Maths M.C.Q (1 Marks)

$a+(n-1) d=0$

$\Rightarrow 50+(n-1)(-2)=0 \Rightarrow n=26$

So $S_{26}=\frac{26}{2}[2 \times 50+25 \times(-2)]=650$

.Hence ${2^{n - 1}} > 100$ are in A.P.

and ${m^{th}}$ mean between $2a,\;b$ is $2a + \frac{{m(b - 2a)}}{{n + 1}}$......$(ii)$

Accordingly, $a + \frac{{m(2b - a)}}{{n + 1}} = 2a + \frac{{m(b - 2a)}}{{n + 1}}$

$ \Rightarrow $ $m(2b - a) = a(n + 1) + m(b - 2a)$

$ \Rightarrow $ $a(n - m + 1) = bm$

$ \Rightarrow $ $\frac{a}{b} = \frac{m}{{n - m + 1}}$.

Hence a worker can do ${\left( {\frac{1}{{150n}}} \right)^{th}}$ part of the work in a day.
Accordingly,

$[150 + 146 + 142 + ....... + {\rm{upto}}\;(n + 8)\,{\rm{terms}}] \times \frac{1}{{150n}} = 1$

$ \Rightarrow $$n = 17$

Therefore number of total days in completion $ = 17 + 8 = 25$.

New value of $\Sigma {x_i} = 1900 - 55 - 45$$ = 1800$, $n = 48$

New mean $ = \frac{{1800}}{{48}}$$ = 37.5$.

Then $(a - d) + a + (a + d) = 12$ and $(a - d)\,a\,(a + d) = 28$

==> $3a = 12$ and $a\,({a^2} - {d^2}) = 28$

==> $a = 4$ and $a\,({a^2} - {d^2}) = 28$

==> $16 - {d^2} = 7$

$\Rightarrow d = \pm \,3$.

$ \Rightarrow $ $({a_1} + {a_{24}}) + ({a_5} + {a_{20}}) + ({a_{10}} + {a_{15}}) = 225$

$ \Rightarrow $ $3({a_1} + {a_{24}}) = 225$

$ \Rightarrow $${a_1} + {a_{24}} = 75$

( In an $A.P.$ the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term)

${a_1} + {a_2} + ...... + {a_{24}} = \frac{{24}}{2}({a_1} + {a_{24}}) = 12 \times 75 = 900$.

and $n = n$, then ${S_1} + {S_2} + ....... + {S_m} = \frac{1}{2}mn(mn + 1)$

$\left[ {{\rm{Using}}\;S\; = \frac{m}{2}(a + l).\;{\rm{Since}}\;{S_1},\;{S_2},\;{S_3},......{S_m}\;{\rm{form}}\;{\rm{an}}\;{\rm{A}}{\rm{.P}}{\rm{.}}} \right]$

He pays $10\%$ annual interest on remaining amount

$\therefore $ Money given in first year

$ = 1000 + \frac{{10000 \times 10}}{{100}} = {\rm{Rs}}.2000$

Money given in second year $= 1000 +$ interest of$ (10000 -1000)$ with interest rate $10\%$ per annum $ = 1000 + \frac{{9000 \times 10}}{{100}} = {\rm{Rs}}.\,1900$

Money paid in third year = Rs. $1800$ etc.

So money given by Jairam in $10$ years will be Rs. $2000$, Rs. $1900$, Rs. $1800$, Rs. $1700$,....,

which is in arithmetic progression, whose first term $a = 2000$and $d = - 100$

Total money given in $10$ years = sum of $10$ terms of arithmetic progression

$ = \frac{{10}}{2}[2(2000) + (10 - 1)( - 100)]$= Rs. $15500$

Therefore, total money given by Jairam

$ = 5000 + 15500 = {\rm{Rs}}{\rm{. }}\,{\rm{20500}}{\rm{.}}$

$\angle C = x + {20^o}$and $\angle D = x + {30^o}$

So, we know that $\angle A + \angle B + \angle C + \angle D = 2\pi $

Putting these values, we get

$({x^o}) + ({x^o} + {10^o}) + ({x^o} + {20^o}) + ({x^o} + {30^o}) = {360^o}$

$ \Rightarrow x = {75^o}$

Hence the angles of the quadrilateral are ${75^o},\;{85^o},\;{95^o},\;{105^o}$.

