Solution

Let the common difference of an AP is $d$. According to the question,

$S_p=0$

$\Rightarrow \frac{p}{2}[2a+(p-1)d]=0[\because S_n=\frac{n}{2}2a+(n-1)d]$

$\Rightarrow 2a+(p-1)d=0$

$\therefore d=\frac{-2a}{p-1}$

$\text{ Now, sum of next }q\text{ terms }=S_{p+q}-S_p=S_{p+q}-0$

$=\frac{p+q}{2}[2a+(p+q-1)d]$

$=\frac{p+q}{2}[2a+(p-1)d+qd]$

$=\frac{p+q}{2}[2a+(p-1)\cdot \frac{-2a}{p-1}+\frac{q(-2a)}{p-1}]$

$=\frac{p+q}{2}[2a+(-2a)-\frac{2aq}{p-1}]$

$=\frac{p+q}{2}[\frac{-2aq}{p-1}]$

$=\frac{-a(p+q)q}{(p-1)}$

$\frac{p}{2}[2a+(p-1)d]=0\because S_n=\frac{n}{2}2a+(n-1)d$

Solution

Let saved in first year ₹ a. Since, each succeeding year an increment ₹ 200 has made. So,it forms an AP whose

First term $=a$, common difference $(d)=200$ and $n=20yr$

$\begin{array}{rl}& \therefore S_{20}=\frac{20}{2}[2a+(20-1)d][\because S_n=\frac{n}{2}2a+(n-1)d]\\ & \Rightarrow 66000=10[2a+19d]\\ & \Rightarrow 66000=20a+190d\\ & \Rightarrow 66000=20a+190\times 200\\ & \Rightarrow 20a=66000-38000\\ & \Rightarrow 20a=28000\\ & \therefore a=\frac{28000}{20}=1400\end{array}$

Hence, he saved ₹ 1400 in the first year.

Solution

Since, the man get a fixed increment of ₹ 320 each month. Therefore, this forms an AP whose First term $=5200$ and Common difference $(d)=320$

(i) Salary for tenth month i.e., for $n=10$,

$\begin{array}{cc}\Rightarrow & a_{10}=a+(n-1)d\\ \Rightarrow & a_{10}=5200+(10-1)\times 320\\ \therefore & a_{10}=5200+9\times 320\\ \therefore & a_{10}=5200+2880\\ a_{10}=8080\end{array}$

(ii) Total earning during the first year.

In a year there are 12 month i.e., $n=12$,

$\begin{array}{rl}{S}_{12}& =\frac{12}{2}[2\times 5200+(12-1)320]\\ & =6[10400+11\times 320]\\ & =6[10400+3520]=6\times 13920=83520\end{array}$

Solution

Let the first term and common ratio of GP be a and $r$, respectively.

According to the question, $p$ th term $=q$

$\Rightarrow a\cdot r^{p-1}=q\dots (i)$

and

$q$ th term $=p$

$\Rightarrow$

$a{r}^{q-1}=p\dots (ii)$

On dividing Eq. (i) by Eq. (ii), we get

$\begin{array}{rl}& \frac{a{r}^{p-1}}{a{r}^{q-1}}=\frac{q}{p}\\ & \Rightarrow r^{p-1-q+1}=\frac{q}{p}\\ & \Rightarrow r^{p-q}=\frac{q}{p}\Rightarrow r=(\frac{q}{p}{)}^{\frac{1}{p-q}}\end{array}$

On substituting the value of $r$ in Eq. (i), we get

$\begin{array}{rl}& a(\frac{q}{p}{)}^{\frac{p-1}{p-q}}=q\Rightarrow a=\frac{q}{(\frac{q}{p}{)}^{\frac{p-1}{p-q}}}=q\cdot (\frac{p}{q}{)}^{\frac{p-1}{p-q}}\\ & \therefore (p+q)\text{ th term, }{T}_{p+q}=a\cdot r^{p+q-1}=q\cdot {\frac{p}{q}}^{\frac{p-1}{p-q}}\cdot (r{)}^{p+q-1}\\ & =q\cdot (\frac{p}{q}{)}^{\frac{p-1}{p-q}}(\frac{q}{p}{)}^{\frac{1}{p-q}}{}^{p+q-1}=q\cdot (\frac{p}{q}{)}^{\frac{p-1}{p-q}}\frac{q}{\frac{p}{p}}\\ & =q\cdot (\frac{p}{q}{)}^{\frac{p-q-1}{p-q}}{\frac{p}{q}}^{\frac{-(p+q-1)}{p-q}}=q\cdot (\frac{p}{q}{)}^{\frac{p-1}{p-q}-\frac{(p+q-1)}{p-q}}\\ & =q\cdot (\frac{p}{q}{)}^{\frac{p-1-p-q+1}{p-q}}=q\cdot (\frac{p}{q}{)}^{\frac{-q}{p-q}}\\ & a=q\cdot (\frac{p}{q}{)}^{\frac{p-1}{p-q}}\end{array}$

