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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 = t_{1} – t_{1} = t_{3} – t_{2}

= 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}

∴ t_{4} = 8 × 3^{4-1}

= 8 × 3^{3}

= 8 × 27

= 216

t_{5} = 8 × 3^{5-1}

= 8 × 3^{4}

= 8 × 81

= 648

t_{6} = 8 × 3^{6-1}

= 8 × 3^{5}

= 8 × 243

= 1944

The next 3 terms are 8, 24, 72, 216, 648, 1944.

(ii) 5, 1, -3, …

d = t_{2} – t_{1} = t_{3} – t_{2}

⇒ 1 – 5 = -3-1

-4 = -4 ∴ It is an A.P.

t_{n} = a+(n – 1)d

t_{4} = 5 + 3 × – 4

= 5 – 12

= -7

15 = a + 4d

= 5 + 4 × -4

= 5 – 16

= -11

t_{6} = 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} = n^{3} -2

Answer:

a_{n} = n^{3} – 2

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 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)

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

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

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

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

The four terms are 2, -6, 12 and -20

(iii) a_{n} = 2n^{2} – 6

Answer:

a_{n} = 2 n^{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 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

= 1^{2} + 1, 2^{2} + 1, 3^{2} + 1, 4^{2} + 1 ……….

∴ n^{th} term is n^{2}+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} = \(\frac { 5n }{ n+2 } \) ; a_{6} and a_{13}

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

Solution:

Question 5.

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

Solution:

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.

Answer:

a_{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 = 6 + 1 = 7

a_{5} = 2 a_{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 sequence is 1, 1, 3, 7, 17,41, …