Trick : In these type of questions, students should satisfy the conditions through options.

Here $(b)$ satisfies both the conditions

$i.e.$ angles are in $A.P.$ with common difference ${10^o}$ and sum of angles is ${360^o}$.

Then ${b^2} - {a^2} = {c^2} - {b^2}$

$ \Rightarrow $ $(b - a)(b + a) = (c - b)(c + b)$

==> $\frac{{b - a}}{{c + b}} = \frac{{c - b}}{{b + a}}$

==> $\frac{{(b - a)(a + b + c)}}{{(c + a)(b + c)}} = \frac{{(c - b)(a + b + c)}}{{(a + b)(c + a)}}$

$ \Rightarrow $ $\frac{{{b^2} + bc - ac - {a^2}}}{{(c + a)(b + c)}} = \frac{{{c^2} + ac - ab - {b^2}}}{{(a + b)(c + a)}}$

$ \Rightarrow $ $\frac{b}{{c + a}} - \frac{a}{{b + c}} = \frac{c}{{a + b}} - \frac{b}{{c + a}}$

Hence $\frac{a}{{b + c}},\;\frac{b}{{c + a}},\;\frac{c}{{a + b}}$ be in $A.P.$

==> $\frac{{2c}}{b} = \frac{b}{a} + \frac{a}{c}$

$ \Rightarrow \frac{{2c}}{b} = \frac{{bc + {a^2}}}{{ac}}$

==> $2a{c^2} = {b^2}c + b{a^2}$

$\therefore \,{a^2}b,\,{c^2}a$ and ${b^2}c$ are in $A.P.$

Add $1$ to each term, we get

$\frac{{a + b + c}}{{b + c}},\frac{{b + c + a}}{{c + a}},\frac{{c + a + b}}{{a + b}}$ are in $A.P.$

Divide each term by $(a + b + c),$

$\frac{1}{{b + c}},\frac{1}{{c + a}},\frac{1}{{a + b}}$ are in $A.P.$

Hence $b + c,\,\,c + a,\,\,a + b$ are in $H.P.$

which is given in question

Therefore, $\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}}$ are in $A. P.$

==> ${\tan ^{ - 1}}\left( {\frac{{2y}}{{1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{x + z}}{{1 - xz}}} \right)$

==> $\frac{{2y}}{{1 - {y^2}}} = \frac{{x + z}}{{1 - xz}}$

But $2y = x + z$

$1 - {y^2} = 1 - xz$

==> ${y^2} = xz$

$xyz$ are both in $G.P. $ and $A.P.$, 

$x = y = z$.

$11^{th}$ term of $A.P. =$ $a + 10d$

$21^{st}$ term of $A.P. = a+ 20d$

$2(a + 10d) = 7(a + 20d)$

==> $2a + 20d = 7a + 140d$

$5a + 120d = 0$

==> $a + 24d = 0$

Hence $25^{th}$ term is $0.$

${A_1} + {A_4} = 8$ ..$(i)$

and ${A_2}.\,{A_3} = 15$ ..$(ii)$

The sum of terms equidistant from the beginning and end is constant and is equal to

sum of first and last terms.

Hence, ${A_2} + {A_3} = {A_1} + {A_4} = 8$ ..$(iii)$

From $(ii)$ and $(iii),$

${A_2} + \frac{{15}}{{{A_2}}} = 8$

==> $A_2^2 - 8{A_2} + 15 = 0$

${A_2} = 3\,\,{\rm{or}}\,\,5$ and ${A_3} = 5\,\,\,{\rm{or}}\,\,{\rm{3}}$.

