Arithmetic Progression (CBSE GRADE 10),To find the sum of n terms of an AP

Arithmetic Progression (CBSE GRADE 10)

1. Learning Objectives: To understand finding the sum of n terms of an AP To find the sum of n terms in Real Life situation.

2. TO FIND THE nth TERM: a n = S n – S n-1

3. Ex1:

4. Ex2:

5. OR Here n= 100, a = 2 , = 200 = x 100 [2 + 200] = 50 x 202 = 10100 Ans.

6. Try this…. Do Q 1 ( iii and iv)

7. TRY YOURSELF…. HW : Ex 5.3

8. HW - EX 5.3 Q3 HW

9. Q4:

10. Q5:

11. Q6: Q7: Q13: Q14:

12. Q8:

13. Q9:

14. (ii)  We have,  an =9−5 n                   Substituting n = 1, 2, 3, 4, .... n, we get                   The sequence 4, – 1, – 6, – 11, .... (9 – 5n), which is an AP with common difference – 5.                   Putting a = 4, d = – 5 and n = 15 in                    Sn = [2 a +( n −1) d ],  we get                    S 15= [2×4+(15−1)(−5)]                    = (8+14×−5)                    = (8−70)=152×−62                   = 15 × – 31 = – 465  

15. Q 11:

16. Here a = 200 , d = 50 and n = 30                    S = [2×200+(30−1)50]                   [ Since Sn = (2 a +( n −1) d ]                  = 15(400 + 29 × 50)                  = 15(400 + 1450)                  = 15 × 1850                  = 27750                  Hence, a delay of 30 days costs the contractor Rs 27750 .  

17. Q.16      A sum of Rs 700 is to be used to give seven each prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes. Sol. Let the respective prizes be a + 60, a + 40, a+20, a, a–20, a–40, a–60           Therefore, The sum of the prizes is                   a + 60 + a + 40 + a + 20 + a + a – 20 + a – 40 + a – 60 = 700                    ⇒  7a = 700                    ⇒   a =700 / 7 a =100                   Thus The seven prizes are 100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60                   or 160, 140, 120, 100, 80, 60, 40 (in Rs)

18. Q.17     In a school, students thought of planting trees in an 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? Sol.              Since there are three sections of each class, so the number of trees planted by class I, class II, class III,... class XII are 1 × 3, 2 × 3, 3 × 3, .... 12 × 3 respectively.                      i.e., 3, 6, 9, ... 36. it forms an AP.        The sum of the number of the trees planted by these classes.                        = (3+36)=6×39=234  

19. Q.18      A spiral is made up of successive semicircles , with centres alternately at A and B, starting with cenre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in fig. What is the total length of such a spiral made up of thirteen consecutive semicircles ? ( Takeπ =22/7 ) Sol.        Length of a semi- circum ference =  πr  where r is the radius of the circle.              Therefore, Length of spiral made up of thirteen consecutive semicircles.                =( π ×0.5+ π ×1.0+ π +1.5+ π ×2.0+....+ π ×6.5) cm                = π ×0.5(1+2+3+....+13) cm                = π ×0.5× (2×1+[13−1)×1] cm                = × × ×14 cm =143 cm  

20. Q.19      200 logs are stacked in the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row? Sol.        Clearly logs stacked in each row form a sequence 20 + 19 + 18 + 17 + .... It is an AP with a = 20, d = 19 – 20 = – 1.               Let  Sn =200 . Then                [2×20+( n −1)(−1)]=200                ⇒  n(40 – n + 1) = 400                ⇒   n 2 −41 n +400=0                ⇒   ( n −16)( n −25)=0                ⇒  n = 16 or 25   The terms go on diminishing and 21st term becomes zero. All terms after 21st term are negative. These negative terms when added to positive terms from 17th term to 20th term, cancel out each other and the sum remains the same.  Thus n = 25 is not valid for this problem. So we take n = 16.            Thus, 200 logs are placed in 16 rows.            Number of logs in the 16th row             = a 16            = a + 15d            = 20 + 15(–1)            = 20 – 15 = 5

21. Q.20      In a potato race, a bucket is placed at the starting point, which is 5 cm 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 earest 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? Sol.       To pick up the first potato second potato, third potato, fourth potato, ....              The distance (in metres ) run by the competitor are              2 × 5 ; 2 × (5 + 3), 2 × (5 + 3 + 3), 2 × (5 + 3 + 3 + 3), ....              i.e., 10, 16, 22, 28, ....              which is in AP with a = 10, d = 16 – 10 = 6              Therefore, The sum of first ten terms,               S 10=102[(2×10+(10−1)×6)]              = 5(20 + 54) = 5 × 74 = 370             Therefore, The total distance the competitor has to run is 370 m.

22. HW - EX 5.3 Q3 HW

23. SUCCESS CRITERIA: I can find the sum of n terms of an AP I can apply the sum of n terms in Real Life situation.

24. THANK YOU