Sequences and Series MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {\text{Let, }}{S_n} = 2 + 5 + 14 + 41 + ..... + {x_n} \cr & {S_n} = 2 + 5 + 14 + ..... + {x_{n - 1}} + {x_n} \cr & 0 = 2 + \left[ {3 + 9 + 27 + .....\,{\text{to}}\left( {n - 1} \right)\,{\text{terms}}} \right] - {x_n} \cr & \therefore {x_n} = 2 + \frac{{3\left( {{3^{n - 1}} - 1} \right)}}{{3 - 1}} = \frac{1}{2} + \frac{1}{2} \cdot {3^n} \cr & \therefore {S_n} = \sum {{x_n}} = \frac{1}{2}\sum 1 + \frac{1}{2}\sum {{3^n}} \cr & = \frac{n}{2} + \frac{1}{2} \cdot \frac{{3\left( {{3^{n - 1}} - 1} \right)}}{{3 - 1}} = \frac{n}{2} + \frac{3}{4}\left( {{3^n} - 1} \right) \cr} $$

We know that $${\log _2}$$ $$4 = 2$$  and $${\log _2} 8 = 3$$
∴ $${\log _2}$$ $$7$$ lies between 2 and 3
∴ $${\log _2}$$ $$7$$ is either rational or irrational but not integer or prime number
If possible let $${\log _2}$$ $$7 = \frac{p}{q}$$  (a rational number)
$$\eqalign{ & \Rightarrow \,{2^{\frac{p}{q}}} = 7 \cr & \Rightarrow \,{2^p} = {7^q} \cr} $$
$$ \Rightarrow $$ even number = odd number
∴ We get a contradiction, so assumption is wrong.
Hence $${\log _2}$$ $$7$$ must be an irrational number.

If $$a, b, c$$  are in G.P. then,
$$\eqalign{ & {b^2} = ac \cr & \Rightarrow b = {\left( {ac} \right)^{\frac{1}{2}}}\,\,\,\,\,.....\left( 1 \right) \cr} $$
Taking $${\log _n}$$ on both the sides of eq. (1).
$$\eqalign{ & {\log _n}b = \frac{1}{2}\left[ {{{\log }_n}\left( {ac} \right)} \right] = \frac{{{{\log }_n}a + {{\log }_n}c}}{2} \cr & {\text{or, }}\frac{{{{\log }_n}a + {{\log }_n}c}}{2} = {\log _n}b \cr & {\text{So,}}\,\,{\log _n}a,{\log _n}b\,\,{\text{and }}{\log _n}c\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & {\text{Hence, }}\frac{1}{{{{\log }_n}a}},\frac{1}{{{{\log }_n}b}},\frac{1}{{{{\log }_n}c}}\,\,{\text{are in H}}{\text{.P}}. \cr & {\log _a}n = \frac{1}{{{{\log }_n}a}} \cr & {\log _b}n = \frac{1}{{{{\log }_n}b}} \cr & {\log _c}n = \frac{1}{{{{\log }_n}c}} \cr & {\text{i}}{\text{.e}}{\text{.,}}\,\,{\log _a}n,{\log _b}n{\text{ and lo}}{{\text{g}}_c}\,n{\text{ are in H}}{\text{.P}}{\text{.}} \cr} $$

$$\eqalign{ & {x^{\ln \left( {\frac{y}{z}} \right)}} \cdot {y^{\ln {{\left( {xz} \right)}^2}}} \cdot {z^{\ln \left( {\frac{x}{y}} \right)}} = {y^{4\,\ln \,y}} \cr & \Rightarrow \ln \,\left[ {{x^{\ln \left( {\frac{y}{z}} \right)}}} \right] + \ln \left[ {{y^{\ln {{\left( {xz} \right)}^2}}}} \right] + \ln \left[ {{z^{\ln \left( {\frac{x}{y}} \right)}}} \right] = \ln \left[ {{y^{4\,\ln \,y}}} \right] \cr & \Rightarrow \left[ {\ln \left( {\frac{y}{z}} \right)\ln \,x} \right] + \left[ {2\,\ln \,\left( {xz} \right)\ln \,y} \right] + \left[ {\ln \left( {\frac{x}{y}} \right)\ln \,z} \right] = 4{\left[ {\ln \,y} \right]^2} \cr & \Rightarrow \ln \,x\left[ {\ln \,y - \ln \,z} \right] + 2\,\ln \,y\left[ {\ln \,x + \ln \,z} \right] + \ln \,z\left[ {\ln \,x - \ln \,y} \right] = 4{\left[ {\ln \,y} \right]^2} \cr & \Rightarrow 3\,\ln \,x + \ln \,z = 4\,\ln \,y \cr & \Rightarrow \frac{{\ln \,x + \ln \,x + \ln \,x + \ln \,z}}{4} = \ln \,y \cr} $$
∴ $$\ln\,y$$  is the AM of $$\ln\,x, \ln\,x, \ln\,x\, $$   & $$\ln\,z$$

