Sequences and Series MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & A{M^2} = A{C^2} - M{C^2} \cr & = {\left( {a + c} \right)^2} - {\left( {a - c} \right)^2} = 4ac \cr & \Rightarrow \,\,A{M^2} = X{Y^2} = 4ac \cr & \Rightarrow \,\,XY = 2\sqrt {ac} \cr & {\text{Similarly,}}\,\,YZ = 2\sqrt {ba} \,\,{\text{and}}\,XZ = 2\sqrt {bc} \cr & {\text{Then,}}\,\,\,\,\,\,\,\,\,\,\,XZ = XY + YZ \cr & \Rightarrow \,\,2\sqrt {bc} = 2\sqrt {ac} + 2\sqrt {ba} \cr & \Rightarrow \,\,\frac{1}{{\sqrt a }} = \frac{1}{{\sqrt b }} + \frac{1}{{\sqrt c }} \cr} $$

Let us take G.P. with three terms $$\frac{a}{r},\,a,\,ar$$
Then,
$$\eqalign{ & S = \frac{a}{r} + a + ar = \frac{{a\left( {{r^2} + r + 1} \right)}}{r} \cr & P = {a^3}, \cr & R = \frac{r}{a} + \frac{1}{a} + \frac{1}{{ar}} = \frac{1}{a}\left( {\frac{{{r^2} + r + 1}}{r}} \right) \cr & \frac{{{P^2}{R^3}}}{{{S^3}}} = \frac{{{a^6}.\frac{1}{{{a^3}}}{{\left( {\frac{{{r^2} + r + 1}}{r}} \right)}^3}}}{{{a^3}{{\left( {\frac{{{r^2} + r + 1}}{r}} \right)}^3}}} = 1 \cr} $$
Therefore, the ratio is $$1:1$$

By $${\text{A}}{\text{.M}}{\text{.}} \geqslant {\text{G}}{\text{.M}}{\text{.}}$$
$$\eqalign{ & {x^4} + {y^4} \geqslant 2{x^2}{y^2}{\text{ and }}2{x^2}{y^2} + {z^2} \geqslant \sqrt 8 \,xyz. \cr & \Rightarrow \frac{{{x^4} + {y^4} + {z^2}}}{{xyz}} \geqslant \sqrt 8 \cr} $$

$$\eqalign{ & m = \frac{{l + n}}{2}{\text{ and common ratio of G}}{\text{.P}}{\text{. }} = r = {\left( {\frac{n}{l}} \right)^{\frac{1}{4}}} \cr & \therefore \,\,{{{G}}_1} = {l^{\frac{3}{4}}}{n^{\frac{1}{4}}},{{{G}}_2} = {l^{\frac{1}{2}}}{n^{\frac{1}{2}}},{{{G}}_3} = {l^{\frac{1}{4}}}{n^{\frac{3}{4}}} \cr & \,\,\,\,{{G}}_1^4 + 2{{G}}_2^4 + {{G}}_3^4 = {l^3}n + 2{l^2}{n^2} + l{n^3} \cr & = ln{\left( {l + n} \right)^2} \cr & = \,ln \times 2{m^2} \cr & = 4l{m^2}n \cr} $$

$$\eqalign{ & \sum\limits_{n = 1}^{10} {\sum\limits_{m = 1}^{10} {\left( {{m^2} + {n^2}} \right)} } \cr & = \sum\limits_{n = 1}^{10} {\left[ {\left( {{1^2} + {n^2}} \right) + \left( {{2^2} + {n^2}} \right) + ..... + \left( {{{10}^2} + {n^2}} \right)} \right]} \cr & = 10\left[ {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + ..... + {{\left( {10} \right)}^2}} \right] + 10\left[ {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + ..... + {{\left( {10} \right)}^2}} \right] \cr & = \frac{{20 \cdot 10 \cdot 11 \cdot 21}}{6} = 7700 \cr} $$

$$\eqalign{ & {\text{Let, }}\frac{1}{{ab}},\frac{1}{{ca}},\frac{1}{{bc}}{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \frac{1}{{ca}} - \frac{1}{{ab}} = \frac{1}{{bc}} - \frac{1}{{ca}} \cr & \Rightarrow \frac{1}{a}\left( {\frac{1}{c} - \frac{1}{b}} \right) = \frac{1}{c}\left( {\frac{1}{b} - \frac{1}{a}} \right) \cr & \Rightarrow \frac{{b - c}}{{abc}} = \frac{{a - b}}{{abc}} \cr & \Rightarrow b - c = a - b \cr & \Rightarrow 2b = a + c \cr} $$
⇒ $$a, b, c$$  are in A.P. Which is true
$$\eqalign{ & {\text{Now, }}\frac{1}{{\sqrt b + \sqrt c }},\frac{1}{{\sqrt c + \sqrt a }},\frac{1}{{\sqrt a + \sqrt b }}{\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore \frac{2}{{\sqrt c + \sqrt a }} = \frac{1}{{\sqrt b + \sqrt c }} + \frac{1}{{\sqrt a + \sqrt b }} \cr & \Rightarrow 2\left( {\sqrt b + \sqrt c } \right)\left( {\sqrt a + \sqrt b } \right) = \left( {\sqrt c + \sqrt a } \right)\left( {\sqrt a + 2\sqrt b + \sqrt c } \right) \cr & \Rightarrow 2\left( {\sqrt {ab} + b + \sqrt {ac} + \sqrt {bc} } \right) = \sqrt {ac} + 2\sqrt {bc} + c + a + 2\sqrt {ab} + \sqrt {ac} \cr & \Rightarrow 2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} = 2\sqrt {ac} + 2\sqrt {bc} + 2\sqrt {ab} + c + a \cr & \Rightarrow 2b = a + c \cr} $$
⇒ $$a, b, c$$  are in A.P. Which is true.
Hence, both the statements are correct

