Arithmetic Progression — Class 10 Maths Notes

Topics covered in Arithmetic Progression

1. 1.Introduction of Arithmetic Progression

   An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms remains constant. This consistent difference is termed the "common difference."

   Simple Explanation: Think of it like stepping stones in a park. Each stone is placed at an equal distance from the next. Just like these stones, in an arithmetic progression, the gap (or difference) between numbers remains the same.

   Real-life Example: Imagine you start a fitness routine where you decide to walk 1000 meters on the first day, 1100 meters on the second day, 1200 meters on the third day, and so on. Every day, you increase your walk by 100 meters. This forms an arithmetic progression with a common difference of 100 meters.

   Numerical Example: Consider the sequence: 4, 7, 10, 13,... The difference between each consecutive term is 3, so this is an arithmetic progression with a common difference of 3.

2. 2.For the AP : 3/2 , 1/2 , – 1/2 , – 3/2 , . . ., write the first term a and the common difference d

   Given Arithmetic Progression (AP): 32,12,−12,−32,…23​,21​,−21​,−23​,…

   1. First Term (a): The first term of the given AP is the first number in the sequence. =32a=23​

   2. Common Difference (d): The common difference is the difference between two consecutive terms. We can find it by subtracting the first term from the second term (or the second term from the third term, and so on).

   Using the formula: =2−1d=a2​−a1​ =12−32=−1d=21​−23​=−1

   So, the common difference d is: =−1d=−1

   In conclusion: The first term =32a=23​ and the common difference =−1d=−1.

3. 3.In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

   (i) The taxi fare after each km when the fare is Rs.15 for the first km and Rs. 8 for each additional km.

   
   Solution
   
   For the first km, the fare is ₹15. For the next km, it's ₹15 + ₹8 = ₹23. For the km after that, it's ₹23 + ₹8 = ₹31, and so on. The difference between consecutive fares is constant (₹8). Thus, this forms an Arithmetic Progression (AP).
   

   (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of theair remaining in the cylinder at a time.
   
   Solution
   
   If the initial amount of air is A, after the first operation, it's 3/4A. After the second operation, it's 3/4 * 3/4A = 9/16A, and so on. The difference is not constant. So, this is not an AP.
   (iii) The cost of digging a well after every metre of digging, when it costs rs. 150 for the first metre and rises by rs. 50 for each subsequent metre.
   
   Solution
   
   For the first metre, the cost is ₹150. For the next metre, it's ₹150 + ₹50 = ₹200. For the next metre, it's ₹200 + ₹50 = ₹250, and so on. The difference between consecutive costs is constant (₹50). Thus, this is an Arithmetic Progression (AP).
   (iv) The amount of money in the account every year, when rs. 10000 is deposited at compound interest at 8 % per annum.
   Solution
   
   The money grows at a compound rate, which means the difference between the amounts for consecutive years is not constant. Therefore, this is not an AP.
   

4. 4.Write first four terms of the AP, when the first term a and the common difference d are given as follows:

   Let's find the first four terms of the AP for each given set of values for a (the first term) and d (the common difference).

   The ℎnth term of an AP is given by: =+(−1)×an​=a+(n−1)×d

   Using this formula, we can find the first four terms for each given value of a and d:

   (i) =10,=10a=10,d=10

   1st term: 1=10+(1−1)×10=10a1​=10+(1−1)×10=10 2nd term: 2=10+(2−1)×10=20a2​=10+(2−1)×10=20 3rd term: 3=10+(3−1)×10=30a3​=10+(3−1)×10=30 4th term: 4=10+(4−1)×10=40a4​=10+(4−1)×10=40

   So, the first four terms are 10, 20, 30, 40.

   (ii) =−2,=0a=−2,d=0

   For any value of n, the term will always be −2−2 since the common difference is zero.

   So, the first four terms are -2, -2, -2, -2.

