Sequences and Series MCQ Questions & Answers in Algebra | Maths

$$360 = {2^3} \times {3^2} \times 5.$$
Sum of all odd proper divisors $$ = \left( {{3^0} + {3^1} + {3^2}} \right)\left( {{5^0} + {5^1}} \right) - 1$$
$$ = 77$$

$$\eqalign{ & \frac{2}{{\sqrt c + \sqrt a }} = \frac{1}{{\sqrt b + \sqrt c }} + \frac{1}{{\sqrt a + \sqrt b }} \cr & = \frac{{2\sqrt b + \sqrt a + \sqrt c }}{{\left( {\sqrt b + \sqrt c } \right)\left( {\sqrt a + \sqrt b } \right)}} \cr & \Rightarrow 2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} \cr & = 2\sqrt {bc} + 2\sqrt {ac} + c + 2\sqrt {ab} + a \cr & \Rightarrow 2b = a + c \cr & \therefore a,b,c{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow ax,bx,cx{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow ax + 1,bx + 1,cx + 1{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow {{\text{9}}^{ax + 1}},{9^{bx + 1}},{9^{cx + 1}}{\text{ are in G}}{\text{.P}}{\text{.}} \cr} $$
[See the properties of A.P & G.P.]

Taking A.M. and G.M. of 7 numbers $$\frac{a}{2},\frac{a}{2},\frac{b}{3},\frac{b}{3},\frac{b}{3},\frac{c}{2},\frac{c}{2}$$
we get,
$$\eqalign{ & \frac{{2 \cdot \frac{a}{2} + 3 \cdot \frac{b}{3} + 2 \cdot \frac{c}{2}}}{7} \geqslant {\left\{ {{{\left( {\frac{a}{2}} \right)}^2}{{\left( {\frac{b}{3}} \right)}^3}{{\left( {\frac{c}{2}} \right)}^2}} \right\}^{\frac{1}{7}}} \cr & \Rightarrow \frac{{{3^7}}}{{{7^7}}} \geqslant \frac{{{a^2}{b^3}{c^2}}}{{{2^2} \cdot {3^3} \cdot {2^2}}} \cr & \Rightarrow {a^2}{b^3}{c^2} \leqslant \frac{{{3^{10}} \cdot {2^4}}}{{{7^7}}} \cr} $$
∴ greatest value of $${a^2}{b^3}{c^2} = \frac{{{3^{10}} \cdot {2^4}}}{{{7^7}}}$$

$$\eqalign{ & x = \sum\limits_{n = 0}^\infty {{a^n}} = \frac{1}{{1 - a}}\,\,\,\,\,\,\,\,\,\,a = 1 - \frac{1}{x} \cr & y = \sum\limits_{n = 0}^\infty {{b^n}} = \frac{1}{{1 - b}}\,\,\,\,\,\,\,\,\,\,b = 1 - \frac{1}{y} \cr & z = \sum\limits_{n = 0}^\infty {{c^n}} = \frac{1}{{1 - c}}\,\,\,\,\,\,\,\,\,\,\,c = 1 - \frac{1}{z} \cr & a,b,c{\text{ are in A}}{\text{.P}}{\text{. OR 2}}b = a + c \cr & 2\left( {1 - \frac{1}{y}} \right) = 1 - \frac{1}{x} + 1 - \frac{1}{y} \cr & \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \cr & \Rightarrow \,\,x,y,z{\text{ are in H}}{\text{.P}}{\text{.}} \cr} $$

We have,
$$\eqalign{ & 9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right) \cr & \Rightarrow \,225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc \cr & \Rightarrow \,{\left( {15a} \right)^2} + {\left( {3b} \right)^2} + {\left( {5c} \right)^2} - 75ac - 45ab - 15bc = 0 \cr & \frac{1}{2}\left[ {{{\left( {15\,a - 3b} \right)}^2} + {{\left( {3b - 5c} \right)}^2} + {{\left( {5c - 15a} \right)}^2}} \right] = 0 \cr} $$
it is possible when $$15a - 3b = 0,3b - 5c = 0$$      and
$$\eqalign{ & 5c - 15a = 0 \cr & \Rightarrow \,15a = 3b = 5c \cr & \Rightarrow \,b = \frac{{5c}}{3},a = \frac{c}{3} \cr & \Rightarrow \,a + b = \frac{c}{3} + \frac{{5c}}{3} = \frac{{6c}}{3} \cr & \Rightarrow \,a + b = 2c \cr} $$
$$ \Rightarrow \,b,\,c,\,a$$   are in A.P.