Let the G.P. be $$a,ar,a{r^2},.....$$
$$S = a + ar + a{r^2} + ..... +\, {\text{to }}2n{\text{ term}} = \frac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}}$$
The series formed by taking term occupying odd places is $${S_1} = a + a{r^2} + a{r^4} + .....\,{\text{to }}n{\text{ terms}}$$
$$\eqalign{ & {S_1} = \frac{{a\left[ {{{\left( {{r^2}} \right)}^n} - 1} \right]}}{{{r^2} - 1}} \cr & \Rightarrow {S_1} = \frac{{a\left( {{r^{2n}} - 1} \right)}}{{{r^2} - 1}} \cr & {\text{Now, }}S = 5{S_1}{\text{ or }}\frac{{a\left( {{r^{2n}} - 1} \right)}}{{r - 1}} = 5\frac{{a\left( {{r^{2n}} - 1} \right)}}{{{r^2} - 1}} \cr & \Rightarrow 1 = \frac{5}{{r + 1}} \cr & \Rightarrow r + 1 = 5 \cr & \therefore r = 4 \cr} $$

Let, $$S = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ..... + \frac{1}{{n\left( {n + 1} \right)}}$$
Now, $$n^{th}$$ term of above series $$ = {a_n} = \frac{1}{{n\left( {n + 1} \right)}}$$
$$\eqalign{ & \Rightarrow {a_n} = \frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}\left( {{\text{by fraction}}} \right) \cr & {\text{Now, }}S = \sum {{a_n}} = \sum {\frac{1}{n} - \sum {\frac{1}{{n + 1}}} } \cr & = \left( {1 + \frac{1}{2} + \frac{1}{3} + ..... + \frac{1}{n}} \right) - \left( {\frac{1}{2} + \frac{1}{3} + ..... + \frac{1}{n} + \frac{1}{{n + 1}}} \right) \cr & = 1 + \left( {\frac{1}{2} - \frac{1}{2}} \right) + \left( {\frac{1}{3} - \frac{1}{3}} \right) + ..... + \left( {\frac{1}{n} - \frac{1}{n}} \right) - \frac{1}{{n + 1}} \cr & = 1 - \frac{1}{{n + 1}} = \frac{{n + 1 - 1}}{{n + 1}} = \frac{n}{{n + 1}} \cr} $$

$$\pi \left[ {\left( {{r_2}^2 - {r_1}^2} \right) + \left( {{r_4}^2 - {r_3}^2} \right) + ..... + \left( {{r_{100}}^2 - {r_{99}}^2} \right)} \right]$$

$$\eqalign{ & \therefore {r_2} - {r_1} = {r_4} - {r_3} = ..... \cr & = {r_{100}} - {r_{99}} = 1 \cr & = \pi \left[ {{r_1} + {r_2} + {r_3} + {r_4} + ..... + {r_{100}}} \right] \cr & = \pi \left[ {1 + 2 + 3 + ..... + 100} \right] \cr & = 5050\,\pi \,sq\,cm \cr} $$

$$\eqalign{ & {S_n} = \left( {1 + {3^{ - 1}}} \right)\left( {1 + {3^{ - 2}}} \right)\left( {1 + {3^{ - 4}}} \right)\left( {1 + {3^{ - 8}}} \right).....\left( {1 + {3^{ - {2^n}}}} \right) \cr & \Rightarrow \left( {1 - {3^{ - 1}}} \right){S_n} = \left( {1 - {3^{ - 1}}} \right)\left( {1 + {3^{ - 1}}} \right)\left( {1 + {3^{ - 2}}} \right)\left( {1 + {3^{ - 4}}} \right)\left( {1 + {3^{ - 8}}} \right).....\left( {1 + {3^{ - {2^n}}}} \right) \cr & \Rightarrow \frac{2}{3}{S_n} = \left( {1 - {3^{ - 2}}} \right)\left( {1 + {3^{ - 2}}} \right)\left( {1 + {3^{ - 4}}} \right)\left( {1 + {3^{ - 8}}} \right).....\left( {1 + {3^{ - {2^n}}}} \right) \cr & = \left( {1 - {3^{ - 4}}} \right)\left( {1 + {3^{ - 4}}} \right)\left( {1 + {3^{ - 8}}} \right).....\left( {1 + {3^{ - {2^n}}}} \right) \cr & = \left( {1 - {3^{ - 8}}} \right)\left( {1 + {3^{ - 8}}} \right).....\left( {1 + {3^{ - {2^n}}}} \right) \cr & ............................................................ \cr & = \left( {1 - {3^{ - {2^n}}}} \right)\left( {1 + {3^{ - {2^n}}}} \right) = 1 - {\left( {{3^{ - {2^n}}}} \right)^2} = 1 - {3^{ - {2^{n + 1}}}} \cr & \Rightarrow {S_n} = \frac{3}{2}\left( {1 - {3^{ - {2^{n + 1}}}}} \right) \cr & \therefore {S_\infty } = \frac{3}{2}\left( {1 - 0} \right) = \frac{3}{2} \cr} $$