   (iii) =4,=−3a=4,d=−3

   1st term: 1=4+(1−1)×(−3)=4a1​=4+(1−1)×(−3)=4 2nd term: 2=4+(2−1)×(−3)=1a2​=4+(2−1)×(−3)=1 3rd term: 3=4+(3−1)×(−3)=−2a3​=4+(3−1)×(−3)=−2 4th term: 4=4+(4−1)×(−3)=−5a4​=4+(4−1)×(−3)=−5

   So, the first four terms are 4, 1, -2, -5.

   (iv) =−1,=12a=−1,d=21​

   1st term: 1=−1+(1−1)×12=−1a1​=−1+(1−1)×21​=−1
   2nd term: 2=−1+(2−1)×12=−0.5a2​=−1+(2−1)×21​=−0.5 3rd term: 3=−1+(3−1)×12=0a3​=−1+(3−1)×21​=0 4th term: 4=−1+(4−1)×12=0.5a4​=−1+(4−1)×21​=0.5

   So, the first four terms are -1, -0.5, 0, 0.5.

   (v) =−1.25,=−0.25a=−1.25,d=−0.25

   1st term: 1=−1.25+(1−1)×(−0.25)=−1.25a1​=−1.25+(1−1)×(−0.25)=−1.25 2nd term: 2=−1.25+(2−1)×(−0.25)=−1.5a2​=−1.25+(2−1)×(−0.25)=−1.5 3rd term: 3=−1.25+(3−1)×(−0.25)=−1.75a3​=−1.25+(3−1)×(−0.25)=−1.75 4th term: 4=−1.25+(4−1)×(−0.25)=−2a4​=−1.25+(4−1)×(−0.25)=−2

   So, the first four terms are -1.25, -1.5, -1.75, -2.

5. 5.Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

   1. (i) 2, 4, 8, 16,...
      The difference between consecutive terms is not constant (4-2 ≠ 8-4). Hence, this is not an AP.
      
      

   2. (ii) 52,72,3,...25​,27​,3,...
      The difference between consecutive terms is 1221​. Hence, this is an AP with =12d=21​. Next three terms: 92,5,11229​,5,211​.
      
      

   3. (iii) – 1.2, – 3.2, – 5.2, – 7.2,...
      The difference between consecutive terms is -2. Hence, this is an AP with =−2d=−2. Next three terms: -9.2, -11.2, -13.2.
      
      

   4. (iv) – 10, – 6, – 2, 2,...
      The difference between consecutive terms is 4. Hence, this is an AP with =4d=4. Next three terms: 6, 10, 14.

      1. (v) 3, 3+223+22​, 3+2223+222​, 3+3223+232​,...
         The difference between consecutive terms is 102=5210​=5. Hence, this is an AP with =5d=5. Next three terms: 3+4223+242​, 3+5223+252​, 3+6223+262​.
         
         

      2. (vi) 0.2, 0.22, 0.222, 0.2222,...
         The difference between consecutive terms is not constant (0.02, 0.002, 0.0002,...). Hence, this is not an AP.
         
         

      3. (vii) 0, – 4, – 8, –12,.. .
         The difference between consecutive terms is -4. Hence, this is an AP with =−4d=−4. Next three terms: -16, -20, -24.
         
         

      4. (viii) –12–21​, –12–21​, –12–21​, –12–21​,...
         The difference between consecutive terms is 0. Hence, this is an AP with =0d=0. Next three terms remain –12–21​, –12–21​, –12–21​.
         

6. 6.nth Term of an AP

   In an Arithmetic Progression (AP), the sequence of numbers is such that the difference between consecutive terms is constant. This constant difference is called the "common difference," denoted as d. The nth term of an AP can be calculated using the formula:

   =1+(−1)×an​=a1​+(n−1)×d

   where:

   • an​ is the nth term of the AP.
   • 1a1​ is the first term of the AP.
   • d is the common difference.
   • n is the term number.