$$\eqalign{ & {\text{Let, }}b = ar,c = a{r^2} \cr & {\text{Given that, }}12 = \frac{{2ab}}{{a + b}} = \frac{{2a.ar}}{{a + ar}} = \frac{{2ar}}{{1 + r}} \cr & {\text{or, }}ar = 6\left( {1 + r} \right)\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Also, }}36 = \frac{{2bc}}{{b + c}} = \frac{{2.ar.a{r^2}}}{{ar + a{r^2}}} = \frac{{2a{r^2}}}{{1 + r}} \cr & \Rightarrow a{r^2} = 18\left( {1 + r} \right)\,\,\,\,\,.....\left( 2 \right) \cr} $$
Dividing (2) by (1), we have
$$\eqalign{ & \frac{{a{r^2}}}{{ar}} = \frac{{18\left( {1 + r} \right)}}{{6\left( {1 + r} \right)}} \cr & \Rightarrow r = 3 \cr & \therefore {\text{From}}\left( 1 \right),a \times 3 = 6\left( {1 + 3} \right) \cr & \Rightarrow a = \frac{{6 \times 4}}{3} = 8 \cr} $$
∴ First five numbers are $$8, 24, 72, 216, 648.$$

$$\eqalign{ & {\text{Since }}{T_p} = a + \left( {p - 1} \right)d = \frac{1}{q}\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{and }}{T_q} = a + \left( {q - 1} \right)d = \frac{1}{p}\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii), we get $$a = \frac{1}{{pq}}\,\,{\text{and }}\,d = \frac{1}{{pq}}$$
Now sum of $$pq$$  terms $$ = \frac{{pq}}{2}\left[ {\frac{2}{{pq}} + \left( {pq - 1} \right)\frac{1}{{pq}}} \right]$$
$$\eqalign{ & = \frac{{pq}}{2} \cdot \frac{2}{{pq}}\left[ {1 + \frac{1}{2}\left( {pq - 1} \right)} \right] \cr & = \left[ {\frac{{2 + pq - 1}}{2}} \right] \cr & = \frac{{pq + 1}}{2} \cr} $$

$$\eqalign{ & X = {6^{\left( {\frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + ....} \right)}} \cr & = {6^{\left[ {\left( {1 \times \frac{1}{2}} \right) + \left( {2 \times \frac{1}{4}} \right) + \left( {3 \times \frac{1}{8}} \right) + \left( {4 \times \frac{1}{{16}}} \right) + ....} \right]}} \cr} $$
∵ It is arithmetic-geometric progression,
$$\eqalign{ & \therefore a = \frac{1}{2};d = 1\,\,\& \,\,r = \frac{1}{2} \cr & \Rightarrow X = {6^{\left[ {\frac{a}{{1 - r}} + \frac{{dr}}{{{{\left( {1 - r} \right)}^2}}}} \right]}} = {6^{\left[ {\frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} + \frac{{1 \times \frac{1}{2}}}{{{{\left( {1 - \frac{1}{2}} \right)}^2}}}} \right]}} = {6^3} = 216 \cr} $$

$$\eqalign{ & \frac{{{e^x} + {e^{ - x}}}}{2} = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} + \frac{{{x^6}}}{{6!}}...... \cr & {\text{Putting }}\,x = \frac{1}{2}{\text{ we get}} \cr & 1 + \frac{1}{{4.2!}} + \frac{1}{{16.4!}} + \frac{1}{{64.4!}} + ...... \cr & \infty = \frac{{{e^{\frac{1}{2}}} + {e^{\frac{{ - 1}}{2}}}}}{2} \cr & \,\,\,\, = \frac{{\sqrt e + \frac{1}{{\sqrt e }}}}{2} \cr & \,\,\,\, = \frac{{e + 1}}{{2\sqrt e }} \cr} $$

$$\eqalign{ & {n^{th}}{\text{ term of series }} = \frac{{{{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]}^2}}}{{{n^2}}} = \frac{1}{4}{\left( {n + 1} \right)^2} \cr & {\text{Sum of }}n{\text{ term }} = \sum {\frac{1}{4}{{\left( {n + 1} \right)}^2}} \cr & = \frac{1}{4}\left[ {\sum {{n^2} + 2\sum n + n} } \right] \cr & = \frac{1}{4}\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{2n\left( {n + 1} \right)}}{2} + n} \right] \cr & {\text{Sum of }}9{\text{ terms}} \cr & = \frac{1}{4}\left[ {\frac{{9 \times 10 \times 19}}{6} + \frac{{18 \times 10}}{2} + 9} \right] \cr & = \frac{{384}}{4} = 96 \cr} $$