Class 11 Maths NCERT Solutions Chapter 8 Ex 8.1 – Sequences and Series

Substituting n=1,2,3,4,5 into a_n=n(n+2):

a_1=1(3)=3 \quad a_2=2(4)=8 \quad a_3=3(5)=15 \quad a_4=4(6)=24 \quad a_5=5(7)=35

Substituting n=1,2,3,4,5 into a_n=\dfrac{n}{n+1}:

a_1=\dfrac{1}{2} \quad a_2=\dfrac{2}{3} \quad a_3=\dfrac{3}{4} \quad a_4=\dfrac{4}{5} \quad a_5=\dfrac{5}{6}

Substituting n=1,2,3,4,5 into a_n=2^n:

a_1=2 \quad a_2=4 \quad a_3=8 \quad a_4=16 \quad a_5=32

Substituting n=1,2,3,4,5 into a_n=\dfrac{2n-3}{6}:

a_1=\dfrac{-1}{6} \quad a_2=\dfrac{1}{6} \quad a_3=\dfrac{3}{6}=\dfrac{1}{2} \quad a_4=\dfrac{5}{6} \quad a_5=\dfrac{7}{6}

Substituting n=1,2,3,4,5 into a_n=(-1)^{n-1}5^{n+1}, keeping careful track of the alternating sign:

a_1=(-1)^{0}5^{2}=25 \qquad a_2=(-1)^{1}5^{3}=-125

a_3=(-1)^{2}5^{4}=625 \qquad a_4=(-1)^{3}5^{5}=-3125 \qquad a_5=(-1)^{4}5^{6}=15625

Substituting n=1,2,3,4,5 into a_n=n\cdot\dfrac{n^2+5}{4}:

a_1=1\cdot\dfrac{6}{4}=\dfrac{3}{2} \qquad a_2=2\cdot\dfrac{9}{4}=\dfrac{9}{2} \qquad a_3=3\cdot\dfrac{14}{4}=\dfrac{21}{2}

a_4=4\cdot\dfrac{21}{4}=21 \qquad a_5=5\cdot\dfrac{30}{4}=\dfrac{75}{2}

Substituting n=17 and n=24 into a_n=4n-3:

a_{17}=4(17)-3=68-3=65

a_{24}=4(24)-3=96-3=93

Substituting n=7 into a_n=\dfrac{n^2}{2^n}:

a_7=\dfrac{7^2}{2^7}=\dfrac{49}{128}

Substituting n=9 into a_n=(-1)^{n-1}n^3. Since n-1=8 is even, (-1)^{8}=1:

a_9=(-1)^{8}(9)^3=1\times729=729

Substituting n=20 into a_n=\dfrac{n(n-2)}{n+3}:

a_{20}=\dfrac{20(20-2)}{20+3}=\dfrac{20\times18}{23}=\dfrac{360}{23}

Building each term from the one before it, starting at a_1=3:

a_1=3

a_2=3a_1+2=3(3)+2=11

a_3=3a_2+2=3(11)+2=35

a_4=3a_3+2=3(35)+2=107

a_5=3a_4+2=3(107)+2=323

Building each term from the one before it, starting at a_1=-1:

a_1=-1

a_2=\dfrac{a_1}{2}=\dfrac{-1}{2}

a_3=\dfrac{a_2}{3}=\dfrac{-1/2}{3}=\dfrac{-1}{6}

a_4=\dfrac{a_3}{4}=\dfrac{-1/6}{4}=\dfrac{-1}{24}

a_5=\dfrac{a_4}{5}=\dfrac{-1/24}{5}=\dfrac{-1}{120}

The first two terms are given directly, and each later term is 1 less than the one before it:

a_1=2 \qquad a_2=2

a_3=a_2-1=2-1=1

a_4=a_3-1=1-1=0

a_5=a_4-1=0-1=-1

First, generating enough Fibonacci terms using a_1=a_2=1 and a_n=a_{n-1}+a_{n-2}:

a_1=1,\ a_2=1,\ a_3=2,\ a_4=3,\ a_5=5,\ a_6=8,\ a_7=13

Now computing \dfrac{a_{n+1}}{a_n} for each value of n from 1 to 5:

n=1:\ \dfrac{a_2}{a_1}=\dfrac{1}{1}=1 \qquad n=2:\ \dfrac{a_3}{a_2}=\dfrac{2}{1}=2 \qquad n=3:\ \dfrac{a_4}{a_3}=\dfrac{3}{2}

n=4:\ \dfrac{a_5}{a_4}=\dfrac{5}{3} \qquad n=5:\ \dfrac{a_6}{a_5}=\dfrac{8}{5}