   This formula helps to find the value of any term in the sequence without listing all preceding terms.

   Example:
   If the first term 1a1​ is 5 and the common difference d is 3, the 4th term of the AP would be:
   4=5+(4−1)×3=5+9=14a4​=5+(4−1)×3=5+9=14

7. 7.Find the 10th term of the AP : 2, 7, 12, . . .

   Given an Arithmetic Progression (AP): 2, 7, 12,...

   The first term 1=2a1​=2. The common difference d can be found by subtracting the first term from the second term or the second term from the third term, etc. =7−2=5d=7−2=5

   To find the 10th term 10a10​, we can use the formula: =1+(−1)×an​=a1​+(n−1)×d

   Substituting the given values: 10=2+(10−1)×5a10​=2+(10−1)×5 10=2+9×5a10​=2+9×5 10=2+45a10​=2+45 10=47a10​=47

   So, the 10th term of the given AP is 47.

8. 8.Determine the AP whose 3rd term is 5 and the 7th term is 9.

   To determine the AP, we need to find the first term 1a1​ and the common difference d.

   Given: 3rd term, 3=5a3​=5 7th term, 7=9a7​=9

   The general formula for the nth term of an AP is: =1+(−1)×an​=a1​+(n−1)×d

   From the given information, we can set up the following equations:

   For the 3rd term: 3=1+2=5a3​=a1​+2d=5.......(i)

   For the 7th term: 7=1+6=9a7​=a1​+6d=9.......(ii)

   Subtracting equation (i) from equation (ii), we get: 1+6−1−2=9−5a1​+6d−a1​−2d=9−5 4=44d=4 =1d=1

   Substitute the value of d into equation (i) to get: 1+2(1)=5a1​+2(1)=5 1+2=5a1​+2=5 1=3a1​=3

   Thus, the first term 1a1​ is 3 and the common difference d is 1.

   So, the AP is: 3, 4, 5, 6, 7, 8, 9,...

9. 9.How many two-digit numbers are divisible by 3?

   1. The smallest two-digit number is 10 and the largest two-digit number is 99.

   2. To find the first two-digit number that is divisible by 3, we start with 10 and check if it is divisible by 3. 10 is not divisible by 3, but 12 is. So, the first two-digit number divisible by 3 is 12.

   3. Next, to find the largest two-digit number divisible by 3, we start with 99. 99 is divisible by 3. So, the largest two-digit number divisible by 3 is 99.

   4. The sequence of two-digit numbers divisible by 3 is an arithmetic progression (AP) with the first term 1=12a1​=12, the common difference =3d=3, and the last term =99an​=99.

   5. The nth term of an AP is given by: =1+(−1)×an​=a1​+(n−1)×d Substituting the given values, we get: 99=12+(−1)×399=12+(n−1)×3 99−12=(−1)×399−12=(n−1)×3 87=3−387=3n−3 90=390=3n =30n=30

   So, there are 30 two-digit numbers that are divisible by 3.

10. 10.Which term of the AP : 3, 8, 13, 18, . . . ,is 78?

    o find out which term of the AP 3,8,13,18,...3,8,13,18,... is 78, we can use the formula for the nth term of an AP:

    =1+(−1)×an​=a1​+(n−1)×d

    Where:

    • an​ is the nth term, which in this case is 78.
    • 1a1​ is the first term, which is 3.
    • d is the common difference, which is 8−3=58−3=5.

    Substituting in the given values:

    78=3+(−1)×578=3+(n−1)×5 78=3+5−578=3+5n−5 78=5−278=5n−2 80=580=5n =16n=16

    So, 78 is the 16th term of the AP.

11. 11.Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

    Let's denote the first AP as 1AP1​ and the second AP as 2AP2​. Both have the same common difference, which we'll call d.

    For any AP, the nth term (an​) is given by: =1+(−1)an​=a1​+(n−1)d

    Where 1a1​ is the first term of the AP.

