NCERT Solutions · Class 11 Mathematics Sequences and Series

Split the sum. Separate the constant part from the geometric part: \[\sum_{k=1}^{11}\left(2 + 3^{k}\right) = \sum_{k=1}^{11} 2 \;+\; \sum_{k=1}^{11} 3^{k}.\]The first sum is \(\displaystyle 2\) added \(\displaystyle 11\) times: \[\sum_{k=1}^{11} 2 = 22 .\]The second is a G.P. with first term \(\displaystyle 3\), common ratio \(\displaystyle 3\) and \(\displaystyle 11\) terms, so by \(\displaystyle \mathrm{S}_n = \dfrac{a\left(r^{\,n} - 1\right)}{r - 1}\), \[\sum_{k=1}^{11} 3^{k} = \frac{3\left(3^{11} - 1\right)}{3 - 1} = \frac{3\left(177147 - 1\right)}{2} = \frac{3 \times 177146}{2} = 265719 .\]Adding, \[\sum_{k=1}^{11}\left(2 + 3^{k}\right) = 22 + 265719 = 265741 .\]\(\displaystyle \displaystyle\sum_{k=1}^{11}\left(2 + 3^{k}\right) = 265741\).

Symmetric choice of terms. When the product of three terms of a G.P. is given, take them as \(\displaystyle \dfrac{a}{r},\ a,\ ar\); the ratio cancels in the product and leaves a single unknown.Product. \[\frac{a}{r}\cdot a \cdot ar = a^{3} = 1 \;\Longrightarrow\; a = 1 .\]Sum. With \(\displaystyle a = 1\), \[\frac{1}{r} + 1 + r = \frac{39}{10} \;\Longrightarrow\; \frac{1}{r} + r = \frac{29}{10}.\]Multiply through by \(\displaystyle 10r\) (\(\displaystyle r \neq 0\)): \[10r^{2} - 29r + 10 = 0 .\]Factorising, \(\displaystyle 10r^{2} - 25r - 4r + 10 = 5r(2r - 5) - 2(2r - 5) = (2r - 5)(5r - 2) = 0\), so \[r = \frac{5}{2} \quad\text{or}\quad r = \frac{2}{5}.\]The two roots are reciprocals, so they give the same three numbers in opposite orders: Check: \(\displaystyle \dfrac{2}{5} + 1 + \dfrac{5}{2} = \dfrac{4 + 10 + 25}{10} = \dfrac{39}{10}\) and \(\displaystyle \dfrac{2}{5}\times 1 \times \dfrac{5}{2} = 1\). ✓Common ratio \(\displaystyle r = \dfrac{5}{2}\) or \(\displaystyle \dfrac{2}{5}\); the terms are \(\displaystyle \dfrac{2}{5},\ 1,\ \dfrac{5}{2}\) (in one order or the other).

Sum of \(\displaystyle n\) terms of a G.P. Here \(\displaystyle 3,\ 3^{2},\ 3^{3},\ \ldots\) has \(\displaystyle a = 3\) and \(\displaystyle r = 3\), so \[\mathrm{S}_n = \frac{a\left(r^{\,n} - 1\right)}{r - 1} = \frac{3\left(3^{n} - 1\right)}{3 - 1} = \frac{3\left(3^{n} - 1\right)}{2}.\]Set this equal to $\displaystyle 120$: \[\frac{3\left(3^{n} - 1\right)}{2} = 120 \;\Longrightarrow\; 3^{n} - 1 = 80 \;\Longrightarrow\; 3^{n} = 81 = 3^{4} \;\Longrightarrow\; n = 4 .\]Check: \(\displaystyle 3 + 9 + 27 + 81 = 120\). ✓$\displaystyle 4$ terms are needed.

Divide one sum by the other. The "next three terms" are the first three multiplied by \(\displaystyle r^{3}\), so dividing kills \(\displaystyle a\) and leaves \(\displaystyle r\) alone.Let the G.P. be \(\displaystyle a,\ ar,\ ar^{2},\ \ldots\). Then \[a + ar + ar^{2} = a\left(1 + r + r^{2}\right) = 16, \tag{1}\] \[ar^{3} + ar^{4} + ar^{5} = ar^{3}\left(1 + r + r^{2}\right) = 128. \tag{2}\]Dividing ($\displaystyle 2$) by ($\displaystyle 1$) — legitimate since the left side of ($\displaystyle 1$) is \(\displaystyle 16 \neq 0\): \[r^{3} = \frac{128}{16} = 8 \;\Longrightarrow\; r = 2 .\]Substituting \(\displaystyle r = 2\) in ($\displaystyle 1$): \[a\left(1 + 2 + 4\right) = 16 \;\Longrightarrow\; 7a = 16 \;\Longrightarrow\; a = \frac{16}{7}.\]Hence \[\mathrm{S}_n = \frac{a\left(r^{\,n} - 1\right)}{r - 1} = \frac{\dfrac{16}{7}\left(2^{n} - 1\right)}{2 - 1} = \frac{16}{7}\left(2^{n} - 1\right).\]Check: \(\displaystyle \mathrm{S}_3 = \frac{16}{7}(7) = 16\) and \(\displaystyle \mathrm{S}_6 - \mathrm{S}_3 = \frac{16}{7}(63) - 16 = 144 - 16 = 128\). ✓\(\displaystyle a = \dfrac{16}{7}\), \(\displaystyle r = 2\), and \(\displaystyle \mathrm{S}_n = \dfrac{16}{7}\left(2^{n} - 1\right)\).