Now, $(p+q)$ th term i.e., $a_{p+q}=a{r}^{p+q-1}$

$\begin{array}{rl}& =q\cdot (\frac{p}{q}{)}^{\frac{p-1}{p-q}}\cdot (\frac{q}{p}{)}^{\frac{p+q-1}{p-q}}\\ & =q\cdot \frac{q\frac{p+q-1-p+1}{p-q}}{p\frac{p+q-1-p+1}{p-q}}=q\cdot (\frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}})\end{array}$

Solution

Here, $a=5$ and $d=2$

Let he finished the job in $n$ days.

Then,

$\begin{array}{rl}& S_n=192\\ & S_n=\frac{n}{2}[2a+(n-1)d]\end{array}$

$\begin{array}{cc}\Rightarrow & 192=\frac{n}{2}[2\times 5+(n-1)2]\\ \Rightarrow & 192=\frac{n}{2}[10+2n-2]\end{array}$

Solution

We know that, sum of interior angles of a polygon of side $n=(2n-4)\times {90}^{\circ }=(n-2)\times {180}^{\circ }$ Sum of interior angles of a polygon with sides 3 is 180 .

Sum of interior angles of polygon with side $4=(4-2)\times {180}^{\circ }={360}^{\circ }$

Similarly, sum of interior angles of polygon with side $5,6,7\dots$ are ${540}^{\circ },{720}^{\circ },{900}^{\circ },\dots$

The series will be ${180}^{\circ },{360}^{\circ }{540}^{\circ },{720}^{\circ },{900}^{\circ },.$.

Here, $a={180}^{\circ }$

and $d={360}^{\circ }-{180}^{\circ }={180}^{\circ }$

Since, common difference is same between two consecutive terms of the series.

So, it form an AP.

We have to find the sum of interior angles of a 21 sides polygon.

It means, we have to find the 19th term of the above series.

$\begin{array}{rl}& \therefore a_{19}=a+(19-1)d\\ & =180+18\times 180=3420\end{array}$

Solution

Side of equilateral $\mathrm{△}ABC=20cm$. By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of $\mathrm{△}ABC$.

Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.

$\therefore$ Perimeter of first triangle $=20\times 3=60cm$

Perimeter of second triangle $=10\times 3=30cm$

Perimeter of third triangle $=5\times 3=15cm$

Now, the series will be $60,30,15,\dots$

Here,

$a=60$

$\therefore r=\frac{30}{60}=\frac{1}{2}[\because \frac{\text{ second term }}{\text{ first term }}=r$]

We have, to find perimeter of sixth inscribed triangle. It is the sixth term of the series.

$\begin{array}{rlr}\therefore & a_6& =a{r}^{6-1}\\ & =60\times {\frac{1}{2}}^5=\frac{60}{32}=\frac{15}{8}cm& [\because a_n=a{r}^{n-1}]\end{array}$

Solution

According to the given information, we have following diagram.

Distance travelled to bring first potato $=24+24=2\times 24=48m$

Distance travelled to bring second potato $=2(24+4)=2\times 28=56m$

Distance travelled to bring third potato $=2(24+4+4)=2\times 32=64m$

Then, the series of distances are $48,56,64,\dots$

Here,

$\begin{array}{rl}& a=48\\ & d=56-48=8\end{array}$

$\text{ and }n=20$

To find the total distance that he run in bringing back all potatoes, we have to find the sum of 20 terms of the above series.

$\begin{array}{cc}\therefore S_{20}& =\frac{20}{2}[2\times 48+19\times 8]\because S_n=\frac{n}{2}2a+(n-1)d\\ & =10[96+152]& \\ & =10\times 248=2480m\end{array}$

Solution

Let the first place team got ₹ a.

Since, award money increases by the same amount for successive finishing places. Therefore series is an AP.

Let the constant amount be $d$.

