋯ up to 12 terms.
This series is an arithmetic series.
Here, a=4, d=8-4=4 and n=12.
Using the formula, S_n=½n[2a+(n-1)d], we get

S₁₂=½⋅12⋅[2⋅4+(12-1)⋅4]
=6⋅(8+44)
=6⋅52
=312.
Hence, the total number of trees planted by the students is 312.
The values shown in the question are social responsibility and awareness for conserving nature.

Ex2. A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:

(i) the production in the 1st year

(ii) the production in the 10th year

(iii) the total production in first 7 years

Solution:
(i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, …, years will form an arithmetic sequence.
Let us denote the number of TV sets manufactured in the nth year by a_n.
Then, a₃=600 and a₇=700 or, a+2d=600 and a+6d=700.
Solving these equations, we get d=25 and a=550.
Therefore, production of TV sets in the first year is 550.
(ii) a₁₀=a+9d=550+9×25=775. So, production of TV sets in the 10th year is 775.
(iii) Also, S₇=½⋅7⋅[2⋅550+(7 -1)⋅25]=½⋅7⋅[100+150]=4375
Thus, the total production of TV sets in first 7 years is 4375.