NCERT Solutions · Class 11 Mathematics Sequences and Series

General term of a G.P. For a G.P. with first term \(\displaystyle a\) and common ratio \(\displaystyle r\), \(\displaystyle a_n=ar^{\,n-1}\).Here \(\displaystyle a=\dfrac{5}{2}\) and \[r=\frac{5/4}{5/2}=\frac{5}{4}\times\frac{2}{5}=\frac{1}{2}.\]Hence \[a_n=\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}=\frac{5}{2}\cdot\frac{1}{2^{\,n-1}}=\frac{5}{2^{\,n}},\] and putting \(\displaystyle n=20\), \[a_{20}=\frac{5}{2^{20}}=\frac{5}{1048576}.\]\(\displaystyle a_{n}=\dfrac{5}{2^{\,n}}\) and \(\displaystyle a_{20}=\dfrac{5}{2^{20}}=\dfrac{5}{1048576}\).

General term of a G.P. Use \(\displaystyle a_n=ar^{\,n-1}\). Two terms of a G.P. are related by \(\displaystyle a_{12}=a_8\,r^{4}\), since the index rises by \(\displaystyle 4\).Given \(\displaystyle a_8=192\) and \(\displaystyle r=2\), \[a_{12}=192\times 2^{4}=192\times 16=3072.\](Equivalently, \(\displaystyle ar^{7}=192\Rightarrow a=\dfrac{192}{128}=\dfrac{3}{2}\), so \(\displaystyle a_{12}=\dfrac{3}{2}\cdot 2^{11}=3072\).)The \(\displaystyle 12^{\text{th}}\) term is \(\displaystyle 3072\).

General term of a G.P. Let the G.P. have first term \(\displaystyle a\) and common ratio \(\displaystyle r\), so \(\displaystyle a_n=ar^{\,n-1}\). Then \[p=a_5=ar^{4},\qquad q=a_8=ar^{7},\qquad s=a_{11}=ar^{10}.\]Now compute the two sides separately. \[q^{2}=\left(ar^{7}\right)^{2}=a^{2}r^{14},\] \[ps=\left(ar^{4}\right)\left(ar^{10}\right)=a^{2}r^{\,4+10}=a^{2}r^{14}.\]Both equal \(\displaystyle a^{2}r^{14}\).Hence \(\displaystyle q^{2}=ps\).

General term of a G.P. Let the common ratio be \(\displaystyle r\); the first term is \(\displaystyle a=-3\), so \(\displaystyle a_n=-3\,r^{\,n-1}\). In a G.P. \(\displaystyle r\neq 0\).The condition "\(\displaystyle 4^{\text{th}}\) term \(\displaystyle =\) square of the \(\displaystyle 2^{\text{nd}}\) term" gives \[a_4=(a_2)^{2}\;\Longrightarrow\; ar^{3}=\left(ar\right)^{2}=a^{2}r^{2}.\]Dividing by \(\displaystyle r^{2}\ (\neq 0)\): \[ar=a^{2}\;\Longrightarrow\; r=a=-3.\]Therefore \[a_7=ar^{6}=(-3)(-3)^{6}=(-3)(729)=-2187.\]The \(\displaystyle 7^{\text{th}}\) term is \(\displaystyle -2187\).

\(\displaystyle n\)th term of a G.P. For a G.P. with first term \(\displaystyle a\) and common ratio \(\displaystyle r\), the \(\displaystyle n\)th term is \(\displaystyle a_n = a r^{\,n-1}\). Set \(\displaystyle a_n\) equal to the given value and solve for \(\displaystyle n\).(a) \(\displaystyle 2,\ 2\sqrt{2},\ 4,\ \ldots\) is $\displaystyle 128$?Here \(\displaystyle a = 2\) and \(\displaystyle r = \dfrac{2\sqrt{2}}{2} = \sqrt{2}\) (check: \(\displaystyle 2\sqrt{2}\cdot\sqrt{2} = 4\), so the ratio is constant).Write everything as a power of $\displaystyle 2$, using \(\displaystyle \sqrt{2} = 2^{1/2}\): \[a_n = 2\left(2^{1/2}\right)^{n-1} = 2^{\,1 + \frac{n-1}{2}}.\]Equating to \(\displaystyle 128 = 2^{7}\) and comparing exponents of the same base, \[1 + \frac{n-1}{2} = 7 \;\Longrightarrow\; \frac{n-1}{2} = 6 \;\Longrightarrow\; n = 13.\](c) \(\displaystyle \dfrac{1}{3},\ \dfrac{1}{9},\ \dfrac{1}{27},\ \ldots\) is \(\displaystyle \dfrac{1}{19683}\)?Here \(\displaystyle a = \dfrac{1}{3}\) and \(\displaystyle r = \dfrac{1/9}{1/3} = \dfrac{1}{3}\), so \[a_n = \frac{1}{3}\left(\frac{1}{3}\right)^{n-1} = \left(\frac{1}{3}\right)^{n} = \frac{1}{3^{\,n}}.\]Since \(\displaystyle 19683 = 3^{9}\), \[\frac{1}{3^{\,n}} = \frac{1}{3^{9}} \;\Longrightarrow\; n = 9.\](a) $\displaystyle 128$ is the 13th term. (c) \(\displaystyle \dfrac{1}{19683}\) is the 9th term.

