11th NCERT/CBSE Sequences and series Exercise 9.1 Questions 14

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Write the first five terms of each of the sequences in Exercise 1 to 6 whose n^th terms are

Question (1)

a_n = n(n+2)

Solution

a_n = n(n+2)
Substituting n = 1, 2, 3, 4, and 5, we obtain a₁ = 1(1+2)=3
a₂ = 2(2+2)=8
a₃ = 3(3+2)=15
a₄ = 4(4+2)=24
a₅ = 5(5+2)=35
Therefore, the required terms are 3, 8, 15, 24, and 35.

Question (2)

${a_n} = \frac{n}{{n + 1}}$

Solution

${a_n} = \frac{n}{{n + 1}}$
Substituting n = 1, 2, 3, 4, 5, we obtain \[{a_1} = \frac{1}{{1 + 1}} = \frac{1}{2}\] \[{a_2} = \frac{2}{{2 + 1}} = \frac{2}{3}\] \[{a_3} = \frac{3}{{3 + 1}} = \frac{3}{4}\] \[{a_4} = \frac{4}{{4 + 1}} = \frac{4}{5}\] \[{a_5} = \frac{5}{{5 + 1}} = \frac{5}{6}\] Therefore, the required terms are \[\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}\;and \;\frac{5}{6}\]

Question (3)

a_n = 2ⁿ

Solution

a_n = 2ⁿ
Substituting n = 1, 2, 3, 4, 5, we obtain
a₁ = 2¹=2
a₂ = 2²=4
a₃ = 2³=8
a₄ = 2⁴=16
a₅ = 2⁵=32

Question (4)

${a_n} = \frac{{2n - 3}}{6}$

Solution

${a_n} = \frac{{2n - 3}}{6}$
Substituting n = 1, 2, 3, 4, 5, we obtain \[{a_1} = \frac{{2 \times 1 - 3}}{6} = \frac{{ - 1}}{6}\] \[{a_2} = \frac{{2 \times 2 - 3}}{6} = \frac{1}{6}\] \[{a_3} = \frac{{2 \times 3 - 3}}{6} = \frac{3}{6} = \frac{1}{2}\] \[{a_4} = \frac{{2 \times 4 - 3}}{6} = \frac{5}{6}\] \[{a_5} = \frac{{2 \times 5 - 3}}{6} = \frac{7}{6}\] Therefore, the required terms are $\frac{{ - 1}}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6}and\frac{7}{6}$

Question (5)

a_n = (-1)^n-1 5ⁿ⁺¹

Solution

Substituting n = 1, 2, 3, 4, 5, we obtain
a₁ = (-1)^1-1 5¹⁺¹=5² = 25
a₂ = (-1)^2-1 5²⁺¹= -5³ = -125
a₃ = (-1)^3-1 5³⁺¹=5⁴ = 625
a₄ = (-1)^4-1 5⁴⁺¹=-5⁵ = -3125
a₅ = (-1)^4-1 5⁵⁺¹=5⁶ = 15625
Therefore, the required terms are 25, –125, 625, –3125, and 15625.

Question (6)

${a_n} = n\frac{{{n^2} + 5}}{4}$

Solution

Substituting n = 1, 2, 3, 4, 5, we obtain
\[{a_1} = 1\times\frac{{{1^2} + 5}}{4} = \frac{6}{4} = \frac{3}{2}\] \[{a_2} = 2 \times \frac{{{2^2} + 5}}{4} = 2 \times \frac{9}{4} = \frac{9}{2}\] \[{a_3} = 3 \times \frac{{{3^2} + 5}}{4} = 3 \times \frac{{14}}{4} = \frac{{21}}{2}\] \[{a_4} = 4 \times \frac{{{4^2} + 5}}{4} = 4 \times \frac{{21}}{4} = 21\] \[{a_5} = 5 \times \frac{{{5^2} + 5}}{4} = 5 \times \frac{{30}}{4} = \frac{{75}}{2}\] Therefore, the required terms are $\frac{3}{2},\frac{9}{2}.\frac{{21}}{2},21, \; and \;\frac{{75}}{2}$

Find the indicated terms in each of the sequences in Exercise 7 to 10 whose n^th terms are

Question (7)

a_n = 4n -3; a₁₇, a₂₄

Solution

Substituting n = 17, we obtain
a₁₇ = 4(17) -3 =68-3=65
Substituting n = 24, we obtain
a₂₄ = 4(24) -3 = 96-3 =93