As we know, ${A_2} = \frac{{{A_1} + {A_3}}}{2}$

==> ${A_1} = 2{A_2} - {A_3}$

==> ${A_1} = 2 \times 3 - 5 = 1$ and ${A_4} = 8 - {A_1} = 7$

Hence the series is, $1, 3, 5, 7.$

So that least number of series is $1.$

$\therefore $ $\frac{1}{{r + q}} - \frac{1}{{p + q}} = \frac{1}{{q + r}} - \frac{1}{{r + p}}$

$ \Rightarrow $ $\frac{{p + q - r - p}}{{(r + p)(p + q)}} = \frac{{r + p - q - r}}{{(q + r)(r + p)}}$

$ \Rightarrow $ $\frac{{q - r}}{{p + q}} = \frac{{p - q}}{{q + r}}$ or ${q^2} - {r^2} = {p^2} - {q^2}$

$\therefore $ $2{q^2} = {r^2} + {p^2}$

Therefore ${p^2},\;{q^2},\;{r^2}$ are in $A.P.$

==> $x = \frac{{u + v}}{2},y = \frac{{u - v}}{2}$,

$\therefore f(u,v) = \left( {\frac{{u + v}}{2}} \right).\left( {\frac{{u - v}}{2}} \right)$

Now,$\frac{{f(x,y) + f(y,x)}}{2} = \frac{{\left( {\frac{{x + y}}{2}.\frac{{x - y}}{2}} \right) + \left( {\frac{{y + x}}{2}.\frac{{y - x}}{2}} \right)}}{2} = 0$.

therefore, ${a_2} - {a_1} = {a_4} - {a_3} = ....... = {a_{2n}} - {a_{2n - 1}} = d$

Here $a_1^2 - a_2^2 + a_3^2 - a_4^2 +$$ ....... + a_{2n - 1}^2 - a_{2n}^2$

$ = ({a_1} - {a_2})({a_1} + {a_2}) + ({a_3} - {a_4})({a_3} + {a_4}) +$$ ...... + ({a_{2n - 1}} - {a_{2n}})({a_{2n - 1}} + {a_{2n}})$

$ = - d({a_1} + {a_2} + ....... + {a_{2n}}) = - d\left\{ {\frac{{2n}}{2}({a_1} + {a_{2n}})} \right\}$

Also we know ${a_{2n}} = {a_1} + (2n - 1)d$$ \Rightarrow $$d = \frac{{{a_{2n}} - {a_1}}}{{2n - 1}}$

$ \Rightarrow $ $ - d = \frac{{{a_1} - {a_{2n}}}}{{2n - 1}}$.

$\therefore $ Therefore the sum is

= $\frac{{n({a_1} - {a_{2n}}).({a_1} + {a_{2n}})}}{{2n - 1}} = \frac{n}{{2n - 1}}(a_1^2 - a_{2n}^2)$.

and common difference $= d$

Let $S = \frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + .......... + \frac{1}{{{a_n}{a_{n + 1}}}}$

$⇒$  $S = \frac{1}{d}\,\left\{ {\frac{d}{{{a_1}{a_2}}} + \frac{d}{{{a_2}{a_3}}} + ...... + \frac{d}{{{a_n}\,\,{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\,\left\{ {\frac{{{a_2} - {a_1}}}{{{a_1}{a_2}}} + \frac{{{a_3} - {a_2}}}{{{a_2}{a_3}}} + ...... + \frac{{{a_{n + 1}} - {a_n}}}{{{a_n}\,\,\,{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\left\{ {\frac{1}{{{a_1}}} - \frac{1}{{{a_2}}} + \frac{1}{{{a_2}}} - \frac{1}{{{a_3}}} + ....... + \frac{1}{{{a_n}}} - \frac{1}{{{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\left\{ {\frac{1}{{{a_n}}} - \frac{1}{{{a_{n + 1}}}}} \right\} = \frac{1}{d}\left\{ {\frac{{{a_{n + 1}} - {a_1}}}{{{a_1}{a_{n + 1}}}}} \right\}$

$⇒$  $S = \frac{1}{d}\left( {\frac{{nd}}{{{a_1}{a_{n + 1}}}}} \right) = \frac{n}{{{a_1}{a_{n + 1}}}}$.

Trick: Check for $n = 2$.