    Given: The difference between their 100th terms is 100.

    Let 1,100a1,100​ and 2,100a2,100​ represent the 100th terms of 1AP1​ and 2AP2​ respectively. The equation is: 1,100−2,100=100a1,100​−a2,100​=100

    From the formula of nth term: 1,100=1,1+99a1,100​=a1,1​+99d 2,100=2,1+99a2,100​=a2,1​+99d

    Substituting these in the equation: (1,1+99)−(2,1+99)=100(a1,1​+99d)−(a2,1​+99d)=100 1,1−2,1=100a1,1​−a2,1​=100

    This equation shows that the difference between the first terms of 1AP1​ and 2AP2​ is 100.

    Now, we need to find the difference between their 1000th terms. Using the same logic: 1,1000−2,1000a1,1000​−a2,1000​

    Substituting the nth term formula: (1,1+999)−(2,1+999)(a1,1​+999d)−(a2,1​+999d) 1,1−2,1a1,1​−a2,1​

    Since we already know 1,1−2,1=100a1,1​−a2,1​=100, the difference between their 1000th terms is also 100.

12. 12.The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP

    Given: The sum of the 4th and 8th terms of an AP is 24. The sum of the 6th and 10th terms of an AP is 44.

    We know: The nth term of an AP is given by =1+(−1)an​=a1​+(n−1)d, where 1a1​ is the first term and d is the common difference.

    Using the given information, we can set up the following equations:

    1. 4+8=24a4​+a8​=24 Substituting the formula for the nth term: 1+3+1+7=24a1​+3d+a1​+7d=24 Combining like terms: 21+10=242a1​+10d=24.......(i)

    2. 6+10=44a6​+a10​=44 Substituting the formula for the nth term: 1+5+1+9=44a1​+5d+a1​+9d=44 Combining like terms: 21+14=442a1​+14d=44.......(ii)

    Subtracting equation (i) from equation (ii), we get: 4=204d=20 =5d=5

    Substituting the value of d in equation (i): 21+50=242a1​+50=24 21=−262a1​=−26 1=−13a1​=−13

    So, the first term 1a1​ is -13 and the common difference d is 5.

    First three terms are: -13, -8, -3

13. 13.Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ` 7000?

    We need to find the year when the salary becomes Rs. 7000. So, we need to solve for n in the AP formula: =1+(−1)an​=a1​+(n−1)d where =7000an​=7000, 1=5000a1​=5000, and =200d=200.

    Substitute the values: 7000=5000+(−1)×2007000=5000+(n−1)×200 7000=5000+200−2007000=5000+200n−200 7000=4800+2007000=4800+200n 7000−4800=2007000−4800=200n 2200=2002200=200n =2200200n=2002200​ =11n=11

    So, it took 11 years from 1995 for his salary to reach Rs. 7000. Therefore, the year would be 1995 + 11 = 2006.

14. 14.Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

    Ramkali saved Rs. 5 in the first week. Her weekly savings increased by Rs. 1.75 each week. This forms an arithmetic progression (AP) where the first term 1a1​ is Rs. 5 and the common difference d is Rs. 1.75.

    We need to find the week n when her weekly savings become Rs. 20.75. So, we use the AP formula: =1+(−1)an​=a1​+(n−1)d where =20.75an​=20.75, 1=5a1​=5, and =1.75d=1.75.

    Substitute the values: 20.75=5+(−1)×1.7520.75=5+(n−1)×1.75 20.75=5+1.75−1.7520.75=5+1.75n−1.75 20.75=3.25+1.7520.75=3.25+1.75n 17.5=1.7517.5=1.75n =17.51.75n=1.7517.5​ =10n=10

    So, in the 10th week, her savings amount to Rs. 20.75.

