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### 2e. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.5)

• 1. PEDAGOGY OF MATHEMATICS – PART II BY Dr. I. UMA MAHESWARI Principal Peniel Rural College of Education,Vemparali, Dindigul District iuma_maheswari@yahoo.co.in
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• 22. Solution: To prove it is an A.P, we have to show d = t2 – t1 = t3 – t2. (i) a – 3, a – 5, a – 7……… t1,t2,t3 d = t2 – t1 = a – 5 – (a – 3) = a – 5 – a + 3 = -2 ∴ d = -2 ∴ It is an A.P. d = t3 – t2 = a – 7 – (a – 5) = a – 7 – a + 5 = -2
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• 26. (v) 1,-1, 1,-1, 1,-1,… d = t2 – t1 = -1 -1 = -2 d = t3 – t2 = 1 – (-1) = 2 -2 ≠ 2 ∴ It is not an A.P.
• 27. Solution: (i) a = 5, d = 6 A.P a, a + d, a + 2d, ……… = 5, 5 + 6, 5 + 2 × 6, ……… = 5, 11, 17,… (ii) a = 7,d = -5 A.P. = a,a + d,a + 2d,… = 7,7 + (-5), 7 + 2(-5), ………. = 7, 2, -3, …….,…
• 28. (iii) a = 3/4, d = 1/2
• 29. (i) tn = -3 + 2n Answer: tn = -3 + 2 n t1 = -3 + 2(1) = -3 + 2 = -1 t2 = -3 + 2(2) = -3 + 4 = 1 First term (a) = -1 and Common difference (d) = 1 – (-1) = 1 + 1 = 2
• 30. (ii) tn = 4 – 7n Answer: tn = 4 – 7n t1 = 4 – 7(1) = 4 – 7 = -3 t2 = 4 – 7(2) = 4 – 14 = -10 First term (a) = – 3 and Common difference (d) = 10 – (-3) = – 10 + 3 = – 7
• 31. Solution: A.P = -11, -15, -19, …….. a = -11 d = t2 – t1 =-15-(-11) = -15 + 11 = -4 n = 19 ∴ tn = a + (n – 1)d t19 = -11 + (19 – 1)(-4) = -11 + 18 × -4 = -11 – 72 = -83
• 32. Answer: First term (a) = 16 Common difference (d) = 11 – 16 = -5 tn = – 54 a + (n – 1) d = -54 16 + (n – 1) (-5) = -54 54 + 21 = -54 54 + 21 = 5n 75 = 5n n = 75/5 = 15 The 15th term is – 54
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• 34. Answer: tn = a + (n – 1)d 9 times 9th term = 15 times 15th term 9t9 = 15 t15 9[a + 8d] = 15[a + 14d] 9a + 72d = 15a + 210d 9a – 15a + 72 d – 210 d = 0 -6a – 138 d = 0 6a + 138 d = 0 6 [a + 23 d] = 0 6 [a + (24 – 1)d] = 0 6 t24 = 0 ∴ Six times 24th terms is 0
• 35. Solution: 3 + k, 18 – k, 5k + 1 are in A.P ⇒ 2b = a + c if a, b, c are inA.P
• 36. Answer: x, 10, y, 24, z are in A.P t2 – t1 = 10 – x d = 10 – x …..(1) t3 – t2 = y – 10 d = y – 10 ……(2) t4 – t3 = 24 – y d = 24 – y …..(3) t5 – t4 = z – 24 d = z – 24 …..(4)
• 37. From (2) and (3) we get y – 10 = 24 – y 2y = 24 + 10 2y = 34 y = 17 From (1) and (2) we get 10 – x = y – 10 – x – y = -10 -10 -x -y = -20 x + y = 20 x + 17 = 20(y = 17)
• 38. x = 20 – 17 = 3 From (1) and (4) we get z – 24 = 10 – x z – 24 = 10 – 3 (x = 3) z – 24 = 7 z = 7 + 24 z = 31 The value of x = 3, y = 17 and z = 31
• 39. Solution: t1 = a = 20 t2 = a + 2 = 22 t3 = a + 2 + 2 = 24 ⇒ d = 2 ∴There are 30 rows. t30 = a + 29d = 20 + 29 × 2 = 20 + 58 = 78 ∴There will be 78 seats in the last row
• 40. When a = 9, d = 7 a + d = 9 + 7 = 16 a = 9 a – d = 9 – 7 = 2 When a = 9, d = -7 a + d = 9 – 7 = 2 a = 9 a – d = 9 – (-7) = 9 + 7 = 16 The three terms are 2, 9, 16 (or) 16, 9, 2
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• 43. Solution: Let the five days temperature be (a – d), a, a + d, a + 2d, a + 3d. The three days sum = a – d + a + a + d = 0 ⇒ 3a = 0 ⇒ a = 0. (given) a + d + a + 2d + a + 3d = 18 3a + 6d = 18 3(0) + 6 d = 18 6d = 18 d = 18/6 = 3 ∴The temperature of each five days is a – d, a, a + d, a + 2d, a + 3d 0 – 3, 0, 0 + 3, 0 + 2(3), 0 + 3(3) = -3°C, 0°C, 3°C, 6°C, 9°C
• 45. We find that the yearly savings is in A.P with a1 = 2000 and d = 600. We are required to find how many years are required to save 20,000 a year ………….. an = 20,000 an = a + (n – 1)d 20000 = 2000 + (n – 1)600 (n – 1)600 = 18000 n – 1 = 18000/600 = 30 n = 31 years
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