$ = \frac{{100}}{2}(1 + 100) = 50(101) = 5050$

Let ${S_1} = 3 + 6 + 9 + 12 + ......... + 99$

=$3(1 + 2 + 3 + 4 + ......... + 33)$

=$3.\frac{{33}}{2}(1 + 33) = 99 \times 17 = 1683$

Let ${S_2} = 5 + 10 + 15 + ........ + 100$

= $5(1 + 2 + 3 + ........ + 20)$

= $5.\frac{{20}}{2}(1 + 20) = 50 \times 21 = 1050$

Let ${S_3} = 15 + 30 + 45 + ........ + 90$

= $15(1 + 2 + 3 + ........ + 6)$

= $15.\frac{6}{2}(1 + 6) = 45 \times 7 = 315$

Required sum =$S - {S_1} - {S_2} + {S_3}$

= $5050 - 1683 - 1050 + 315= 2632.$

First term of the series $a = \frac{1}{x} + y$,

Second term =$\frac{2}{x} + y$

$d = \left( {\frac{2}{x} + y} \right) - \left( {\frac{1}{x} + y} \right) = \frac{1}{x}$

Sum of $r$ terms of the series

$ = \frac{r}{2}\left[ {2\left( {\frac{1}{x} + y} \right) + (r - 1)\frac{1}{x}} \right]$

$ = \frac{r}{2}\left[ {\frac{2}{x} + 2y + \frac{r}{x} - \frac{1}{x}} \right]$

$ = \frac{{{r^2} - r + 2r}}{{2x}} + ry$

$ = \left\{ {\frac{{r{\mkern 1mu} (r + 1)}}{{2x}}} \right\} + ry$.

$ \Rightarrow $$\frac{1}{{{{\log }_x}\sqrt 3 \,}} + \frac{1}{{{{\log }_x}\sqrt[4]{3}}} + \frac{1}{{{{\log }_x}\sqrt[6]{3}}} + ... + \frac{1}{{{{\log }_x}\sqrt[{16}]{3}}} = 36$

$ \Rightarrow $ $\frac{1}{{(1/2){{\log }_x}3}} + \frac{1}{{(1/4){{\log }_x}3}} + \frac{1}{{(1/6){{\log }_x}3}} + ..... + \frac{1}{{(1/16){{\log }_x}3}} = 36$

$ \Rightarrow $ $({\log _3}x)(2 + 4 + 6 + ..... + 16) = 36$

$ \Rightarrow $ $({\log _3}x)\frac{8}{2}[2 + 16] = 36$

$ \Rightarrow $${\log _3}x = \frac{1}{2}$

$ \Rightarrow $$x = {3^{1/2}}$

$ \Rightarrow x = \sqrt 3 $.

$ = \frac{n}{2}\{ 4a + 4nd - 2d - 2a - nd + d\} = \frac{n}{2}\{ 2a + (3n - 1)d\} $

$ = \frac{1}{3}.\frac{{3n}}{2}\{ 2a + (3n - 1)d\} = \frac{1}{3}{S_{3n}}$.

This is an $AP$ with first term $13$ and common difference $4$.

Let the number of terms be $n$.

Then $97 = 13 + (n - 1)4$

$ \Rightarrow $ $4n = 88$

$ \Rightarrow $ $n = 22$

Therefore the sum of the numbers

$ = \frac{{22}}{2}[13 + 97] = 11(110) = 1210$.

$\log a + \log \left( {\frac{{{a^2}}}{b}} \right) + \log \left( {\frac{{{a^3}}}{{{b^2}}}} \right) + \log \left( {\frac{{{a^4}}}{{{b^3}}}} \right) + ...... + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$

This is an $A.P.$ with first term $\log a$

and the common difference $\log \left( {\frac{{{a^2}}}{b}} \right) - \log a = \log \left( {\frac{a}{b}} \right)$

Therefore the sum of $n$ terms is

$\frac{n}{2}\left[ {\log a + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)} \right] = \frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$.

Trick : Check for $n = 1,\;2$.

