Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.4

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.4

Question 1.
Find the next three terms of the following sequence.
(i) 8, 24, 72, …….
(ii) 5, 1, -3, …….
(iii) \(\frac { 1 }{ 4 } \),\(\frac { 2 }{ 9 } \),\(\frac { 3 }{ 16 } \)………..
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,
r = \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}\)
⇒ \(\frac{24}{8}=\frac{72}{24}\)
⇒ 3 = 3
∴ It is a geometric sequence
∴ The n^th term of a G.P is t_n = ar^n-1
∴ t₄ = 8 × 3^4-1
= 8 × 3³
= 8 × 27
= 216

t₅ = 8 × 3^5-1
= 8 × 3⁴
= 8 × 81
= 648

t₆ = 8 × 3^6-1
= 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.
t_n = 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.

(iii) \(\frac { 1 }{ 4 } \),\(\frac { 2 }{ 9 } \),\(\frac { 3 }{ 16 } \),………..
Here a_n = Numerators are natural numbers and denominators are squares of the next numbers
\(\frac { 1 }{ 4 } \),\(\frac { 2 }{ 9 } \),\(\frac { 3 }{ 16 } \),\(\frac { 4 }{ 25 } \),\(\frac { 5 }{ 36 } \),\(\frac { 6 }{ 49 } \)………….

Question 2.
Find the first four terms of the sequences whose n^th terms are given by

(i) a_n = n³ -2
Answer:
a_n = 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) a_n = (-1)ⁿ⁺¹ n(n + 1)
Answer:
a_n = (-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) a_n = 2n² – 6
Answer:
a_n = 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

Question 3.
Find the n^th term of the following sequences
(i) 2, 5, 10, 17, ……….
(ii) 0, \(\frac { 1 }{ 2 } \), \(\frac { 2 }{ 3 } \),…..
(iii) 3, 8, 13, 18, ………
Solution:
(i) 2, 5, 10, 17
= 1² + 1, 2² + 1, 3² + 1, 4² + 1 ……….
∴ n^th term is n²+1
(ii) 0, \(\frac { 1 }{ 2 } \),\(\frac { 2 }{ 3 } \),………….
= \(\frac { 1-1 }{ 1 } \),\(\frac { 2-1 }{ 2 } \),\(\frac { 3-1 }{ 3 } \)…..
⇒ \(\frac { n-1 }{ n } \)
∴ nth term is \(\frac { n-1 }{ n } \)
(iii) 3, 8, 13, 18
a = 3
d = 5
t_n = a + (n – 1)d
= 3 + (n – 1)5
= 3 + 5n – 5
= 5n – 2
∴ n^th term is 5n – 2

Question 4.
Find the indicated terms of the sequences whose n^th terms are given by
(i) a_n = \(\frac { 5n }{ n+2 } \) ; a₆ and a₁₃
(ii) a_n = -(n² – 4); a₄ and a₁₁
Solution:

Question 5.
Find a₈ and a₁₅ whose n^th term is

Solution:

Question 6.
If a₁ = 1, a₂ = 1 and a_n = 2a_n-1 + a_n-2 n > 3, n ∈ N. Then find the first six terms of the sequence.
Answer:
a₁ = a₂ = 1
a_n = 2a_n-1 + a_n-2
a₃ = 2a_3-1 + a_3-2 = 2a₂ + a₁
= 2(1) + 1 = 3
a₄ = 2a_4-1 + a_4-2
= 2a₃ + a₂
= 2(3) + 1 = 6 + 1 = 7
a₅ = 2 a_5-1 + a_5-2
= 2a₄ + a₃
= 2(7) + 3 = 17
a₆ = 2a_6-1 + a_6-2
= 2a₅ + a₄
= 2(17) + 7
= 34 + 7 = 41
The sequence is 1, 1, 3, 7, 17,41, …