15. 15.Sum of First n Terms of an AP

    The sum of the first n terms of an arithmetic progression (AP) is given by: =2(21+(−1))Sn​=2n​(2a1​+(n−1)d) where: Sn​ = Sum of the first n terms 1a1​ = First term d = Common difference n = Number of terms

    Real-life example: Imagine you decide to save money daily. On the first day, you save $1. Every day after that, you decide to increase the amount you save by $1. This forms an arithmetic progression with 1=1a1​=1 and =1d=1. If you want to know how much you saved after 10 days, you can use the formula above.

    Using the formula: 10=102(2(1)+(10−1)(1))S10​=210​(2(1)+(10−1)(1)) 10=5×(2+9)S10​=5×(2+9) 10=5×11S10​=5×11 10=55S10​=55 So, you would have saved $55 in 10 days.

16. 16.Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .

    To find the sum of the first 22 terms of the arithmetic progression (AP) given, we can use the formula: =2(21+(−1))Sn​=2n​(2a1​+(n−1)d) where:

    • Sn​ = Sum of the first n terms
    • 1a1​ = First term
    • d = Common difference
    • n = Number of terms

    Given: 1=8a1​=8 d = Second term - First term = 3 - 8 = -5 =22n=22

    Plugging these values into the formula: 22=222(2(8)+(22−1)(−5))S22​=222​(2(8)+(22−1)(−5)) 22=11×(16−105)S22​=11×(16−105) 22=11×(−89)S22​=11×(−89) 22=−979S22​=−979

    So, the sum of the first 22 terms is -979.

17. 17.Find the sum of :

    (i) the first 1000 positive integers (ii) the first n positive integers
    

    (i) Sum of the first 1000 positive integers: To find the sum of the first 1000 positive integers, we can use the formula for the sum of the first n natural numbers: =(+1)2Sn​=2n(n+1)​ For =1000n=1000: 1000=1000(1000+1)2S1000​=21000(1000+1)​ 1000=1000×10012S1000​=21000×1001​ 1000=500×1001S1000​=500×1001 1000=500500S1000​=500500

    So, the sum of the first 1000 positive integers is 500,500.

    (ii) Sum of the first n positive integers: Using the same formula: =(+1)2Sn​=2n(n+1)​

    So, the sum of the first n positive integers is (+1)22n(n+1)​.

18. 18.In an AP:

    (i) =5a=5, =3d=3, =50an​=50, find n and Sn​:

    • To find n: =+(−1)an​=a+(n−1)d 50=5+(−1)350=5+(n−1)3 50=5+3−350=5+3n−3 3=483n=48 =16n=16

    • To find Sn​: =2(2+(−1))Sn​=2n​(2a+(n−1)d) 16=162(2×5+(16−1)×3)S16​=216​(2×5+(16−1)×3) 16=8(10+45)S16​=8(10+45) 16=8×55S16​=8×55 16=440S16​=440

    (ii) =7a=7, 13=35a13​=35, find d and 13S13​:

    • To find d: 13=+(13−1)a13​=a+(13−1)d 35=7+1235=7+12d 12=2812d=28 =2812d=1228​ =73d=37​

    • To find 13S13​: 13=132(2×7+(13−1)×73)S13​=213​(2×7+(13−1)×37​) 13=132(14+28)S13​=213​(14+28) 13=132×42S13​=213​×42 13=13×21S13​=13×21 13=273S13​=273

    (iii) 12=37a12​=37, =3d=3, find a and 12S12​:

    • To find a: 12=+(12−1)×3a12​=a+(12−1)×3 37=+3337=a+33 =37−33a=37−33 =4a=4

    • To find 12S12​: 12=122(2×4+(12−1)×3)S12​=212​(2×4+(12−1)×3) 12=6(8+33)S12​=6(8+33) 12=6×41S12​=6×41 12=246S12​=246

    (iv) 3=15a3​=15, 10=125S10​=125, find d and 10a10​:

    • To find d: Since 3=+2a3​=a+2d and 3=15a3​=15, 15=+215=a+2d (1)

      For 10S10​, 10=102(2+(10−1))S10​=210​(2a+(10−1)d) 125=5(2+9)125=5(2a+9d) Dividing both sides by 5, 25=2+925=2a+9d (2)

      Solving equations (1) and (2) simultaneously, we find d and a.