$ \Rightarrow $ $\frac{{2a + (m - 1)d}}{{2a + (n - 1)d}} = \frac{m}{n}$

$ \Rightarrow $ $\frac{{a + \frac{1}{2}(m - 1)d}}{{a + \frac{1}{2}(n - 1)d}} = \frac{m}{n}$

$ \Rightarrow $ $an + \frac{1}{2}(m - 1)nd = am + \frac{1}{2}(n - 1)md$

$ \Rightarrow $ $a(n - m) + \frac{d}{2}[mn - n - mn + m] = 0$

$ \Rightarrow $ $a(n - m) + \frac{d}{2}(m - n) = 0$

$ \Rightarrow $ $ a = \frac{d}{2}$ or $d = 2a$

So, required ratio, $\frac{{{T_m}}}{{{T_n}}} = \frac{{a + (m - 1)d}}{{a + (n - 1)d}} = \frac{{a + (m - 1)2a}}{{a + (n - 1)2a}}$

$ = \frac{{1 + 2m - 2}}{{1 + 2n - 2}} = \frac{{2m - 1}}{{2n - 1}}$.

Trick : Replace $m$ by $2m - 1$ and $n$ by $2n - 1$.

Obviously if ${S_m}$ is of degree $2$, then ${T_m}$ is of $1$.  $i.e.$ linear.

$\therefore $ $\sin d\,\{ co{\rm{sec}}\;{a_1}co{\rm{sec}}\;{a_2} + ..... + {\rm{cosec}}\;{a_{n - 1}}{\rm{cosec}}\;{a_n}\} $

$ = \frac{{\sin ({a_2} - {a_1})}}{{\sin {a_1}.\;\sin {a_2}}} + ...... + \frac{{\sin ({a_n} - {a_{n - 1}})}}{{\sin {a_{n - 1}}\sin {a_n}}}$

$ = (\cot {a_1} - \cot {a_2}) + (\cot {a_2} - \cot {a_3}) + .... + (\cot {a_{n - 1}} - \cot {a_n})$

$ = \cot {a_1} - \cot {a_n}$.

${T_{11}}$ and $T{'_{11}}$ be the respective ${11^{th}}$ terms, then

$\frac{{{S_n}}}{{S{'_n}}} = \frac{{\frac{n}{2}[2a + (n - 1)d]}}{{\frac{n}{2}[2a' + (n - 1)d']}} = \frac{{7n + 1}}{{4n + 27}}$

$ \Rightarrow $ $\frac{{a + \frac{{(n - 1)}}{2}d}}{{a' + \frac{{(n - 1)}}{2}d'}} = \frac{{7n + 1}}{{4n + 27}}$

Now put $n = 21$,

we get $\frac{{a + 10d}}{{a' + 10d'}} = \frac{{{T_{11}}}}{{T{'_{11}}}} = \frac{{148}}{{111}} = \frac{4}{3}$.

Note : If ratio of sum of $n$ terms of two $A.P.'s$ are given in terms of $n$ and ratio of their ${p^{th}}$ terms are to be found then put $n = 2p - 1$.

Here we put $n = 11 \times 2 - 1 = 21$.

==> $\frac{{\frac{n}{2}[2{a_1} + (n - 1){d_1}]}}{{\frac{n}{2}[2{a_2} + (n - 1){d_2}]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{2\left[ {{a_1} + \left( {\frac{{n - 1}}{2}} \right)\,{d_1}} \right]}}{{2\left[ {{a_2} + \left( {\frac{{n - 1}}{2}} \right)\,{d_2}} \right]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{{a_1} + \left( {\frac{{n - 1}}{2}} \right)\,{d_1}}}{{{a_2} + \left( {\frac{{n - 1}}{2}} \right)\,{d_2}}} = \frac{{2n + 3}}{{6n + 5}}$

Put $n = 25$ then $\frac{{{a_1} + 12{d_1}}}{{{a_2} + 12{d_2}}} = \frac{{2(25) + 3}}{{6(25) + 3}}$

==> $\frac{{{T_{{{13}_1}}}}}{{{T_{{{13}_2}}}}} = \frac{{53}}{{155}}$.

