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PEDAGOGY OF
MATHEMATICS – PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in
X STD
Ex 2.4
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
Solution:
(i) 8, 24, 72…
In an arithmetic sequence a = 8,
d = t1 – t1 = t3 – t2
= 24 – 8 72 – 24
= 16 ≠ 48
So, it is not an arithmetic sequence. In a
geometric sequence
⇒ 3 = 3
∴ It is a geometric sequence
∴The nth term of a G.P is tn = arn-1
∴ t4 = 8 × 34-1
= 8 × 33
= 8 × 27
= 216
t5 = 8 × 35-1
= 8 × 34
= 8 × 81
= 648
t6 = 8 × 36-1
= 8 × 35
= 8 × 243
= 1944
The next 3 terms are 8, 24, 72, 216, 648, 1944.
(ii) 5, 1, -3, …
d = t2 – t1 = t3 – t2
⇒ 1 – 5 = -3-1
-4 = -4 ∴ It is an A.P.
tn = a+(n – 1)d
t4 = 5 + 3 × – 4
= 5 – 12
= -7
15 = a + 4d
= 5 + 4 × -4
= 5 – 16
= -11
t6 = a + 5d
= 5 + 5 × – 4
= 5 – 20
= – 15
∴The next three terms are 5, 1, -3, -7, -11, -15.
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
(i) an = n3 -2
Answer:
an = n3 – 2
a1 = 13 – 2 = 1 – 2 = -1
a2 = 23 – 2 = 8 – 2 = 6
a3 = 33 – 2 = 27 – 2 = 25
a4 = 43 – 2 = 64 – 2 = 62
The four terms are -1, 6, 25 and 62
(ii) an = (-1)n+1 n(n + 1)
Answer:
an = (-1)n+1 n(n + 1)
a1 = (-1)2 (1) (2) = 1 × 1 × 2 = 2
a2 = (-1)3 (2) (3) = -1 × 2 × 3 = -6
a3 = (-1)4 (3) (4) = 1 × 3 × 4 = 12
a4 = (-1)5 (4) (5) = -1 × 4 × 5 = -20
The four terms are 2, -6, 12 and -20
(iii) an = 2n2 – 6
Answer:
an = 2 n2 – 6
a1 = 2(1)2 – 6 = 2 – 6 = -4
a2 = 2(2)2 – 6 = 8 – 6 = 2
a3 = 2(3)2 – 6 = 18 – 6 = 12
a4 = 2(4)2 – 6 = 32 – 6 = 26
The four terms are -4, 2, 12, 26
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
(iii) 3, 8, 13, 18
a = 3
d = 5
tn = a + (n – 1)d
= 3 + (n – 1)5
= 3 + 5n – 5
= 5n – 2
∴ nth term is 5n – 2
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
Answer:
a1 = a2 = 1
an = 2an-1 + an-2
a3 = 2a3-1 + a3-2 = 2a2 + a1
= 2(1) + 1 = 3
a4 = 2a4-1 + a4-2
= 2a3 + a2
= 2(3) + 1 = 6 + 1 = 7
a5 = 2 a5-1 + a5-2
= 2a4 + a3
= 2(7) + 3 = 17
a6 = 2a6-1 + a6-2
= 2a5 + a4
= 2(17) + 7
= 34 + 7 = 41
The sequence is 1, 1, 3, 7, 17,41, …
2d. Pedagogy of Mathematics -  Part II (Numbers and Sequence - Ex 2.4)
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2d. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.4)

  • 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
  • 14. Solution: (i) 8, 24, 72… In an arithmetic sequence a = 8, d = t1 – t1 = t3 – t2 = 24 – 8 72 – 24 = 16 ≠ 48 So, it is not an arithmetic sequence. In a geometric sequence
  • 15. ⇒ 3 = 3 ∴ It is a geometric sequence ∴The nth term of a G.P is tn = arn-1 ∴ t4 = 8 × 34-1 = 8 × 33 = 8 × 27 = 216
  • 16. t5 = 8 × 35-1 = 8 × 34 = 8 × 81 = 648 t6 = 8 × 36-1 = 8 × 35 = 8 × 243 = 1944 The next 3 terms are 8, 24, 72, 216, 648, 1944.
  • 17. (ii) 5, 1, -3, … d = t2 – t1 = t3 – t2 ⇒ 1 – 5 = -3-1 -4 = -4 ∴ It is an A.P. tn = a+(n – 1)d t4 = 5 + 3 × – 4 = 5 – 12 = -7 15 = a + 4d = 5 + 4 × -4 = 5 – 16 = -11 t6 = a + 5d = 5 + 5 × – 4 = 5 – 20 = – 15 ∴The next three terms are 5, 1, -3, -7, -11, -15.
  • 19. (i) an = n3 -2 Answer: an = n3 – 2 a1 = 13 – 2 = 1 – 2 = -1 a2 = 23 – 2 = 8 – 2 = 6 a3 = 33 – 2 = 27 – 2 = 25 a4 = 43 – 2 = 64 – 2 = 62 The four terms are -1, 6, 25 and 62
  • 20. (ii) an = (-1)n+1 n(n + 1) Answer: an = (-1)n+1 n(n + 1) a1 = (-1)2 (1) (2) = 1 × 1 × 2 = 2 a2 = (-1)3 (2) (3) = -1 × 2 × 3 = -6 a3 = (-1)4 (3) (4) = 1 × 3 × 4 = 12 a4 = (-1)5 (4) (5) = -1 × 4 × 5 = -20 The four terms are 2, -6, 12 and -20
  • 21. (iii) an = 2n2 – 6 Answer: an = 2 n2 – 6 a1 = 2(1)2 – 6 = 2 – 6 = -4 a2 = 2(2)2 – 6 = 8 – 6 = 2 a3 = 2(3)2 – 6 = 18 – 6 = 12 a4 = 2(4)2 – 6 = 32 – 6 = 26 The four terms are -4, 2, 12, 26
  • 23. (iii) 3, 8, 13, 18 a = 3 d = 5 tn = a + (n – 1)d = 3 + (n – 1)5 = 3 + 5n – 5 = 5n – 2 ∴ nth term is 5n – 2
  • 26. Answer: a1 = a2 = 1 an = 2an-1 + an-2 a3 = 2a3-1 + a3-2 = 2a2 + a1 = 2(1) + 1 = 3 a4 = 2a4-1 + a4-2 = 2a3 + a2 = 2(3) + 1 = 6 + 1 = 7
  • 27. a5 = 2 a5-1 + a5-2 = 2a4 + a3 = 2(7) + 3 = 17 a6 = 2a6-1 + a6-2 = 2a5 + a4 = 2(17) + 7 = 34 + 7 = 41 The sequence is 1, 1, 3, 7, 17,41, …
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