Find \(\displaystyle r\) from the given term, then sum. With \(\displaystyle a = 729\), the 7th term is \(\displaystyle a_7 = a r^{6}\): \[729\,r^{6} = 64 \;\Longrightarrow\; r^{6} = \frac{64}{729} = \left(\frac{2}{3}\right)^{6} \;\Longrightarrow\; r = \frac{2}{3}\ \ \left(\text{taking the positive ratio}\right).\]Now apply \(\displaystyle \mathrm{S}_n = \dfrac{a\left(1 - r^{\,n}\right)}{1 - r}\) with \(\displaystyle n = 7\): \[\mathrm{S}_7 = \frac{729\left(1 - \left(\dfrac{2}{3}\right)^{7}\right)}{1 - \dfrac{2}{3}} = 3 \times 729\left(1 - \frac{128}{2187}\right) = 2187 - 2187\cdot\frac{128}{2187} = 2187 - 128 = 2059 .\](The equation \(\displaystyle r^{6} = \left(\frac{2}{3}\right)^{6}\) also admits \(\displaystyle r = -\frac{2}{3}\), an alternating G.P. whose seventh term is likewise $\displaystyle 64$; that branch gives \(\displaystyle \mathrm{S}_7 = 463\). The intended G.P. is the one with positive terms.)\(\displaystyle \mathrm{S}_7 = 2059\).

Turn each condition into an equation in \(\displaystyle a\) and \(\displaystyle r\). Let the G.P. be \(\displaystyle a,\ ar,\ ar^{2},\ \ldots\)Fifth term is $\displaystyle 4$ times the third term: \[ar^{4} = 4\,ar^{2} \;\Longrightarrow\; r^{2} = 4 \;\Longrightarrow\; r = \pm 2\] (here \(\displaystyle a \neq 0\) and \(\displaystyle r \neq 0\), since a G.P. has no zero term).Sum of the first two terms is \(\displaystyle -4\): \(\displaystyle a + ar = a(1 + r) = -4\).Case \(\displaystyle r = 2\): \(\displaystyle a(3) = -4 \Rightarrow a = -\dfrac{4}{3}\), giving \[-\frac{4}{3},\ -\frac{8}{3},\ -\frac{16}{3},\ -\frac{32}{3},\ \ldots\]Case \(\displaystyle r = -2\): \(\displaystyle a(-1) = -4 \Rightarrow a = 4\), giving \[4,\ -8,\ 16,\ -32,\ \ldots\]Check (second case): \(\displaystyle 4 + (-8) = -4\) ✓ and fifth term \(\displaystyle 64 = 4 \times 16 =\) $\displaystyle 4$ × third term ✓. The first case checks similarly: \(\displaystyle -\frac43 - \frac83 = -4\) and \(\displaystyle -\frac{64}{3} = 4\times\left(-\frac{16}{3}\right)\). ✓Two G.P.s satisfy the conditions: \(\displaystyle -\dfrac{4}{3},\ -\dfrac{8}{3},\ -\dfrac{16}{3},\ \ldots\) (with \(\displaystyle a = -\dfrac{4}{3},\ r = 2\)) and \(\displaystyle 4,\ -8,\ 16,\ -32,\ \ldots\) (with \(\displaystyle a = 4,\ r = -2\)).

Three numbers are in G.P. iff the middle one is their geometric mean, i.e. iff \(\displaystyle y^{2} = xz\).Let the given G.P. have first term \(\displaystyle A\) and common ratio \(\displaystyle R\), so its \(\displaystyle n\)th term is \(\displaystyle AR^{\,n-1}\). Then \[x = a_4 = AR^{3}, \qquad y = a_{10} = AR^{9}, \qquad z = a_{16} = AR^{15}.\]Compute the two sides: \[y^{2} = \left(AR^{9}\right)^{2} = A^{2}R^{18},\] \[xz = \left(AR^{3}\right)\left(AR^{15}\right) = A^{2}R^{3+15} = A^{2}R^{18}.\]Hence \(\displaystyle y^{2} = xz\).Note that the indices $\displaystyle 4$, $\displaystyle 10$, $\displaystyle 16$ are themselves in A.P. with common difference $\displaystyle 6$; that is what makes the exponents \(\displaystyle 3, 9, 15\) an A.P. and therefore the terms \(\displaystyle x, y, z\) a G.P. — with common ratio \(\displaystyle \dfrac{y}{x} = R^{6}\).Since \(\displaystyle y^{2} = xz\), the numbers \(\displaystyle x, y, z\) are in G.P. (common ratio \(\displaystyle R^{6}\)). Hence proved.

