Sequences and Series MCQ Questions & Answers in Algebra | Maths

$$a, b, c$$  are in A.P.
⇒ $$2b = a + c$$
Now,
$$\eqalign{ & {e^{\frac{1}{c}}} \times {e^{\frac{1}{a}}} = {e^{\frac{{\left( {a + c} \right)}}{{ac}}}} = {e^{\frac{{2b}}{{ac}}}} = {\left( {{e^{\frac{b}{{ac}}}}} \right)^2} \cr & \therefore {e^{\frac{1}{c}}},{e^{\frac{b}{{ac}}}},{e^{\frac{1}{a}}}{\text{ in G}}{\text{.P}}{\text{. with common ratio}} \cr & = \frac{{{e^{\frac{b}{{ac}}}}}}{{{e^{\frac{1}{c}}}}} = {e^{\frac{{\left( {b - a} \right)}}{{ac}}}} = {e^{\frac{d}{{\left( {b - d} \right)\left( {b + d} \right)}}}} \cr & = {e^{\frac{d}{{\left( {{b^2} - {d^2}} \right)}}}} \cr & \left[ {\because a,b,c{\text{ are in A}}{\text{.P}}{\text{. with common difference }}d\,\,\therefore b - a = c - b = d} \right] \cr} $$

$$\eqalign{ & {1^3} - {2^3} + {3^3} - {4^3} + ..... + {9^3} \cr & = \,\,{1^3} + {2^3} + {3^3} + ..... + {9^3} - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right) \cr & = {\left[ {\frac{{9 \times 10}}{2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right] \cr & = {\left( {45} \right)^2} - 16.{\left[ {\frac{{4 \times 5}}{2}} \right]^2} \cr & = 2025 - 1600 \cr & = 425 \cr} $$

On simplification, $${\left( {{b^2} - ac} \right)^2} + {\left( {{c^2} - bd} \right)^2} + {\left( {ad - bc} \right)^2} \leqslant 0$$
$$\eqalign{ & \Rightarrow \,\,{b^2} = ac,{c^2} = bd,ad = bc \cr & \Rightarrow \,\,\frac{b}{a} = \frac{c}{b} = \frac{d}{c}. \cr} $$

$${t_n} = \frac{{\left( {n + 1} \right) - 1}}{{\left( {n + 1} \right)!}} = \frac{1}{{n!}} - \frac{1}{{\left( {n + 1} \right)!}}.$$

$${\text{AM}} \geqslant {\text{GM}}$$   for positive numbers. So, $$\frac{x}{3} \geqslant \sqrt {{\log_5}3 \cdot {{\log }_7}5 \cdot {{\log }_9}7} = \,\root 3 \of {{{\log }_9}3} $$
$$\therefore \,\,\frac{x}{3} \geqslant \root 3 \of {\frac{1}{{{{\log }_3}9}}} = \root 3 \of {\frac{1}{2}} .$$

$$\eqalign{ & 2\log \frac{{3b}}{{5c}} = \log \frac{{5c}}{a} + \log \frac{a}{{3b}} \cr & \Rightarrow \,\,{\left( {\frac{{3b}}{{5c}}} \right)^2} = \frac{{5c}}{a}.\frac{a}{{3b}} \cr & \Rightarrow \,\,3b = 5c. \cr & {\text{Also, }}{b^2} = ac.\,\,{\text{So, }}9ac = 25{c^2}\,{\text{or, }}9a = 25c. \cr & \therefore \,\,\frac{{9a}}{5} = 5c = 3b \cr & \Rightarrow \,\,\frac{a}{5} = \frac{b}{3} = \frac{c}{{\frac{9}{5}}} \cr & \Rightarrow \,\,b + c < a. \cr} $$

Let $$S_n$$ denote the sum of $$n$$ terms of an A.P.
According to given
$$\eqalign{ & \frac{{{S_n}}}{{{S_{2n}} - {S_n}}} = k\,\,\,\forall n \geqslant 1 \cr & \Rightarrow \frac{{{S_1}}}{{{S_2} - {S_1}}} = \frac{{{S_2}}}{{{S_4} - {S_2}}} \cr & \Rightarrow {S_1}{S_4} - {S_1}{S_2} = S_2^2 - {S_1}{S_2} \cr & \Rightarrow {S_1}{S_4} = S_2^2 \cr & \Rightarrow a\frac{4}{2}\left[ {2a + \left( {4 - 1} \right)d} \right] = {\left( {a + a + d} \right)^2} \cr & \Rightarrow a\left( {4a + 6d} \right) = {\left( {2a + d} \right)^2} \cr & \Rightarrow 2ad = {d^2} \cr & \Rightarrow 2a = d \cr} $$
Since $$a = 1 ,$$  we get $$d = 2$$