Putting $n = 1,\;2,\;3,\;.............,$ we get,

${S_1} = A + B,\,{S_2} = 2A + 4B,\,\,{S_3} = 3A + 9B$

Therefore ${T_1} = {S_1} = A + B,\;{T_2} = {S_2} - {S_1} = A + 3B,$

${T_3} = {S_3} - {S_2} = A + 5B$,

Hence the sequence is $(A + B),(A + 3B),\;(A + 5B),...$

Here $a = A + B$ and common difference $d = 2B$.

and $3 + 10 + 17 + 24 + ......$….. $(ii)$

Now from $(i),$ ${m^{th}}$ term $ = (2m + 61)$

and ${m^{th}}$ term of $(ii)$ series $ = (7m - 4)$

Under condition,

$ \Rightarrow 7m - 4 = 2m + 61$

$\Rightarrow 5m = 65$

$\Rightarrow m = 13$.

$ \Rightarrow $$n\theta = N\pi + (m\theta )$

$ \Rightarrow $ $\theta = \frac{{N\pi }}{{n - m}}$,

putting $N = 1,\;2,\;3.........,$ we get

$\frac{\pi }{{n - m}},\;\frac{{2\pi }}{{n - m}},\;\frac{{3\pi }}{{n - m}}.........$

which are obviously in $A.P.$

Since common difference $d = \frac{\pi }{{n - m}}$.

$ = \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ......\infty } \right)$
+ $\left( {1 + \frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ......\infty } \right)$

$ = \left( {\frac{1}{{1 - (1/2)}}} \right) + \left( {\frac{1}{{1 - (1/3)}}} \right) = 2 + \frac{3}{2} = \frac{7}{2}$.

$\therefore $$a + b = 3,\;ab = p$

$c,\;d$ are roots of ${x^2} - 12x + q = 0$

$\therefore $$c + d = 12,\;cd = q$

$a,\;b,\;c,\;d$ are in GP.

$\therefore $$\frac{b}{a} = \frac{d}{c}$$ \Rightarrow $$\frac{{a + b}}{{a - b}} = \frac{{c + d}}{{c - d}}$

$ \Rightarrow $$\frac{{{{(a - b)}^2}}}{{{{(a + b)}^2}}} = \frac{{{{(c - d)}^2}}}{{{{(c + d)}^2}}}$$ \Rightarrow $$1 - \frac{{4ab}}{{{{(a + b)}^2}}} = 1 - \frac{{4cd}}{{{{(c + d)}^2}}}$

$ \Rightarrow $$\frac{{ab}}{{{{(a + b)}^2}}} = \frac{{cd}}{{{{(c + d)}^2}}}$$ \Rightarrow $$\frac{p}{9} = \frac{q}{{144}}$

$ \Rightarrow $$\frac{p}{1} = \frac{q}{{16}}$$ \Rightarrow $$\frac{p}{q} = \frac{1}{{16}}$$ \Rightarrow $$\frac{{p + q}}{{q - p}} = \frac{{17}}{{15}}$.

Trick : Let $a = 1,\;b = 2,\;c = 4,\;d = 8$, then
$p = 2,\;q = 32$ $ \Rightarrow $ $(q + p):(q - p) = 17:15$.

$⇒$  $2b = a + c,b -a = c -b$

${a^2},{b^2},{c^2}$ are in $H.P.$

$\frac{1}{{{b^2}}} - \frac{1}{{{a^2}}} = \frac{1}{{{c^2}}} - \frac{1}{{{b^2}}}$ $ \Rightarrow \frac{{{a^2} - {b^2}}}{{{a^2}{b^2}}} = \frac{{{b^2} - {c^2}}}{{{b^2}{c^2}}}$

$⇒$  $(a - b)[{c^2}(a + b) - {a^2}(b + c)] = 0$,

$[\because \,(b - c) = (a - b)]$

$⇒$  $a = b$ or ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$

$⇒$  ${c^2}a + {c^2}b - {a^2}b - {a^2}c = 0$

$⇒$  $ac\,(c - a) = b\,({a^2} - {c^2})$

$⇒$  $ac = - b\,(c + a)$

$⇒$  $ - ac = b.2b$

$⇒$  ${b^2} = ( - a/2)\,c$,

$\therefore - a/2,b,c$ are in $G.P.$

==> $\frac{1}{{y - x}},\frac{1}{{2(y - a)}},\frac{1}{{y - z}}$ are in $A.P.$

==> $\frac{1}{{2(y - a)}} - \frac{1}{{(y - x)}} = \frac{1}{{y - z}} - \frac{1}{{2(y - a)}}$