Here, l = 275, n = 16 and $S_{16}=8000$

$\therefore$ l = a+(n-)d

$\Rightarrow$ l = a+(16-1)(-d)

[we take common difference (−ve) because series is decreasing]

$\Rightarrow$ 275 = a-15d

and

$\Rightarrow 8000=8[2a+(16-1)(-d)]$

$\Rightarrow 8000=8[2a-15d]$

$\Rightarrow 1000=2a-15d$

On subtracting Eq. (i) from Eq. (ii), we get

$(2a-15d)-(a-15d)=1000-275$

$\Rightarrow 2a-15d-a+15d=725$

$\therefore a=725$

Hence, first place team receive ₹ 725 .

Solution

Since, $a_1,a_2,a_3,\dots ,a_n$ are in AP.

$\begin{array}{rl}& \Rightarrow a_2-a_1=a_3-a_2=\dots =a_n-a_{n-1}=d\text{ [common difference] }\\ & \text{ If }{a}_2-a_1=d\text{, then }(\sqrt{a_2}{)}^2-(\sqrt{a_1}{)}^2=d\\ & \Rightarrow (\sqrt{a_2}-\sqrt{a_1})(\sqrt{a_2}+\sqrt{a_1})=d\\ & \Rightarrow \frac{1}{\sqrt{a_1}+\sqrt{a_2}}=\frac{\sqrt{a_2}-\sqrt{a_1}}{d}\\ & \frac{1}{\sqrt{a_2}+\sqrt{a_3}}=\frac{\sqrt{a_3}-\sqrt{a_2}}{d}\\ & \frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{d}\end{array}$

On adding these terms, we get

$\begin{array}{rl}& \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\\ =& \frac{1}{d}[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+\dots +\sqrt{a_n}-\sqrt{a_{n-1}}]\text{ [using above relations] }\\ =& \frac{1}{d}[\sqrt{a_n}-\sqrt{a_1}]\end{array}$

Again,

$a_n=a_1+(n-1)d[\because T_n=a+(n-1)d]$

$\Rightarrow$

$a_n-a_1=(n-1)d$

$\begin{array}{cc}\Rightarrow & (\sqrt{a_n}{)}^2-(\sqrt{a_1}{)}^2=(n-1)d\\ \Rightarrow & (\sqrt{a_n}-\sqrt{a_1})(\sqrt{a_n}+\sqrt{a_1})=(n-1)d\Rightarrow \sqrt{a_n}-\sqrt{a_1}=\frac{(n-1)d}{\sqrt{a_n}+\sqrt{a_1}}\end{array}$

On putting this value in Eq. (i), we get

$\begin{array}{c}\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\\ =\frac{(n-1)d}{d(\sqrt{a_n}+\sqrt{a_1})}=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}\end{array}$

Solution

Given series, $(3^3-2^3)+(5^3-4^3)+(7^3-6^3)+$

$=(3^3+5^3+7^3+\dots )-(2^3+4^3+6^3+\dots )$

Let $T_n$ be the $n$th term of the series (i),

then $T_n=(n.$th term of ${.3}^3,5^3,7^3,\dots )-(n.$th term of ${.2}^3,4^3,6^3,\dots )=(2n+1{)}^3-(2n{)}^3$

$\begin{array}{rl}& =(2n+1-2n)[(2n+1{)}^2+(2n+1)2n+(2n{)}^2][\because a^3-b^3=(a-b)(a^2+ab+b^2)]\\ & =[4n^2+1+4n+4n^2+2n+4n^2]=[12n^2+6n]+1\end{array}$

(i) Let $S_n$ denote the sum of $n$ term of series (i). Then,

$\begin{array}{rl}{S}_n& =\mathrm{\Sigma }{T}_n=\mathrm{\Sigma }(12n^2+6n)\\ & =12\mathrm{\Sigma }{n}^2+6\mathrm{\Sigma }n+\mathrm{\Sigma }n\\ & =12\cdot \frac{n(n+1)(2n+1)}{6}+\frac{6n(n+1)}{2}+n\\ & =2n(n+1)(2n+1)+3n(n+1)+n\\ & =2n(n+1)(2n+1)+3n(n+1)+n\\ & =(2n^2+2n)(2n+1)+3n^2+3n+n\\ & =4n^3+2n^2+4n^2+2n+3n^2+3n+n\\ & =4n^3+9n^2+6n\\ S_{10}& =4\times (10{)}^3+9\times (10{)}^2+6\times 10\\ & =4\times 1000+9\times 100+60\\ & =4000+900+60=4960\end{array}$

(ii) Sum of 10 terms,

Solution

Given that, sum of $n$ terms of an AP,

$\begin{array}{rl}{S}_n& =2n+3n^2\\ T_n& =S_n-S_{n-1}\\ & =(2n+3n^2)-[2(n-1)+3(n-1{)}^2]\\ & =(2n+3n^2)-[2n-2+3(n^2+1-2n)]\\ & =(2n+3n^2)-(2n-2+3n^2+3-6n)\\ & =2n+3n^2-2n+2-3n^2-3+6n\\ \therefore & =6n-1\\ \therefore \text{ rth term }{T}_r& =6r-1\end{array}$

Solution

Let the numbers be $a$ and $b$.