$$\eqalign{ & {\text{We know that }}e = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ...... \cr & {\text{and }}{e^{ - 1}} = 1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ...... \cr & \therefore \,\,e + {e^{ - 1}} = 2\left[ {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + ......} \right] \cr & \therefore \,\,\frac{1}{{2!}} + \frac{1}{{4!}} + \frac{1}{{6!}} + ...... = \frac{{e + {e^{ - 1}}}}{2} - 1 \cr & = \frac{{{e^2} + 1 - 2e}}{{2e}} \cr & = \frac{{{{\left( {e - 1} \right)}^2}}}{{2e}} \cr} $$

100 ! contains 5, 10, 15, . . . . . , 100 $$-$$ 20 in number
25, 50, 75, 100 $$-$$ 4 in number
∴ number of zeros $$= 20 + 4 = 24$$
( $$\because $$ the numbers in the $${2^{nd}}$$ line have $$5 \times 5$$  as a factor)

Value $$ = {2^{\frac{1}{4} + \frac{2}{8} + \frac{3}{{16}} + .....\,{\text{to }}\infty }}.$$

$${a_1} + {a_{2n}} = {a_2} + {a_{2n - 1}} = ..... = {a_n} + {a_{n + 1}} = k\left( {{\text{say}}} \right).$$
Expression $$ = k\left\{ {\frac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{{a_1} - {a_2}}} + ..... + \frac{{\sqrt {{a_n}} - \sqrt {{a_{n + 1}}} }}{{{a_n} - {a_{n + 1}}}}} \right\} = \frac{k}{{ - d}}\left( {\sqrt {{a_1}} - \sqrt {{a_{n + 1}}} } \right),$$
where $$d$$ = common difference
$$ = \frac{k}{{ - d}}\frac{{{a_1} - {a_{n + 1}}}}{{\sqrt {{a_1}} + \sqrt {{a_{n + 1}}} }} = \left( {{a_1} + {a_{2n}}} \right).\frac{{ - nd}}{{ - d\left( {\sqrt {{a_1}} + \sqrt {{a_{n + 1}}} } \right)}}.$$

Let the last three numbers in A.P. be $$a, a + 6, a + 12,$$    then the first term is also $$a + 12.$$
But $$a + 12, a, a + 6$$    are in G.P.
$$\eqalign{ & \therefore {a^2} = \left( {a + 12} \right)\left( {a + 6} \right) \cr & \Rightarrow {a^2} = {a^2} + 18a + 72 \cr & \therefore a = - 4. \cr} $$
∴ The numbers are $$8, - 4, 2, 8.$$

$$\eqalign{ & {\left( {\frac{8}{5}} \right)^2} + {\left( {\frac{{12}}{5}} \right)^2} + {\left( {\frac{{16}}{5}} \right)^2} + {\left( {\frac{{20}}{5}} \right)^2} + ..... + {\left( {\frac{{44}}{5}} \right)^2} \cr & {\text{S}} = \frac{{16}}{{25}}\left( {{2^2} + {3^2} + {4^2} + ..... + {{11}^2}} \right) \cr & = \frac{{16}}{{25}}\left( {\frac{{11\left( {11 + 1} \right)\left( {22 + 1} \right)}}{6} - 1} \right) \cr & = \frac{{16}}{{25}} \times 505 \cr & = \frac{{16}}{5} \times 101 \cr & \Rightarrow \,\frac{{16}}{5}m = \frac{{16}}{5} \times 101 \cr & \Rightarrow \,m = 101. \cr} $$