Question (8)

${a_n} = \frac{{{n^2}}}{{{2^n}}};{a_7}$

Solution

Substituting n = 7, we obtain \[{a_7} = \frac{{{7^2}}}{{2^ 7}} = \frac{49}{128}\]

Question (9)

a_n = (-1)^n-1 n³; a₉

Solution

Substituting n = 9, we obtain
a₉ = (-1)^9-1 (9)³ = (9)³ = 729

Question (10)

${a_n} = \frac{{n\left( {n - 2} \right)}}{{n + 3}};{a_{20}}$

Solution

Substituting n = 20, we obtain \[{a_{20}} = \frac{{20\left( {20 - 2} \right)}}{{20 + 3}} = \frac{{20\left( {18} \right)}}{{23}} = \frac{{360}}{{23}}\]

Write the first five terms of each of the dequences in Exercise 11 to 13 and obtain the corresponding series:

Question (11)

a₁ =3, a_n = 3a_n-1 + 2 for all n >1

Solution

a₁ =3, a_n = 3a_n-1 + 2 for all n >1
⇒ a₂=3a₁ +2 = 3(3) +2 =11
a₃ = 3a₂ +2 = 3(11) +2 = 35
a₄ = 3a₃+2 = 3(35) +2 = 107
a₅ = 3a₄+2 = 3(17) +2 = 323
Hence, the first five terms of the sequence are 3, 11, 35, 107, and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Question (12)

\[{a_1} = - 1,{a_n} = \frac{{{a_{n - 1}}}}{n},n \ge 2\]

Solution

\[{a_1} = - 1,{a_n} = \frac{{{a_{n - 1}}}}{n},n \ge 2\] \[ \Rightarrow {a_2} = \frac{{{a_1}}}{2} = \frac{{ - 1}}{2}\] \[ {a_3} = \frac{{{a_2}}}{3} = \frac{{ - 1}}{6}\] \[{a_4} = \frac{{{a_3}}}{4} = \frac{{ - 1}}{{24}}\] \[{a_5} = \frac{{{a_4}}}{5} = \frac{{ - 1}}{{120}}\] Hence, the first five terms of the sequence are \[ - 1,\frac{{ - 1}}{2},\frac{{ - 1}}{6},\frac{{ - 1}}{{24}} \; and \; \frac{{ - 1}}{{120}}\] The corresponding series is \[\left( { - 1} \right) + \left( {\frac{{ - 1}}{2}} \right) + \left( {\frac{{ - 1}}{6}} \right) + \left( {\frac{{ - 1}}{{24}}} \right) + \left( {\frac{{ - 1}}{{120}}} \right) + ...\]

Question (13)

a₁ = a₂ = 2, a_n = a_n-1 -1, n > 2

Solution

a₁ = a₂ = 2, a_n = a_n-1 -1, n > 2
⇒ a₃ = a₂ - 1 = 2 - 1 = 1
a₄ = a₃ - 1 = 1 - 1 = 0
a₅ = a₅ - 1 = 0 - 1 = -1
Hence, the first five terms of the sequence are 2, 2, 1, 0, and –1.
The corresponding series is 2 + 2 + 1 + 0 + (–1) + …

Question (14)

The Fibonacci sequence is defind by
1 = a₁ = a₂ and a_n = a_n-1 + a_n-2, n > 2
Find $\frac{{{a_{n + 1}}}}{{{a_n}}}$, for n=1, 2, 3, 4, 5

Solution

1 = a₁ = a₂
a_n = a_n-1 + a_n-2 , n> 2
∴ a₃ = a₂ + a₁ = 1 + 1 =2
a₄ = a₃ + a₂ = 2 + 1 =3
a₅ = a₄ + a₃ = 3 + 2 =5
a₆ = a₅ + a₄ = 5 + 3 =8
\[\therefore \;For\;n = 1,\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_2}}}{{{a_1}}} = \frac{1}{1} = 1\] \[\therefore \;For\;n = 2,\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_3}}}{{{a_2}}} = \frac{2}{1} = 2\] \[For\;n = 3,\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_4}}}{{{a_3}}} = \frac{3}{2}\] \[\therefore \;For\;n = 4,\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_5}}}{{{a_4}}} = \frac{5}{3}\] \[\therefore \;For\;n = 5,\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_6}}}{{{a_5}}} = \frac{8}{5}\]