SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Sequences and Series

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EXERCISE 8.1 11–14 (part 2 of 7)

  1. Write the first five terms of each of the sequences in Exercises $\displaystyle 11$ to $\displaystyle 13$ and obtain the corresponding series:

    Exercise 11

    a1=3,an=3an−1+2\displaystyle a_{1}=3, a_{n}=3 a_{n-1}+2 for all n>1\displaystyle n>1

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    NCERT’s answer
    $\displaystyle 3,11,35,107,323 ; \quad 3+11+35+107+323+\ldots$
    Recursion, term by term. Each term is built from the one before it, so generate them in order using \(\displaystyle a_n=3a_{n-1}+2\) with \(\displaystyle a_1=3\). \[a_2=3(3)+2=11,\qquad a_3=3(11)+2=35,\] \[a_4=3(35)+2=107,\qquad a_5=3(107)+2=323.\]The corresponding series is obtained by joining the terms with \(\displaystyle +\) signs.First five terms: \(\displaystyle 3,\ 11,\ 35,\ 107,\ 323\); corresponding series: \(\displaystyle 3+11+35+107+323+\ldots\)

  2. Exercise 12

    a1=−1,an=an−1n,n≥2\displaystyle a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n \geq 2

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    NCERT’s answer
    $\displaystyle -1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120} ;-1+\left(\frac{-1}{2}\right)+\left(\frac{-1}{6}\right)+\left(\frac{-1}{24}\right)+\left(\frac{-1}{120}\right)+\ldots$
    Recursion, term by term. Use \(\displaystyle a_n=\dfrac{a_{n-1}}{n}\) for \(\displaystyle n\ge 2\), starting from \(\displaystyle a_1=-1\). \[a_2=\frac{-1}{2}=-\frac{1}{2},\qquad a_3=\frac{-\tfrac12}{3}=-\frac{1}{6},\] \[a_4=\frac{-\tfrac16}{4}=-\frac{1}{24},\qquad a_5=\frac{-\tfrac{1}{24}}{5}=-\frac{1}{120}.\]First five terms: \(\displaystyle -1,\ -\dfrac{1}{2},\ -\dfrac{1}{6},\ -\dfrac{1}{24},\ -\dfrac{1}{120}\); corresponding series: \(\displaystyle -1-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{24}-\dfrac{1}{120}-\ldots\)

  3. Exercise 13

    a1=a2=2,an=an−1−1,n>2\displaystyle a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n>2

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    NCERT’s answer
    $\displaystyle 2,2,1,0,-1 ; \quad 2+2+1+0+(-1)+\ldots$
    Recursion, term by term. Here \(\displaystyle a_1=a_2=2\) are given, and \(\displaystyle a_n=a_{n-1}-1\) applies only from \(\displaystyle n=3\) onward. \[a_3=a_2-1=2-1=1,\qquad a_4=a_3-1=1-1=0,\qquad a_5=a_4-1=0-1=-1.\]First five terms: \(\displaystyle 2,\ 2,\ 1,\ 0,\ -1\); corresponding series: \(\displaystyle 2+2+1+0+(-1)+\ldots\)

  4. Exercise 14

    The Fibonacci sequence is defined by 1=a1=a2 and an=an−1+an−2,n>2.1=a_{1}=a_{2} \text { and } a_{n}=a_{n-1}+a_{n-2}, n>2 . Find an+1an\displaystyle \frac{a_{n+1}}{a_{n}}, for n=1,2,3,4,5\displaystyle n=1,2,3,4,5

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    NCERT’s answer
    $\displaystyle 1,2, \frac{3}{2}, \frac{5}{3}$ and $\displaystyle \frac{8}{5}$
    Generate the terms, then form the ratios. With \(\displaystyle a_1=a_2=1\) and \(\displaystyle a_n=a_{n-1}+a_{n-2}\) for \(\displaystyle n>2\): \[a_3=1+1=2,\qquad a_4=2+1=3,\qquad a_5=3+2=5,\qquad a_6=5+3=8.\]Now take \(\displaystyle \dfrac{a_{n+1}}{a_n}\) for \(\displaystyle n=1,2,3,4,5\): \[\frac{a_2}{a_1}=\frac{1}{1}=1,\quad \frac{a_3}{a_2}=\frac{2}{1}=2,\quad \frac{a_4}{a_3}=\frac{3}{2},\quad \frac{a_5}{a_4}=\frac{5}{3},\quad \frac{a_6}{a_5}=\frac{8}{5}.\]The required ratios are \(\displaystyle 1,\ 2,\ \dfrac{3}{2},\ \dfrac{5}{3},\ \dfrac{8}{5}\).