Let first term = $$a,$$ common ratio = $$r,$$ where $$ – 1 < r < 1$$
Then, $$\frac{a}{{1 - r}} = 2\,\,{\text{and }}\frac{{{a^3}}}{{1 - {r^3}}} = 24$$
$$\eqalign{ & \therefore \frac{{1 - {r^3}}}{{{{\left( {1 - r} \right)}^3}}} = \frac{1}{3} \cr & {\text{i}}{\text{.e}}{\text{., }}1 - 2r + {r^2} = 3\left( {1 + r + {r^2}} \right) \cr & {\text{or, }}2{r^2} + 5r + 2 = 0 \cr & \therefore r = - 2{\text{ or }}\frac{{ - 1}}{2}{\text{ As }} - 1 < r < 1 \cr & \therefore {\text{we have }}r = - \frac{1}{2} \cr & \therefore {\text{The series is }}3 - \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + ..... \cr} $$

$$\eqalign{ & \because \,\,x,y,z{\text{ are the }}{p^{{\text{th}}}},{q^{{\text{th}}}}{\text{ and }}{r^{{\text{th}}}}\,{\text{terms of an A}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\,x = A + \left( {p - 1} \right)D;y = A + \left( {q - 1} \right)D;z = A + \left( {r - 1} \right)D \cr & \Rightarrow \,\,x - y = \left( {p - q} \right)D;y - z = \left( {q - r} \right)D;z - x = \left( {r - p} \right)D\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{where }}A{\text{ is the first term and }}D{\text{ is the common difference}}{\text{.}} \cr & {\text{Also }}x,y,z{\text{ are the }}{p^{{\text{th}}}},{q^{{\text{th}}}}{\text{ and }}{r^{{\text{th}}}}\,{\text{terms of a G}}{\text{.P}}{\text{.}} \cr & \therefore \,\,x = A{R^{p - 1}},y = A{R^{q - 1}},z = A{R^{r - 1}} \cr & \therefore \,\,{x^{y - z}}{y^{z - x}}{z^{x - y}} = {\left( {A{R^{p - 1}}} \right)^{y - z}}{\left( {A{R^{q - 1}}} \right)^{z - x}}{\left( {A{R^{r - 1}}} \right)^{x - y}} \cr & = {A^{y - z + z - x + x - y}}{R^{\left( {p - 1} \right)\left( {y - z} \right) + \left( {q - 1} \right)\left( {z - x} \right) + \left( {r - 1} \right)\left( {x - y} \right)}} \cr & = {A^0}{R^{\left( {p - 1} \right)\left( {q - r} \right)D + \left( {q - 1} \right)\left( {r - p} \right)D + \left( {r - 1} \right)\left( {p - q} \right)D}} \cr & = {A^0}{R^0} = 1 \cr} $$

$$\eqalign{ & \frac{{A{L_1}}}{{AB}} = \frac{{{L_1}{M_1}}}{{BC}}; \cr & \therefore \,\,\frac{1}{{n + 1}} = \frac{{{L_1}{M_1}}}{a}; \cr & \therefore \,\,{L_1}{M_1} = \frac{a}{{n + 1}} \cr & \frac{{A{L_2}}}{{AB}} = \frac{{{L_2}{M_2}}}{{BC}}; \cr & \therefore \,\,\frac{2}{{n + 1}} = \frac{{{L_2}{M_2}}}{a}; \cr & \therefore \,\,{L_2}{M_2} = \frac{{2a}}{{n + 1}},{\text{e}}{\text{.t}}{\text{.c,}} \cr} $$
∴ the required sum $${\text{ = }}\frac{a}{{n + 1}} + \frac{{2a}}{{n + 1}} + \frac{{3a}}{{n + 1}} + ..... + \frac{{na}}{{n + 1}}$$
$$ = \frac{a}{{n + 1}}\left( {1 + 2 + 3 + ..... + n} \right) = \frac{a}{{n + 1}} \cdot \frac{{n\left( {n + 1} \right)}}{2}.$$