==> $\frac{{y - x - 2y + 2a}}{{(y - x)}} = \frac{{2y - 2a - y + z}}{{(y - a) - (z - a)}}$

$ \Rightarrow \frac{{ - x - y + 2a}}{{(y - x)}} = \frac{{y + z - 2a}}{{(y - z)}}$

$ \Rightarrow \frac{{(x - a) + (y - a)}}{{(x - a) - (y - a)}} = \frac{{(y - a) + (z - a)}}{{(y - a) - (z - a)}}$

==> $\frac{{(x - a)}}{{(y - a)}} = \frac{{(y - a)}}{{(z - a)}}$

$(x - a),(y - a),(z - a)$ are in $G.P.$

$\therefore \,\,{G_1} = {p^{2/3}}\,\,{q^{1/3}},\,\,{G_2} = {p^{1/3}}\,\,\,{q^{2/3}}$

$\therefore \,\frac{{G_1^2}}{{{G_2}}} + \frac{{G_2^2}}{{{G_1}}} = \frac{{{p^{4/3}}\,\,{q^{2/3}}}}{{{p^{1/3}}\,\,{q^{2/3}}}} + \frac{{{p^{2/3}}\,{q^{4/3}}}}{{{p^{2/3}}\,{q^{1/3}}}}$

$ = p + q = 2 \times \,\left( {\frac{{p + q}}{2}} \right)\, = 2A$.

Then the middle term of the $A.P.$ is $\frac{{a + b}}{2}$.

The middle term of the $G.P.$ is $\sqrt {ab} $.

The middle term of the $H.P.$ is $\frac{{2ab}}{{a + b}}$.

Obviously, these terms are in $G.P.$

Then ${T_p} = a + (p - 1)d,\;$

${T_q} = a + (q - 1)d$

and ${T_r} = a + (r - 1)d$.

If ${T_p},\;{T_q},\;{T_r}$ are in $G.P.$

Then its common ratio $R = \frac{{{T_q}}}{{{T_p}}} = \frac{{{T_r}}}{{{T_q}}} = \frac{{{T_q} - {T_r}}}{{{T_p} - {T_q}}}$

$ = \frac{{[a + (q - 1)d] - [a + (r - 1)d]}}{{[a + (p - 1)d] - [a + (q - 1)d]}} = \frac{{q - r}}{{p - q}}$

Similarly, we can show that $R = \frac{{q - r}}{{p - q}} = \frac{{r - s}}{{q - r}}$

Hence $(p - q),\;(q - r),\;(r - s)$ be in $G.P.$

$⇒$  ${(a + 2c)^2} - {(2b)^2} = {a^2} + 4{c^2}$

$⇒$  ${a^2} + 4ac + 4{c^2} - 4{b^2} = {a^2} + 4{c^2}$

$⇒$  $4ac - 4{b^2} = 0$

$⇒$  ${b^2} = ac$

Hence $a, b, c$ are in $G.P.$

$= 0.037+0.000037+0.0000000037+…….$

= $\frac{{37}}{{{{10}^3}}} + \frac{{37}}{{{{10}^6}}} + \frac{{37}}{{{{10}^9}}} + ......$

= $37\left[ {\frac{1}{{{{10}^3}}} + \frac{1}{{{{10}^6}}} + \frac{1}{{{{10}^9}}} + ....} \right]$

= $37\left[ {\frac{{1/{{10}^3}}}{{1 - 1/{{10}^3}}}} \right] $

$= 37\left[ {\frac{1}{{{{10}^3}}}.\frac{{{{10}^3}}}{{999}}} \right]$ = $\frac{{37}}{{999}}$.