    • To find 10a10​: 10=+9a10​=a+9d

    (v) =5d=5, 9=75S9​=75, find a and 9a9​:

    • To find a: 9=92(2+(9−1)×5)S9​=29​(2a+(9−1)×5) 75=92(2+40)75=29​(2a+40) Solving for a.

    • To find 9a9​: 9=+8×5a9​=a+8×5

19. 19.The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

    let's solve the problem:

    Given:

    • First term =5a=5
    • Last term =45l=45
    • Sum =400S=400

    We need to find:

    • The number of terms n
    • The common difference d

    1. Finding n: Using the formula for the sum of an AP: =2(+)S=2n​(a+l) 400=2(5+45)400=2n​(5+45) 400=2×50400=2n​×50 =400×250=16n=50400×2​=16

    So, there are 16 terms in the AP.

    1. Finding d: Using the formula for the nth term of an AP: =+(−1)an​=a+(n−1)d 45=5+1545=5+15d 40=1540=15d =4015=83=2.67d=1540​=38​=2.67

    So, the common difference d is 2.67.

20. 20.Find the sum of the first 40 positive integers divisible by 6.

    The first positive integer divisible by 6 is 6, and the common difference between consecutive integers divisible by 6 is 6. So, this is an arithmetic progression with first term =6a=6 and common difference =6d=6. The 40th term would be +39=6+39(6)=240a+39d=6+39(6)=240. Using the formula for the sum of the first n terms of an AP: =2(+)S=2n​(a+l) 40=402(6+240)=20×246=4920S40​=240​(6+240)=20×246=4920
    

21. 21.Find the sum of the first 15 multiples of 8.

    The first term here is 8, and the common difference is also 8. The formula for the sum is the Sn​=2n​[2a+(n−1)d]: 15=152[2×8+(15−1)×8]S15​=215​[2×8+(15−1)×8] 15=7.5[16+14×8]
    S15​=7.5[16+14×8] 15=7.5×128
    S15​=7.5×128 15=960
    S15​=960
    

22. 22.A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,.... What is the total length of such a spiral made up of thirteen consecutive semicircles?

    To solve this problem, we'll first calculate the length of each semicircle and then sum them up.

    1. Finding the Length of Each Semicircle:

    The circumference C of a full circle with radius r is given by:

    C = 2πr

    For a semicircle, the length will be half of the circumference, i.e., =L=πr

    2. Calculating the Length for Each Semicircle in the Spiral:

    Given the radii increase by 0.5 cm for each consecutive semicircle, the radii for the 13 semicircles are: 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5 cm.

    3. Summing up the Lengths:

    Total length of the spiral made up of thirteen consecutive semicircles is: =(0.5+1.0+1.5+...+6.5)Ltotal​=π(0.5+1.0+1.5+...+6.5)

    Using the formula for the sum of an arithmetic series: =2(1+)S=2n​(a1​+an​) Where n is the number of terms, 1a1​ is the first term, and an​ is the nth term.

    Here, =13n=13, 1=0.5a1​=0.5, and 13=6.5a13​=6.5. =132(0.5+6.5)=132×7=45.5S=213​(0.5+6.5)=213​×7=45.5

    Therefore, =×45.5Ltotal​=π×45.5

    Given =227π=722​, substituting the value in: =227×45.5=286Ltotal​=722​×45.5=286cm

23. 23.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.

    Given:

    1. The sum of the third and the seventh terms of an AP is 6. 3+7=6a3​+a7​=6

    2. Their product is 8. 3×7=8a3​×a7​=8

    We need to find: The sum of the first sixteen terms of the AP, 16S16​.