Then,

$A=\frac{a+b}{2}$

$\Rightarrow 2A=a+b$

and $G_1,G_2$ be geometric mean between $a$ and $b$, then $a,G_1,G_2,b$ are in GP.

Let $r$ be the common ratio.

$\begin{array}{rl}& \text{ Then, }\\ & b=a{r}^{4-1}\\ & \Rightarrow b=a{r}^3\Rightarrow \frac{b}{a}=r^3\\ & \therefore r={\frac{b}{a}}^{1/3}\end{array}$

$\begin{array}{rl}& \text{ Now, }\\ & G_1=ar=a{\frac{b}{a}}^{1/3}\\ & \text{ and }\\ & G_2=a{r}^2=a{\frac{b}{a}}^{2/3}\\ & RHS=\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=\frac{a{\frac{b}{a}}^{1/3^2}}{a\frac{b}{a}}+\frac{a{\frac{b}{a}}^{2/3}}{a\frac{b}{a}}\\ & =\frac{a^2\frac{b^{2/3}}{a}}{a\frac{b}{a}}+\frac{a^2\frac{b}{a}}{a\frac{b}{a}}\\ & =a+a\frac{b}{a}=a+b=2A\\ & =LHS\end{array}$

Solution

Since, ${\theta }_1,{\theta }_2,{\theta }_3,\dots ,{\theta }_n$ are in AP.

$\Rightarrow {\theta }_2-{\theta }_1={\theta }_3-{\theta }_2=\cdots ={\theta }_n-{\theta }_{n-1}=d$

$\because r={\frac{b}{a}}^{1/3}$

Now, we have to prove

$\sec q_1\sec q_2+\sec q_2\sec q_3+\cdots +\sec q_{n-1}\sec {\theta }_n=\frac{\tan {\theta }_n-\tan {\theta }_1}{\sin d}$

or it can be written as

sind $[\sec {\theta }_1\sec {\theta }_2+\sec {\theta }_2\sec {\theta }_3+\cdots +\sec {\theta }_{n-1}\sec {\theta }_n]=\tan {\theta }_n-\tan {\theta }_1$

Now, taking only first term of LHS

$\begin{array}{rl}\sin d\sec {\theta }_1\sec {\theta }_2& =\frac{\sin d}{\cos {\theta }_1\cos {\theta }_2}=\frac{\sin ({\theta }_2-{\theta }_1)}{\cos {\theta }_1\cos {\theta }_2}\\ & =\frac{\sin {\theta }_2\cos {\theta }_1-\cos {\theta }_2\sin {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}\\ & =\frac{\sin {\theta }_2\cos {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}-\frac{\cos {\theta }_2\sin {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}=\tan {\theta }_2-\tan {\theta }_1\end{array}$

Similarly, we can solve other terms which will be $\tan {\theta }_3-\tan {\theta }_2,\tan {\theta }_4-\tan {\theta }_3,\cdots$

$\begin{array}{rl}\therefore LHS& =\tan {\theta }_2-\tan {\theta }_1+\tan {\theta }_3-\tan {\theta }_2+\cdots +\tan {\theta }_n-\tan {\theta }_{n-1}\\ & =-\tan {\theta }_1+\tan {\theta }_n=\tan {\theta }_n-\tan {\theta }_1\\ & =\text{ RHS }\text{ Hence proved. }\end{array}$

Solution

Let first term and common difference of the AP be a and $d$, respectively. Then,

$S_p=q$

$\Rightarrow$

$\begin{array}{rl}\frac{p}{2}[2a+(p-1)d]& =q\\ 2a+(p-1)d& =\frac{2q}{p}\end{array}$

and

$S_q=p$

$\begin{array}{cccc}\Rightarrow & & \frac{q}{2}[2a+(q-1)d]& =p\\ \Rightarrow & 2a+(q-1)d& =\frac{2p}{q}\end{array}$