Geometric mean condition. Three numbers \(\displaystyle p, x, q\) are in G.P. exactly when the ratio of consecutive terms is the same, i.e. \(\displaystyle \dfrac{x}{p} = \dfrac{q}{x}\), which gives \(\displaystyle x^{2} = pq\).With \(\displaystyle p = -\dfrac{2}{7}\) and \(\displaystyle q = -\dfrac{7}{2}\), \[x^{2} = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right) = 1 .\]Hence \(\displaystyle x = \pm 1\).Both values are admissible (neither makes a term zero): \(\displaystyle x = \pm 1\).

Sum of \(\displaystyle n\) terms of a G.P. For \(\displaystyle r \neq 1\), \(\displaystyle \mathrm{S}_n = \dfrac{a\left(1 - r^{\,n}\right)}{1 - r}\).For \(\displaystyle 0.15,\ 0.015,\ 0.0015,\ \ldots\) \[a = 0.15, \qquad r = \frac{0.015}{0.15} = 0.1 = \frac{1}{10}, \qquad n = 20 .\]Therefore \[\mathrm{S}_{20} = \frac{0.15\left(1 - (0.1)^{20}\right)}{1 - 0.1} = \frac{0.15}{0.9}\left(1 - 10^{-20}\right) = \frac{1}{6}\left(1 - 10^{-20}\right).\]\(\displaystyle \mathrm{S}_{20} = \dfrac{1}{6}\left(1 - 10^{-20}\right)\), i.e. \(\displaystyle \dfrac{1}{6}\left(1 - \dfrac{1}{10^{20}}\right)\).

Sum of \(\displaystyle n\) terms of a G.P. For \(\displaystyle r \neq 1\), \(\displaystyle \mathrm{S}_n = \dfrac{a\left(r^{\,n} - 1\right)}{r - 1}\).For \(\displaystyle \sqrt{7},\ \sqrt{21},\ 3\sqrt{7},\ \ldots\) \[a = \sqrt{7}, \qquad r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3},\] and the ratio is indeed constant, since \(\displaystyle \dfrac{3\sqrt{7}}{\sqrt{21}} = \dfrac{3\sqrt{7}}{\sqrt{3}\,\sqrt{7}} = \sqrt{3}\).Hence \[\mathrm{S}_n = \frac{\sqrt{7}\left(\left(\sqrt{3}\right)^{n} - 1\right)}{\sqrt{3} - 1}.\]Rationalising the denominator by multiplying numerator and denominator by \(\displaystyle \sqrt{3} + 1\) (and using \(\displaystyle (\sqrt3-1)(\sqrt3+1)=2\)): \[\mathrm{S}_n = \frac{\sqrt{7}\left(\sqrt{3} + 1\right)}{2}\left(3^{\,n/2} - 1\right).\]\(\displaystyle \mathrm{S}_n = \dfrac{\sqrt{7}\left(3^{\,n/2} - 1\right)}{\sqrt{3} - 1} = \dfrac{\sqrt{7}\left(\sqrt{3} + 1\right)}{2}\left(3^{\,n/2} - 1\right).\)

Sum of \(\displaystyle n\) terms of a G.P. For \(\displaystyle r \neq 1\), \(\displaystyle \mathrm{S}_n = \dfrac{a\left(1 - r^{\,n}\right)}{1 - r}\).For \(\displaystyle 1,\ -a,\ a^{2},\ -a^{3},\ \ldots\) \[\text{first term} = 1, \qquad r = \frac{-a}{1} = -a .\]The condition \(\displaystyle a \neq -1\) is exactly what guarantees \(\displaystyle r = -a \neq 1\), so the formula applies: \[\mathrm{S}_n = \frac{1\left(1 - (-a)^{n}\right)}{1 - (-a)} = \frac{1 - (-a)^{n}}{1 + a}.\]\(\displaystyle \mathrm{S}_n = \dfrac{1 - (-a)^{n}}{1 + a}\).

Sum of \(\displaystyle n\) terms of a G.P. For \(\displaystyle r \neq 1\), \(\displaystyle \mathrm{S}_n = \dfrac{a\left(r^{\,n} - 1\right)}{r - 1}\).For \(\displaystyle x^{3},\ x^{5},\ x^{7},\ \ldots\) \[a = x^{3}, \qquad r = \frac{x^{5}}{x^{3}} = x^{2}.\]Since \(\displaystyle x \neq \pm 1\), we have \(\displaystyle r = x^{2} \neq 1\), so \[\mathrm{S}_n = \frac{x^{3}\left(\left(x^{2}\right)^{n} - 1\right)}{x^{2} - 1} = \frac{x^{3}\left(x^{2n} - 1\right)}{x^{2} - 1}.\]\(\displaystyle \mathrm{S}_n = \dfrac{x^{3}\left(x^{2n} - 1\right)}{x^{2} - 1}\).