Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.5, Numbers and Sequence, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Arithmetic progression, definition of arithmetic progression, terms and common difference of an A.P., In an Arithmetic progression, conditions for three numbers to be in A.P.,

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

2. X STD
Ex 2.5

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

23. (ii) 1/2,1/3,1/4,1/4, ………..

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

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

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

44. Solution:

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