On subtracting Eq. (ii) from Eq. (i), we get

$2a+(p-1)d-2a-(q-1)d=\frac{2q}{p}-\frac{2p}{q}$

$\Rightarrow [(p-1)-(q-1)]d=\frac{2q^2-2p^2}{pq}$

$\Rightarrow [p-1-q+1]d=\frac{2(q^2-p^2)}{pq}$

$\therefore d=\frac{-2(p+q)}{pq}$

On substituting the value of $d$ in Eq. (i), we get

$\begin{array}{rl}& 2a+(p-1)\frac{-2(p+q)}{pq}=\frac{2q}{p}\\ & \Rightarrow 2a=\frac{2q}{p}+\frac{2(p+q)(p-1)}{pq}\\ & \Rightarrow a=\frac{q}{p}+\frac{(p+q)(p-1)}{pq}\\ & \text{ Now, }{S}_{p+q}=\frac{p+q}{2}[2a+(p+q-1)d]\\ & =\frac{p+q}{2}\frac{2q}{p}+\frac{2(p+q)(p-1)}{pq}-\frac{(p+q-1)2(p+q)}{pq}\\ & =(p+q)\frac{q}{p}+\frac{(p+q)(p-1)-(p+q-1)(p+q)}{pq}\\ & =(p+q)\frac{q}{p}+\frac{(p+q)(p-1-p-q+1)}{pq}\\ & =p+q\frac{q}{p}-\frac{p+q}{p}=(p+q)\frac{q-p-q}{p}\\ & S_{p+q}=-(p+q)\\ & S_{p-q}=\frac{p-q}{2}[2a+(p-q-1)d]\end{array}$

$\begin{array}{rl}& =\frac{p-q}{2}\frac{2q}{p}+\frac{2(p+q)(p-1)}{pq}-\frac{(p-q-1)2(p+q)}{pq}\\ & =(p-q)\frac{q}{p}+\frac{p+q(p-1-p+q+1)}{pq}\\ & =(p-q)\frac{q}{p}+\frac{(p+q)q}{pq}\\ & =(p-q)\frac{q}{p}+\frac{p+q}{p}=(p-q)\frac{(p+2q)}{p}\end{array}$

Solution

Let $A,d$ are the first term and common difference of AP and $x,R$ are the first term and common ratio of GP, respectively.

According to the given condition,

and

$\begin{array}{rl}A+(p-1)d& =a\\ A+(q-1)d& =b\\ A+(r-1)d& =c\\ a& =x{R}^{p-1}\\ b& =x{R}^{q-1}\\ c& =x{R}^{r-1}\end{array}$

On subtracting Eq. (ii) from Eq. (i), we get

$d(p-1-q+1)=a-b$

$\Rightarrow a-b=d(p-q)$

On subtracting Eq. (iii) from Eq. (ii), we get

$d(q-1-r+1)=b-c$

$\Rightarrow b-c=d(q-r)$

On subtracting Eq. (i) from Eq. (iii), we get

$d(r-1-p+1)=c-a$

$\Rightarrow c-a=d(r-p)$

Now, we have to prove $a^{b-c}{b}^{c-a}{c}^{a-b}=1$

$\text{ Taking LHS }=a^{b-c}{b}^{c-a}{c}^{a-b}$

Using Eqs. (iv), (v), (vi) and (vii), (viii), (ix),

$\begin{array}{rl}LHS& =(x{R}^{p-1}{)}^d(q-r)(x{R}^{q-1}{)}^d(r-p)(x{R}^{r-1}{)}^{d(p-q)}\\ & =x^{d(q-r)+d(r-p)+d(p-q)}{R}^{(p-1)d(q-r)+(q-1)d(r-p)+(r-1)d(p-q)}\\ & =x^{d(q-r+r-p+p-q)}\end{array}$

$\begin{array}{rl}{R}^{d(pq-pr-q+r+qr-pq-r+p+rp-rq-p+q)}& =x^0{R}^0=1\\ & =RHS\end{array}$

Solution

(d) Given, $S_n=3n+2n^2$

First term of the AP,

$\begin{array}{rl}& \therefore T_1=3\times 1+2(1{)}^2=3+2=5\\ & \text{ and }{T}_2=S_2-S_1\\ & =[3\times 2+2\times (2{)}^2]-[3\times 1+2\times (1{)}^2]\\ & =14-5=9\end{array}$

$\therefore$ Common difference $(d)=T_2-T_1=9-5=4$

Solution

(c) It is given that, $T_3=4$

Let $a$ and $r$ the first term and common ratio, respectively.

