## Tamilnadu State Board Class 10 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,

∴ It is a geometric sequence

∴ The n^{th} term of a G.P is tn = ar^{n-1
}

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

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(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 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) 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 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,…

(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 an = 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} = 20_{(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

a6 = 2a_{(6-1)} + a_{(6-2)}

= 2a_{5} + a^{7}_{4}

= 2 × 17 + 7

= 34 + 7

= 41

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