Convert each term to a string of 9s. Each term is a repunit-like number: \(\displaystyle 8 = \dfrac{8}{9}\times 9\), \(\displaystyle 88 = \dfrac{8}{9}\times 99\), \(\displaystyle 888 = \dfrac{8}{9}\times 999\), and in general the \(\displaystyle k\)th term is \(\displaystyle \dfrac{8}{9}\left(10^{k} - 1\right)\), because \(\displaystyle \underbrace{9\ldots9}_{k} = 10^{k} - 1\).So \[\mathrm{S}_n = \sum_{k=1}^{n}\frac{8}{9}\left(10^{k} - 1\right) = \frac{8}{9}\left[\sum_{k=1}^{n}10^{k} \;-\; \sum_{k=1}^{n}1\right].\]The first bracketed sum is a G.P. with \(\displaystyle a = 10\), \(\displaystyle r = 10\), \(\displaystyle n\) terms: \[\sum_{k=1}^{n}10^{k} = \frac{10\left(10^{n} - 1\right)}{10 - 1} = \frac{10\left(10^{n} - 1\right)}{9},\] and the second is simply \(\displaystyle n\). Therefore \[\mathrm{S}_n = \frac{8}{9}\left[\frac{10\left(10^{n} - 1\right)}{9} - n\right] = \frac{80}{81}\left(10^{n} - 1\right) - \frac{8n}{9}.\]Check: \(\displaystyle n = 2\) gives \(\displaystyle \dfrac{80}{81}(99) - \dfrac{16}{9} = \dfrac{7920 - 144}{81} = 96 = 8 + 88\). ✓\(\displaystyle \mathrm{S}_n = \dfrac{8}{81}\left[10\left(10^{n} - 1\right) - 9n\right] = \dfrac{80}{81}\left(10^{n} - 1\right) - \dfrac{8n}{9}\).

Multiply term by term. The two sequences are \[2,\ 4,\ 8,\ 16,\ 32 \qquad\text{and}\qquad 128,\ 32,\ 8,\ 2,\ \tfrac{1}{2},\] both of length 5. Their corresponding products are \[2\times 128 = 256,\quad 4\times 32 = 128,\quad 8\times 8 = 64,\quad 16\times 2 = 32,\quad 32\times\tfrac{1}{2} = 16 .\]These form a G.P. with \(\displaystyle a = 256\) and \(\displaystyle r = \dfrac{128}{256} = \dfrac{1}{2}\) — as expected, since the ratio of a product sequence is the product of the ratios, \(\displaystyle 2 \times \dfrac14 = \dfrac12\).Summing $\displaystyle 5$ terms: \[\mathrm{S}_5 = \frac{a\left(1 - r^{5}\right)}{1 - r} = \frac{256\left(1 - \dfrac{1}{32}\right)}{1 - \dfrac{1}{2}} = 512\times\frac{31}{32} = 496 .\]Check by direct addition: \(\displaystyle 256 + 128 + 64 + 32 + 16 = 496\). ✓The required sum is 496.

A sequence is a G.P. iff the ratio of consecutive terms is a non-zero constant. So form the products and compute that ratio.The corresponding terms multiply to \[a\mathrm{A},\quad (ar)(\mathrm{AR}),\quad \left(ar^{2}\right)\left(\mathrm{AR}^{2}\right),\ \ldots,\ \left(ar^{n-1}\right)\left(\mathrm{AR}^{n-1}\right),\] that is, \[a\mathrm{A},\quad a\mathrm{A}(r\mathrm{R}),\quad a\mathrm{A}(r\mathrm{R})^{2},\ \ldots,\ a\mathrm{A}(r\mathrm{R})^{n-1}.\]The general \(\displaystyle k\)th product is \(\displaystyle \left(ar^{k-1}\right)\left(\mathrm{AR}^{k-1}\right) = a\mathrm{A}\,(r\mathrm{R})^{k-1}\), so \[\frac{(k+1)\text{th product}}{k\text{th product}} = \frac{a\mathrm{A}(r\mathrm{R})^{k}}{a\mathrm{A}(r\mathrm{R})^{k-1}} = r\mathrm{R},\] a constant independent of \(\displaystyle k\) (and non-zero, since \(\displaystyle r \neq 0\) and \(\displaystyle \mathrm{R} \neq 0\) in a G.P.).Hence the products form a G.P. with first term \(\displaystyle a\mathrm{A}\).The products form a G.P. with first term \(\displaystyle a\mathrm{A}\) and common ratio \(\displaystyle r\mathrm{R}\) — the product of the two common ratios.