Then,

$a{r}^2=4$

Product of first 5 terms $=a\cdot ar\cdot a{r}^2\cdot a{r}^3\cdot a{r}^4$

$=a^5{r}^{10}=(a{r}^2{)}^5=(4{)}^5$

Solution

(a) Let the first term be a and common difference be $d$.

According to the question, $9\cdot T_9=13\cdot T_{13}$

$\begin{array}{ccc}\Rightarrow & 9(a+8d)& =13(a+12d)\\ \Rightarrow & 9a+72d& =13a+156d\\ \Rightarrow & (9a-13a)& =156d-72d\\ \Rightarrow & -4a& =84d\\ \Rightarrow & a& =-21d\\ \Rightarrow & a+21d& =0\\ \therefore & & \text{ 22nd term i.e., }{T}_{22}& =[a+21d]\\ & T_{22}& =0\text{[using Eq. (i)]}\end{array}$

Solution

(b) Given, $x,2y$ and $3z$ are in AP.

Then,

$2y=\frac{x+3z}{2}$

$\Rightarrow y=\frac{x+3z}{4}$

$\Rightarrow 4y=x+3z$

and $x,y,z$ are in GP

$\begin{array}{cc}\text{ Then, }& \frac{y}{x}=\frac{z}{y}=\lambda \\ \Rightarrow & y=x\lambda \text{ and }z=\lambda y={\lambda }^{2}x\end{array}$

On substituting these values in Eq. (i), we get

$4(x\lambda )=x+3({\lambda }^{2}x)$

$\Rightarrow 4\lambda x=x+3{\lambda }^{2}x$

$\Rightarrow 4\lambda =1+3{\lambda }^2$

$\Rightarrow 3{\lambda }^2-4\lambda +1=0$

$\Rightarrow (3\lambda -1)(\lambda -1)=0$

$\therefore \lambda =\frac{1}{3},\lambda =1$

Solution

(c) Given, $S_n=q{n}^2$ and $S_m=q{m}^2$

$\begin{array}{rl}& \therefore S_1=q,S_2=4q,S_3=9q\text{ and }{S}_4=16q\\ & \text{ Now, }{T}_1=q\\ & \therefore T_2=S_2-S_1=4q-q=3q\\ & T_3=S_3-S_2=9q-4q=5q\\ & T_4=S_4-S_3=16q-9q=7q\end{array}$

So, the series is $q,3q,5q,7q,\dots$

$\begin{array}{rl}& \text{ Here, }a=q\text{ and }d=3q-q=2q\\ & \therefore S_q=\frac{q}{2}[2\times q+(q-1)2q]\\ & =\frac{q}{2}\times [2q+2q^2-2q]=\frac{q}{2}\times 2q^2=q^3\end{array}$

Solution

(b) Let first term be a and common difference be $d$.

Then, $S_n=\frac{n}{2}[2a+(n-1)d]$

$\therefore S_{2n}=\frac{2n}{2}[2a+(2n-1)d]$

$S_{2n}=n[2a+(2n-1)d]$

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

According to the question, $S_{2n}=3S_n$

$\begin{array}{rlrl}\Rightarrow & & n[2a+(2n-1)d]& =3\frac{n}{2}[2a+(n-1)d]\\ \Rightarrow & & 4a+(4n-2)d& =6a+(3n-3)d\\ \Rightarrow & & -2a+(4n-2-3n+3)d& =0\\ \Rightarrow & & -2a+(n+1)d& =0\\ \Rightarrow & & d& =\frac{2a}{n+1}\end{array}$

Now,

$\begin{array}{rl}\frac{S_{3n}}{S_n}& =\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{6a+(9n-3)\frac{2a}{n+1}}{2a+(n-1)\frac{2a}{n+1}}\\ & =\frac{6an+6a+18an-6a}{2an+2a+2an-2a}\\ & =\frac{24an}{4an}=\frac{S_{3n}}{S_n}=6\end{array}$

Solution

(b) We know that,

$AM\ge GM$

$\begin{array}{cc}\Rightarrow & \frac{4^x+4^{1-x}}{2}\ge \sqrt{4^x\cdot 4^{1-x}}\\ \Rightarrow & 4^x+4^{1-x}\ge 2\sqrt{4}\\ \Rightarrow & 4^x+4^{1-x}\ge 2\cdot 2\\ \Rightarrow & 4^x+4^{1-x}\ge 4\end{array}$