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NCERT Solutions · Class 11 Mathematics Sequences and Series

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EXERCISE 8.1 1–10 (part 1 of 7)

  1. Write the first five terms of each of the sequences in Exercises $\displaystyle 1$ to $\displaystyle 6$ whose \(\displaystyle n^{\text {th }}\) terms are:

    Exercise 1

    \(\displaystyle a_{n}=n(n+2)\)

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    NCERT’s answer
    $\displaystyle 3,8,15,24,35$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. A sequence is completely known once its \(\displaystyle n^{\text{th}}\) term is known: put \(\displaystyle n=1,2,3,4,5\) in turn.For \(\displaystyle a_n=n(n+2)\), \[a_1=1(3)=3,\quad a_2=2(4)=8,\quad a_3=3(5)=15,\quad a_4=4(6)=24,\quad a_5=5(7)=35.\]First five terms: \(\displaystyle 3,\ 8,\ 15,\ 24,\ 35\).

  2. Exercise 2

    \(\displaystyle a_{n}=\frac{n}{n+1}\)

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    NCERT’s answer
    $\displaystyle \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Put \(\displaystyle n=1,2,3,4,5\) in \(\displaystyle a_n=\dfrac{n}{n+1}\): \[a_1=\frac{1}{2},\quad a_2=\frac{2}{3},\quad a_3=\frac{3}{4},\quad a_4=\frac{4}{5},\quad a_5=\frac{5}{6}.\]First five terms: \(\displaystyle \dfrac{1}{2},\ \dfrac{2}{3},\ \dfrac{3}{4},\ \dfrac{4}{5},\ \dfrac{5}{6}\).

  3. Exercise 3

    \(\displaystyle a_{n}=2^{n}\)

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    NCERT’s answer
    $\displaystyle 2$, $\displaystyle 4$, $\displaystyle 8$, $\displaystyle 16$ and $\displaystyle 32$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Put \(\displaystyle n=1,2,3,4,5\) in \(\displaystyle a_n=2^n\): \[a_1=2^1=2,\quad a_2=2^2=4,\quad a_3=2^3=8,\quad a_4=2^4=16,\quad a_5=2^5=32.\]First five terms: \(\displaystyle 2,\ 4,\ 8,\ 16,\ 32\).

  4. Exercise 4

    \(\displaystyle a_{n}=\frac{2 n-3}{6}\)

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    NCERT’s answer
    $\displaystyle -\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}$ and $\displaystyle \frac{7}{6}$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Put \(\displaystyle n=1,2,3,4,5\) in \(\displaystyle a_n=\dfrac{2n-3}{6}\): \[a_1=\frac{-1}{6},\quad a_2=\frac{1}{6},\quad a_3=\frac{3}{6}=\frac{1}{2},\quad a_4=\frac{5}{6},\quad a_5=\frac{7}{6}.\]First five terms: \(\displaystyle -\dfrac{1}{6},\ \dfrac{1}{6},\ \dfrac{1}{2},\ \dfrac{5}{6},\ \dfrac{7}{6}\).

  5. Exercise 5

    \(\displaystyle a_{n}=(-1)^{n-1} 5^{n+1}\)

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    NCERT’s answer
    $\displaystyle 25$, -$\displaystyle 125$, $\displaystyle 625$, -$\displaystyle 3125$, $\displaystyle 15625$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. The factor \(\displaystyle (-1)^{n-1}\) is \(\displaystyle +1\) for odd \(\displaystyle n\) and \(\displaystyle -1\) for even \(\displaystyle n\), so the signs alternate starting with \(\displaystyle +\).For \(\displaystyle a_n=(-1)^{n-1}5^{\,n+1}\), \[a_1=5^2=25,\quad a_2=-5^3=-125,\quad a_3=5^4=625,\quad a_4=-5^5=-3125,\quad a_5=5^6=15625.\]First five terms: \(\displaystyle 25,\ -125,\ 625,\ -3125,\ 15625\).

  6. Exercise 6

    \(\displaystyle a_{n}=n \frac{n^{2}+5}{4}\).

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    NCERT’s answer
    $\displaystyle \frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21$ and $\displaystyle \frac{75}{2}$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Put \(\displaystyle n=1,2,3,4,5\) in \(\displaystyle a_n=n\,\dfrac{n^{2}+5}{4}\): \[a_1=1\cdot\frac{6}{4}=\frac{3}{2},\qquad a_2=2\cdot\frac{9}{4}=\frac{9}{2},\qquad a_3=3\cdot\frac{14}{4}=\frac{21}{2},\] \[a_4=4\cdot\frac{21}{4}=21,\qquad a_5=5\cdot\frac{30}{4}=\frac{75}{2}.\]First five terms: \(\displaystyle \dfrac{3}{2},\ \dfrac{9}{2},\ \dfrac{21}{2},\ 21,\ \dfrac{75}{2}\).
  7. Find the indicated terms in each of the sequences in Exercises $\displaystyle 7$ to $\displaystyle 10$ whose \(\displaystyle n^{\text {th }}\) terms are:

    Exercise 7

    \(\displaystyle a_{n}=4 n-3 ; a_{17}, a_{24}\)

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    NCERT’s answer
    $\displaystyle 65$, $\displaystyle 93$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Substitute the required index directly into \(\displaystyle a_n=4n-3\); there is no need to list the earlier terms. \[a_{17}=4(17)-3=68-3=65,\qquad a_{24}=4(24)-3=96-3=93.\]\(\displaystyle a_{17}=65\) and \(\displaystyle a_{24}=93\).

  8. Exercise 8

    \(\displaystyle a_{n}=\frac{n^{2}}{2^{n}} ; a_{7}\)

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    NCERT’s answer
    $\displaystyle \frac{49}{128}$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Put \(\displaystyle n=7\) in \(\displaystyle a_n=\dfrac{n^{2}}{2^{n}}\): \[a_7=\frac{7^{2}}{2^{7}}=\frac{49}{128}.\]\(\displaystyle a_{7}=\dfrac{49}{128}\).

  9. Exercise 9

    \(\displaystyle a_{n}=(-1)^{n-1} n^{3} ; a_{9}\)

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    NCERT’s answer
    $\displaystyle 729$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Put \(\displaystyle n=9\) in \(\displaystyle a_n=(-1)^{n-1}n^{3}\). Since \(\displaystyle n-1=8\) is even, \(\displaystyle (-1)^{8}=1\): \[a_9=(-1)^{8}\cdot 9^{3}=1\cdot 729=729.\]\(\displaystyle a_{9}=729\).

  10. Exercise 10

    \(\displaystyle a_{n}=\frac{n(n-2)}{n+3} ; a_{20}\).

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    NCERT’s answer
    $\displaystyle \frac{360}{23}$
    Substitution in the \(\displaystyle n^{\text{th}}\) term. Put \(\displaystyle n=20\) in \(\displaystyle a_n=\dfrac{n(n-2)}{n+3}\): \[a_{20}=\frac{20(20-2)}{20+3}=\frac{20\times 18}{23}=\frac{360}{23}.\]\(\displaystyle a_{20}=\dfrac{360}{23}\).