$= 0.5 + 0.073 + 0.00073$

$= 0.5 +$ $\frac{{73}}{{1000}} + \frac{{73}}{{100000}} + ....$

= $0.5 + 73\left[ {\frac{1}{{1000}} + \frac{1}{{100000}} + .....} \right]$

= $0.5 + 73\left[ {\frac{{1/1000}}{{1 - \frac{1}{{100}}}}} \right]$

= $0.5 + \frac{{73}}{{1000}}.\frac{{100}}{{99}} = \frac{5}{{10}} + \frac{{73}}{{990}}$

= $\frac{{495 + 73}}{{990}} = \frac{{568}}{{990}}$.

and $\frac{{{a^2}}}{{1 - {r^2}}} = 3$ .....(ii)

From (i) and (ii), $\frac{a}{{1 + r}} = 1$

$\Rightarrow a = 1 + r$

From (i), $\frac{{1 + r}}{{1 - r}} = 3 $

$\Rightarrow r = \frac{1}{2}$ , from (i), $ a = 3/2$

So, first term $= 3/2$ and common ratio $= 1/2.$

$\therefore $${\left( {\frac{1}{5}} \right)^{{{\log }_{\sqrt 5 }}\left( {\frac{1}{2}} \right)}} = {\left( {\frac{1}{5}} \right)^{{{\log }_5}\left( {\frac{1}{4}} \right)}} = {5^{ - {{\log }_5}{4^{ - 1}}}} = {5^{{{\log }_5}4}} = 4$.

${({S_2})_\infty } = \frac{{{a^2}}}{{1 - {r^2}}} = 3$

or ${a^2} = 3\,(1 - {r^2})$ or $9\,{(1 - r)^2} = 3\,(1 - {r^2})$ [by $(i)$]

or $3\,(1 - 2r + {r^2}) = 1 - {r^2}$ or $2{r^2} - 3r + 1 = 0$

or $(r - 1)\,(2r - 1) = 0$,

$\therefore $$r = 1,\frac{1}{2}$

If $r = 1,$ then $a = 3(1 - 1) = 0$ which is impossible.

If $r = \frac{1}{2},$then $a = 3\,\left( {1 - \frac{1}{2}} \right) = 3/2$

So first series is $3/2, 3/4, 3/8, 3/16,.....$

$ = 0.14 + 0.00189 + 0.00000189 + .......$

$ = \frac{{14}}{{100}} + 189\left[ {\frac{1}{{{{10}^5}}} + \frac{1}{{{{10}^8}}} + ....\infty } \right]$

$ = \frac{7}{{50}} + 189\,\left[ {\frac{{1/{{10}^5}}}{{1 - (1/{{10}^3})}}} \right]$$ = \frac{7}{{50}} + 189\,\left[ {\frac{1}{{{{10}^5}}} \times \frac{{{{10}^3}}}{{999}}} \right]$

$ = \frac{7}{{50}} + \frac{{189}}{{999 \times 100}}$

$ = \frac{7}{{50}} + \frac{7}{{3700}} = \frac{7}{{50}} + \frac{7}{{25 \times 148}}$$ = \frac{{21}}{{148}}$.

and $\frac{{{a^2}}}{{1 - {r^2}}} = \frac{a}{{1 - r}}.\frac{a}{{1 + r}} = y$ …..$(ii)$

$ \Rightarrow $ $y = x.\frac{a}{{1 + r}} = x.\frac{{x(1 - r)}}{{1 + r}}$

$ \Rightarrow $$\frac{y}{{{x^2}}} = \frac{{1 - r}}{{1 + r}}$

$ \Rightarrow $ $\frac{{{x^2}}}{y} = \frac{{1 + r}}{{1 - r}}$

$ \Rightarrow $$\frac{{{x^2}}}{y}(1 - r) = 1 + r$ 

$\Rightarrow r[1+ \frac{{x^2}}{{y}}] = -1 + \frac{{x^2}}{{y}}$

$\Rightarrow r=\frac{{{x^2} + y}}{{{x^2} - y}}$

then $xy = {x^2} - {x^3} + {x^4} - ......\infty $

Adding, $y + xy = x + 0 + 0...... + 0$

$ \Rightarrow $$x - xy = y $

$\Rightarrow x(1 - y) = y$

$\Rightarrow x = \frac{y}{{1 - y}}$.

Aliter : $y = \frac{x}{{1 - ( - x)}} $

$\Rightarrow y = \frac{x}{{1 + x}}$

$ \Rightarrow $$y + yx = x$

$\Rightarrow x = \frac{y}{{1 - y}}$.