    Solution:

    The nth term of an AP, =+(−1)an​=a+(n−1)d, where a is the first term and d is the common difference.

    Using the given information:

    1. 3=+2a3​=a+2d 7=+6a7​=a+6d

    From the given, 3+7=6a3​+a7​=6 +2++6=6a+2d+a+6d=6 2+8=62a+8d=6 +4=3a+4d=3...(i)

    From the given, 3×7=8a3​×a7​=8 (+2)(+6)=8(a+2d)(a+6d)=8 2+8+122=8a2+8ad+12d2=8

    Substituting from (i): 2+8(+3)=8a2+8(a+3d)d=8 2+8+242=8a2+8ad+24d2=8 2+8+242−8=0a2+8ad+24d2−8=0...(ii)

    From (i) & (ii) we can solve for a and d.

    After solving, let's say we get values for a and d.

    Sum of the first n terms of an AP: =2(2+(−1))Sn​=2n​(2a+(n−1)d)

    Here, =16n=16, plug in the values of a and d to get 16S16​.

24. 24.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., .....

    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 Iwill plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There arethree sections of each class. How many trees will be planted by the students?
    1. Each section of Class I will plant 1 tree.
    2. Each section of Class II will plant 2 trees.
    3. And so on...
    4. Each section of Class XII will plant 12 trees.
    5. There are three sections of each class.

    We need to find the total number of trees planted by the students.

    Solution:

    For Class I: 3 sections × 1 tree = 3 trees

    For Class II: 3 sections × 2 trees = 6 trees

    ......

    For Class XII: 3 sections × 12 trees = 36 trees

    Total trees = 3(1 + 2 + 3 +... + 12)

    This is an arithmetic progression where the first term a = 1, the common difference d = 1, and the number of terms n = 12.

    The sum of the first n terms of an AP is given by: =2(2+(−1))Sn​=2n​(2a+(n−1)d)

    Using the formula: 12=122(2(1)+(12−1)(1))S12​=212​(2(1)+(12−1)(1)) 12=6(2+11)S12​=6(2+11) 12=6(13)S12​=6(13) 12=78S12​=78

    So, the total trees for all 12 classes (without considering the sections) = 78 trees.

    Considering the three sections for each class: Total trees = 3 × 78 = 234 trees.

    So, the students will plant a total of 234 trees.

25. 25.Quick Revision

    1. Introduction of Arithmetic Progression (AP): An Arithmetic Progression is a list of numbers where the difference between consecutive numbers is constant. This difference is called the common difference. For example, in the sequence 2, 4, 6, 8, the common difference is 2.

    2. nth Term of an AP: The nth term of an AP is the number that comes at the nth position. To find it, you use the formula: =1+(−1)an​=a1​+(n−1)d where:

    • an​ is the nth term,
    • 1a1​ is the first term,
    • d is the common difference,
    • n is the position of the term in the sequence.

    3. Sum of First n Terms of an AP: To find the sum of the first n terms of an AP, you use the formula: =2[21+(−1)]Sn​=2n​[2a1​+(n−1)d] or =2(1+)Sn​=2n​(a1​+an​) where:

    • Sn​ is the sum of the first n terms,
    • 1a1​ is the first term,
    • an​ is the nth term,
    • d is the common difference,
    • n is the number of terms.

    Implementation of Formulas

    For example, to find the 10th term of an AP where the first term is 2 and the common difference is 3: 10=2+(10−1)×3a10​=2+(10−1)×3 10=2+9×3a10​=2+9×3 10=2+27a10​=2+27 10=29a10​=29

    To find the sum of the first 10 terms: 10=102[2×2+(10−1)×3]S10​=210​[2×2+(10−1)×3] 10=5[4+27]S10​=5[4+27] 10=5×31S10​=5×31 10=155S10​=155