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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) \(\frac { 1 }{ 4 } \), \(\frac { 2 }{ 9 } \), \(\frac { 3 }{ 16 } \)

(i) 216, 648, 1944 (This sequence is multiple of 3)

Next three terms are 216, 648, 1944

(ii) Next three terms are -7, -11, -15

(adding -4 with each term)

(iii) Next three terms are \(\frac { 4 }{ 25 } \),\(\frac { 5 }{ 36 } \) and \(\frac { 6 }{ 49 } \)

[using \(\frac{n}{(n+1)^{2}}\)]

Question 2.

Find the first four terms of the sequences whose nth terms are given by

(i) a_{n} = n^{3} – 2

(ii) a_{n} = (-1)^{n+1} n(n+1)

(iii) a_{n} = 2n^{2} – 6

Solution:

t_{n }= a_{n }= n^{3 }-2

(i) a_{1} = 1^{3} – 2 = 1 – 2 – 1

a_{2} = 2^{3} – 2 = 8 – 2 = 6

a_{3} = 3^{3} – 2 = 27 – 2 = 25

a_{4} = 4^{3} – 2 = 64 – 2 = 62

∴ The first four terms are -1, 6, 25, 62, ……….

(ii) a_{n} = (-1)^{n+1} n(n + 1)

a_{1} = (-1)^{1+1} (1) (1 +1)

= (-1)^{2} (1) (2) = 2

a_{2} = (-1)^{2+1} (2) (2 + 1)

= (-1)^{3} (2) (3)= -6

a_{3} = (-1)^{3+1} (3) (3 + 1)

= (-1)^{4} (3) (4) = 12

a_{4} = (-1)^{4+1} (4) (4 + 1)

= (-1)^{5} (4) (5) = -20

∴ The first four terms are 2, -6, 12, -20,…

(iii) a_{n} = 2n^{2} – 6

a_{1} = 2(1)^{2} – 6 = 2 – 6 = -4

a_{2} = 2(2)^{2} – 6 = 8 – 6 = 2

a_{3} = 2(3)^{2} – 6 = 18 – 6 = 12

a_{4} = 2(4)^{2} – 6 = 32 – 6 = 26

∴ The first four terms are -4, 2, 12, 26, …

Question 3.

Find the n^{th} term of the following sequences

(i) 2, 5, 10, 17, ……

Answer:

(1^{2} + 1);(2^{2} + 1),(3^{2} + 1),(4^{2} + 1)….

n^{th} term is n^{2} + 1

a_{n} = n^{2} + 1

(ii) 0,\(\frac { 1 }{ 2 } \),\(\frac { 2 }{ 3 } \) ……

Answer:

(\(\frac { 1-1 }{ 1 } \)), (\(\frac { 2-1 }{ 2 } \)), (\(\frac { 3-1 }{ 3 } \)) …..

n^{th} term is \(\frac { n-1 }{ n } \)

a_{n} = \(\frac { n-1 }{ n } \)

(iii) 3,8,13,18,…….

Answer:

[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….

The n^{th} term is 5n – 2

a_{n} = 5n – 2

Question 4.

Find the indicated terms of the sequences whose n^{th} terms are given by

(i) a_{n} = \(\frac { 5n }{ n+2 } \) ; a_{6} and a_{13}

Answer:

a_{n} = \(\frac { 5n }{ n+2 } \)

a_{6} = \(\frac { 5(6) }{ 6+2 } \) = \(\frac { 30 }{ 8 } \) = \(\frac { 15 }{ 4 } \)

a_{13} = \(\frac { 5(13) }{ 13+2 } \) = \(\frac{5 \times 13}{15}\) = \(\frac { 13 }{ 3 } \)

a_{6} = \(\frac { 15 }{ 4 } \), a_{13} = \(\frac { 13 }{ 3 } \)

(ii) a_{n} = – (n^{2} – 4); a_{4} and a_{11}

Answer:

a_{n} = -(n^{2} – 4)

a_{4} = -(4^{2} – 4)

= – (16 – 4)

= -12

a_{11} = -(11^{2} – 4)

= – (121 – 4)

= – 117

a_{4} = -12 and a_{11} = -117

Question 5.

Find a_{8} and a_{15} whose n^{th} term is a_{n
}Answer:

Question 6.

If a_{1} = 1, a_{2} = 1 and a_{n} = 2a_{n-1} + a_{n-2}, n > 3, n ∈ N, then find the first six terms of the sequence.

Solution:

a_{1} = 1, a_{2} = 1, a_{n} = 2a_{n-1} + a_{n-2}

a_{3} = 2a_{(3-1)} + a_{(3-2)}

= 2a_{2} + a_{1}

= 2 × 1 + 1 = 3

a_{4} = 2a_{(4-1)} + a_{(4-2)}

= 2a_{3} + a_{2}

= 2 × 3 + 1 = 7

a_{5} = 2a_{(5-1)} + a_{(5-2)}

= 2a_{4} + a_{3}

= 2 × 7 + 3 = 17

a_{6} = 2a_{(6-1)} + a_{(6-2)}

= 2a_{5} + a_{4}

= 2 × 17 + 7

= 34 + 7

= 41

∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 ………..