![Answer:
Number of terns (n) = 28
tn = 4n – 3
t1 = 4(1) – 3 = 4 – 3 = 1
t2 = 4(2) – 3 = 8 – 3 = 5
t3 = 4(3) – 3 = 12 – 3 = 9
Here a = 1, d = 5 – 1 = 4
S28 = n/2 [2a + (n – 1)d]
= 28/2 [2 + (27) (4)]
= 14 [2 + 108]
= 14 × 110
= 1540
Sum of 28 terms = 1540](https://image.slidesharecdn.com/2f-201212153739/85/2f-Pedagogy-of-Mathematics-Part-II-Numbers-and-Sequence-Ex-2-6-21-320.jpg)
![(ii) Number of bricks required
Sn = n/2 [2a + (n-1) d]
S30 = 30/2 [200 + 29 (-2)]
= 15[200 – 58]
= 2130
(i) Number of bricks required for the top most step = 42 bricks
(ii) Number of bricks required = 2130](https://image.slidesharecdn.com/2f-201212153739/85/2f-Pedagogy-of-Mathematics-Part-II-Numbers-and-Sequence-Ex-2-6-35-320.jpg)
2f. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.6)
- 1. BY Dr. I. UMA MAHESWARI Principal Peniel Rural College of Education,Vemparali, Dindigul District iuma_maheswari@yahoo.co.in
- 2. Ex 2.6
- 16. Solution: (i) 3, 7, 11,. . . upto 40 terms. a = 3, d = t2 – t1 = 7 – 3 = 4 n = 40 Sn = n/2 (2a + (n – 1)d) S40 = 20/2 (2× 3 + 39d) = 20(6 + 39 × 4) = 20(6 + 156) = 20 × 162 = 3240
- 17. (ii) 102, 97, 952,… up to 27 terms a = 102, d = t2 – t1 = 97 – 102 = -5 n = 27
- 18. (iii) 6 + 13 + 20 + … + 97 a = 6,d = 7, l = 97
- 19. Answer: 5,7,9, 11, 13,… Sn = 480 a = 5, d = 2, Sn = 480
- 20. 2n2 + 8n – 960 = 0 ⇒ n2 + 4n – 480 = 0 ⇒ n2 + 24n – 20n – 480 = 0 ⇒ n(n + 24) – 20(n + 24) = 0 ⇒ (n – 20)(n + 24) = 0 ⇒ n = 20,-24 No. of terms cannot be -ve. ∴ No. of consecutive odd integers beginning with 5 will sum to 480 is 20.
- 21. Answer: Number of terns (n) = 28 tn = 4n – 3 t1 = 4(1) – 3 = 4 – 3 = 1 t2 = 4(2) – 3 = 8 – 3 = 5 t3 = 4(3) – 3 = 12 – 3 = 9 Here a = 1, d = 5 – 1 = 4 S28 = n/2 [2a + (n – 1)d] = 28/2 [2 + (27) (4)] = 14 [2 + 108] = 14 × 110 = 1540 Sum of 28 terms = 1540
- 22. Solution: Given Sn = 2n2 – 3n S1 = 2(1)2 – 3(1) = 2 – 3 = – 1 ⇒ t1 = a = – 1 S2 = 2(22) – 3(2) = 8 – 6 = 2 t2 = S2 – S1 = 2 – (-1) = 3 ∴ d = t2 – t1 = 3 – (-1) = 4 Consider a, a + d, a + 2d, ….…. -1, -1 + 4, -1 + 2(4), …..… -1, 3, 7,…. Clearly this is an A.P with a = – 1, and d = 4.
- 23. Solution: t104 = 125 t4 = 0 a + (n – 1)d = tn
- 25. Solution: Sum of all odd positive integers less than 450 is given by 1 + 3 + 5 + … + 449 a = 1 d = 2 l = 449
- 26. = 50625 Another method: Sum of all +ve odd integers = n2. We can use the formula n2 = 2252 = 50625
- 27. Answer: Natural numbers between 602 and 902 603,604, …, 901 a = 603, l = 901, d = 1,
- 28. Sum of all natural numbers between 602 and 902 which are not divisible by 4. = Sum of all natural numbers between 602 and 902 = Sum of all natural numbers between 602 and 902 which are divisible by 4. l = 902 – 2 = 900 To make 602 divisible by 4 we have to add 2 to 602. ∴ 602 + 2 = 604 which is divisible by 4. To make 902 divisible by 4, subtract 2 from 902. ∴ 900 is the last number divisible by 4.
- 29. Sum of all natural numbers between 602 and 902 which are not divisible 4. = 224848 – 56400 = 168448
- 32. Solution: LoanAmount = ₹ 65,000 Repayment through installments 400 + 700 + 1000 + 1300 + … a = 400 d = 300 Sn = 65000 Sn = n/2 (2a + (n – 1)d) = 65000 (n/2)(2 × 400 + (n – 1)300) = 65000 n(800 + 300n – 300) = 130000 n(500 + 300n) = 130000 500n + 300n2 = 130000
- 33. Number of terms should be (+ve) and cannot be (-ve) or fractional number. ∴ He will take 20 months to clear the loans.
- 34. Answer: Total number of steps = 30 ∴ n = 30 Number of bricks for the bottom = 100 a = 100 2 bricks is less for each step (i) Number of bricks required for the top most step tn = a + (n – 1)d t30 = 100 + 29 (-2) = 100 – 58 = 42
- 35. (ii) Number of bricks required Sn = n/2 [2a + (n-1) d] S30 = 30/2 [200 + 29 (-2)] = 15[200 – 58] = 2130 (i) Number of bricks required for the top most step = 42 bricks (ii) Number of bricks required = 2130
- 36. Solution:
- 38. Solution: