Solution

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

$S_{p}=0 \\ $

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

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

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

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

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

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

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

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

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

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

$\frac{p}{2}[2 a+(p-1) d]=0 \quad \because S_{n}=\frac{n}{2}{2 a+(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 $n=20 yr$

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

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{matrix} \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{matrix} $

(ii) Total earning during the first year.

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

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

Solution

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

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

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

and

$q$ th term $=p$

$\Rightarrow$

$ a r^{q-1}=p \quad \quad …(ii) $

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

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

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

$ \begin{aligned} & a \Big(\frac{q}p\Big)^{\frac{p-1}{p-q}}=q \Rightarrow a=\frac{q}{\Big(\frac{q}p\Big)^{\frac{p-1}{p-q}}}=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \\ & \therefore \quad(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 \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \Big(\frac{q}p\Big)^{\frac{1}{p-q}}{ }^{p+q-1}=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \frac{q}{\frac{p}{p}} \\ &=q \cdot \Big(\frac{p}q\Big)^{\frac{p-q-1}{p-q}} \frac{p}q^{\frac{-(p+q-1)}{p-q}}=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}-\frac{(p+q-1)}{p-q}} \\ &=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1-p-q+1}{p-q}}=q \cdot \Big(\frac{p}q\Big)^{\frac{-q}{p-q}} \\ & a=q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \end{aligned} $

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

$ \begin{aligned} & =q \cdot \Big(\frac{p}q\Big)^{\frac{p-1}{p-q}} \cdot \Big(\frac{q}p\Big)^{\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 \Big(\frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}}\Big) \end{aligned} $

Solution

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

Let he finished the job in $n$ days.

Then,

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

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

Solution

We know that, sum of interior angles of a polygon of side $n=(2 n-4) \times 90^{\circ}=(n-2) \times 180^{\circ}$ 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 $5,6,7 \ldots$ are $540^{\circ}, 720^{\circ}, 900^{\circ}, \ldots$

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

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

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

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

Solution

Side of equilateral $\triangle A B C=20 cm$. By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of $\triangle A B C$.

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

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

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

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

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

Here,

$ a=60 $

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

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

$ \begin{aligned} \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{aligned} $

Solution

According to the given information, we have following diagram.

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

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

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

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

Here,

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

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

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

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

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

$\Rightarrow \quad 1000=2 a-15 d$

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

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

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

$\therefore a =725 $

Hence, first place team receive ₹ 725 .

Solution

Since, $a_1, a_2, a_3, \ldots, a_{n}$ are in AP.

$ \begin{aligned} & \Rightarrow \quad a_2-a_1=a_3-a_2=\ldots=a_{n}-a_{n-1}=d \quad \text { [common difference] } \\ & \text { If } a_2-a_1=d \text {, then }(\sqrt{a_2})^{2}-(\sqrt{a_1})^{2}=d \\ & \Rightarrow \quad(\sqrt{a_2}-\sqrt{a_1})(\sqrt{a_2}+\sqrt{a_1})=d \\ & \Rightarrow \quad \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{aligned} $

On adding these terms, we get

$ \begin{aligned} & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \\ = & \frac{1}{d}[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+\ldots+\sqrt{a_{n}}-\sqrt{a_{n-1}}] \quad \text { [using above relations] } \\ = & \frac{1}{d}[\sqrt{a_{n}}-\sqrt{a_1}] \end{aligned} $

Again,

$ a_{n}=a_1+(n-1) d \quad[\because T_{n}=a+(n-1) d] $

$\Rightarrow$

$ a_{n}-a_1=(n-1) d $

$ \begin{matrix} \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{matrix} $

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

$ \begin{gathered} \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\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{gathered} $

Solution

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

$ =(3^{3}+5^{3}+7^{3}+\ldots)-(2^{3}+4^{3}+6^{3}+\ldots) $

Let $T_{n}$ be the $n$th term of the series (i),

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

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

(i) Let $S_{n}$ denote the sum of $n$ term of series (i). Then,

$ \begin{aligned} S_{n} & =\Sigma T_{n}=\Sigma(12 n^{2}+6 n) \\ & =12 \Sigma n^{2}+6 \Sigma n+\Sigma n \\ & =12 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{6 n(n+1)}{2}+n \\ & =2 n(n+1)(2 n+1)+3 n(n+1)+n \\ & =2 n(n+1)(2 n+1)+3 n(n+1)+n \\ & =(2 n^{2}+2 n)(2 n+1)+3 n^{2}+3 n+n \\ & =4 n^{3}+2 n^{2}+4 n^{2}+2 n+3 n^{2}+3 n+n \\ & =4 n^{3}+9 n^{2}+6 n \\ 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{aligned} $

(ii) Sum of 10 terms,

Solution

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

$ \begin{aligned} S_{n} & =2 n+3 n^{2} \\ T_{n} & =S_{n}-S_{n-1} \\ & =(2 n+3 n^{2})-[2(n-1)+3(n-1)^{2}] \\ & =(2 n+3 n^{2})-[2 n-2+3(n^{2}+1-2 n)] \\ & =(2 n+3 n^{2})-(2 n-2+3 n^{2}+3-6 n) \\ & =2 n+3 n^{2}-2 n+2-3 n^{2}-3+6 n \\ \therefore \quad & =6 n-1 \\ \therefore \quad \text { rth term } T_{r} & =6 r-1 \end{aligned} $

Solution

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

Then,

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

$ \Rightarrow \quad 2 A=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{aligned} & \text { Then, } \\ & b=a r^{4-1} \\ & \Rightarrow \quad b=a r^{3} \Rightarrow \frac{b}{a}=r^{3} \\ & \therefore \quad r=\frac{b}a^{1 / 3} \end{aligned} $

$ \begin{aligned} & \text { Now, } \\ & G_1=a r=a \frac{b}a^{1 / 3} \\ & \text { and } \\ & G_2=a r^{2}=a \frac{b}a^{2 / 3} \\ & R H S=\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=2 A \\ & =LHS \end{aligned} $

Solution

Since, $\theta_1, \theta_2, \theta_3, \ldots, \theta_{n}$ are in AP.

$ \Rightarrow \quad \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{aligned} \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{aligned} $

Similarly, we can solve other terms which will be $\tan \theta_3-\tan \theta_2, \tan \theta_4-\tan \theta_3, \cdots$

$ \begin{aligned} \therefore \quad 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 } \quad \text { Hence proved. } \end{aligned} $

Solution

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

$ S_{p}=q $

$\Rightarrow$

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

and

$ S_{q}=p $

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

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

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

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

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

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

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

$ \begin{aligned} & 2 a+(p-1) \quad \frac{-2(p+q)}{p q}=\frac{2 q}{p} \\ & \Rightarrow \quad 2 a=\frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q} \\ & \Rightarrow \quad a=\frac{q}{p}+\frac{(p+q)(p-1)}{p q} \\ & \text { Now, } \quad S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d] \\ &= \frac{p+q}{2} \frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q}-\frac{(p+q-1) 2(p+q)}{p q} \\ &=(p+q) \frac{q}{p}+\frac{(p+q)(p-1)-(p+q-1)(p+q)}{p q} \\ &=(p+q) \frac{q}{p}+\frac{(p+q)(p-1-p-q+1)}{p q} \\ &=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}[2 a+(p-q-1) d] \end{aligned} $

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

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{aligned} 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{aligned} $

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

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

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

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

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

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

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{aligned} 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{aligned} $

$ \begin{aligned} R^{d(p q-p r-q+r+q r-p q-r+p+r p-r q-p+q)} & =x^{0} R^{0}=1 \\ & =RHS \end{aligned} $

Solution

(d) Given, $S_{n}=3 n+2 n^{2}$

First term of the AP,

$ \begin{aligned} & \therefore \quad T_1=3 \times 1+2(1)^{2}=3+2=5 \\ & \text { and } \quad T_2=S_2-S_1 \\ & =[3 \times 2+2 \times(2)^{2}]-[3 \times 1+2 \times(1)^{2}] \\ & =14-5=9 \end{aligned} $

$\therefore$ 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 $=a \cdot a r \cdot a r^{2} \cdot a r^{3} \cdot a r^{4}$

$ =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, $\quad 9 \cdot T_9=13 \cdot T_{13}$

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

Solution

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

Then,

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

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

$\Rightarrow 4 y =x+3 z $

and $x, y, z$ are in GP

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

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{aligned} & \therefore \quad S_1=q, S_2=4 q, S_3=9 q \text { and } S_4=16 q \\ & \text { Now, } \quad T_1=q \\ & \therefore \quad T_2=S_2-S_1=4 q-q=3 q \\ & T_3=S_3-S_2=9 q-4 q=5 q \\ & T_4=S_4-S_3=16 q-9 q=7 q \end{aligned} $

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

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

Solution

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

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

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

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

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

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

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

Now,

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

Solution

(b) We know that,

$ AM \geq GM $

$ \begin{matrix} \Rightarrow & \frac{4^{x}+4^{1-x}}{2} \geq \sqrt{4^{x} \cdot 4^{1-x}} \\ \Rightarrow & 4^{x}+4^{1-x} \geq 2 \sqrt{4} \\ \Rightarrow & 4^{x}+4^{1-x} \geq 2 \cdot 2 \\ \Rightarrow & 4^{x}+4^{1-x} \geq 4 \end{matrix} $

Solution

(a) $\quad \sum_{r=1}^{n} \frac{S_{r}}{S_{r}}=\frac{S_1}{S_1}+\frac{S_2}{S_2}+\frac{S_3}{S_3}+\ldots+\frac{S_{n}}{S_{n}}$

Let $T_{n}$ be the $n$th term of the above series.

$ \begin{aligned} \therefore \quad T_{n} & =\frac{S_{n}}{S_{n}}=\frac{\frac{n(n+1)^{2}}{2}}{\frac{n(n+1)}{2}} \\ & =\frac{n(n+1)}{2}=\frac{1}{2}[n^{2}+n] \end{aligned} $

$\therefore$ Sum of the above series $=\Sigma T_{n}=\frac{1}{2}[\Sigma n^{2}+\Sigma n]$

$ \begin{aligned} & =\frac{1}{2} \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}=\frac{1}{2} \cdot \frac{n(n+1)}{2} \frac{(2 n+1)}{3}+1 \\ & =\frac{1}{4} n(n+1) \frac{2 n+1+3}{3}=\frac{1}{4 \times 3} n(n+1)(2 n+4) \\ & =\frac{1}{12} n(n+1)(2 n+4)=\frac{1}{6} n(n+1)(n+2) \end{aligned} $

Solution

(d) Let $S_{n}$ be sum of the series $2+3+6+11+18+\ldots+t_{50}$.

$ \begin{matrix} \therefore & S_{n}=2+3+6+11+18+\ldots+t_{50} \\ \text { and } & S_{n}=0+2+3+6+11+18+\ldots+t_{49}+t_{50} \end{matrix} $

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

$ \begin{aligned} & 0 & =2+1+3+5+7+\cdots+t_{50} \\ \Rightarrow & t_{50} & =2+1+3+5+7+\cdots \text { upto } 49 \text { terms } \\ \therefore & t_{50} & =2+[1+3+5+7+\cdots \text { upto } 49 \text { terms }] \\ & & =2+\frac{49}{2}[2 \times 1+48 \times 2] \\ & & =2+\frac{49}{2} \times[2+96] \\ & & =2+[49+49 \times 48] \\ & & =2+49 \times 49=2+(49)^{2} \end{aligned} $

Solution

(a) Let the length, breadth and height of rectangular solid block is $\frac{a}{r}$, a and ar, respectively.

$ \begin{aligned} & \therefore \quad \text { Volume }=\frac{a}{r} \times a \times a r=216 cm^{3} \\ & \Rightarrow \quad a^{3}=216 \Rightarrow a^{3}=6^{3} \\ & \therefore \quad a=6 \\ & \text { Surface area }=2 \frac{a^{2}}{r}+a^{2} r+a^{2}=252 \\ & \Rightarrow \quad 2 a^{2} \frac{1}{r}+r+1=252 \\ & \Rightarrow \quad 2 \times 36 \frac{1+r^{2}+r}{r}=252 \\ & \Rightarrow \quad \frac{1+r^{2}+r}{r}=\frac{252}{2 \times 36} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 1+r^{2}+r=\frac{126}{36} r \Rightarrow 1+r^{2}+r=\frac{21}{6} r \\ & \Rightarrow \quad 6+6 r^{2}+6 r=21 r \Rightarrow 6 r^{2}-15 r+6=0 \\ & \Rightarrow \quad 2 r^{2}-5 r+2=0 \Rightarrow(2 r-1)(r-2)=0 \\ & \therefore \quad r=\frac{1}{2}, 2 \\ & \text { For } r=\frac{1}{2}: \quad \text { Length }=\frac{a}{r}=\frac{6 \times 2}{1}=12 \\ & \text { Breadth }=a=6 \\ & \text { Height }=a r=6 \times \frac{1}{2}=3 \\ & \text { For } r=2: \quad \text { Length }=\frac{a}{r}=\frac{6}{2}=3 \\ & \text { Breadth }=a=6 \\ & \text { Height }=a r=6 \times 2=12 \end{aligned} $

Solution

Given that, $a, b$ and $c$ are in GP.

$ \begin{aligned} & \text { Then, } \\ & \frac{b}{a}=\frac{c}{b}=r \\ & \Rightarrow \quad b=a r \Rightarrow c=b r \\ & \begin{matrix} \Rightarrow & \frac{a-b}{b-c}=\frac{a-a r}{a r-b r}=\frac{a(1-r)}{r(a-b)}=\frac{a(1-r)}{r(a-a r)} \end{matrix} \\ & =\frac{a(1-r)}{a r(1-r)}=\frac{1}{r} \\ & \therefore \quad \frac{a-b}{b-c}=\frac{1}{r}=\frac{a}{b} \text { or } \frac{b}{c} \end{aligned} $

Solution

Let AP be $a, a+d, a+2 d \cdots a+(n-1) d$

$ \begin{aligned} & \therefore \quad a_1+a_{n}=a+a+(n-1) d \\ & =2 a+(n-1) d \\ & a_2+a_{n-1}=(a+d)+[a+(n-2) d] \\ & =2 a+(n-1) d \\ & a_2+a_{n-1}=a_1+a_{n} \\ & a_3+a_{n-2}=(a+2 d)+[a+(n-3) d] \\ & =2 a+(n-1) d \\ & =a_1+a_{n} \end{aligned} $

an $AP$ is equal to [first term +last term].

Solution

It is given that, $T_3=4$

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

Then,

$ a r^{2}=4 $

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

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

[using Eq. (i)]

Solution

False

Consider an AP $a, a+d, a+2 d, \ldots$

Now,

$ \frac{a_2}{a_1}=\frac{a+d}{a} \neq \frac{a+2 d}{a+d} $

Thus, AP is not a GP.

Solution

True

Consider the progression $a, a+d, a+2 d, \ldots$

and sequence of prime number $2,3,5,7,11, \ldots$

Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern.

Solution

True

Consider an AP $a, a+d, a+2 d, \ldots$

Now,

$ \begin{aligned} a_2+a_4 & =a+d+a+3 d \\ & =2 a+4 d=2 a_3 \\ a_3 & =\frac{a_2+a_4}{2} \\ \frac{a_3+a_5}{2} & =\frac{a+2 d+a+4 d}{2}=\frac{2 a+6 d}{2} \\ & =a+3 d=a_4 \end{aligned} $

$ \Rightarrow \quad a_3=\frac{a_2+a_4}{2} $

Again

Hence, the statement is true.

Solution

False

Let two GP are $a, a r_1, a r_1^{2}, a r_2^{3}, \ldots$ and $b, b r_2, b r_2^{2}, b r_2^{3}, \ldots$

Now, sum of two GP $a+b,(a r_1+b r_2),(a r_1^{2}+b r_2^{2}), \ldots$

Now, $\quad \frac{T_2}{T_1}=\frac{a r_1+b r_2}{a+b}$ and $\frac{T_3}{T_2}=\frac{a r_1^{2}+b r_2^{2}}{a r_1+b r_2}$

$\therefore \quad \frac{T_2}{T_1} \neq \frac{T_3}{T_2}$

Again, difference of two GP is $a-b, a r_1-b r_2, a r_1^{2}-b r_2^{2}, \ldots$

Now,

$ \frac{T_2}{T_1}=\frac{a r_1-b r_2}{a-b} \text { and } \frac{T_3}{T_2}=\frac{a r_1^{2}-b r_2^{2}}{a r_1-b r_2} $

$ \therefore \quad \frac{T_2}{T_1} \neq \frac{T_3}{T_2} $

So, the sum or difference of two GP is not a GP. Hence, the statement is false.

Solution

False

Let

$\begin{aligned} S_n & =a n^2+b n+c \\ S_1 & =a+b+c \\ a_1 & =a+b+c \\ S_2 & =4 a+2 b+c \\ a_2 & =S_2-S_1 \\ & =4 a+4 b+c-(a+b+c)=3 a+b \\ S_3 & =9 a+3 b+c \\ a_3 & =S_3-S_2=5 a+b \end{aligned}$

Now,$\quad a_2-a_1=(3 a+b)-(a+b+c)=2 a-c$

$\quad a_3-a_2=(5 a+b)-(3 a+b)=2 a$

Now,$\quad a_2-a_1 \neq a_3-a_2$

Hence, the statement is false.

Solution

(i) $4,1, \frac{1}{4}, \frac{1}{16}$

$\Rightarrow$

$ \frac{T_2}{T_1}=\frac{1}{4} \Rightarrow \frac{T_3}{T_2}=\frac{1}{4} \Rightarrow \frac{T_4}{T_3}=\frac{1 / 16}{1 / 4}=\frac{1}{4} $

Hence, it is a GP.

(ii) $2,3,5,7$

$ \begin{matrix} \because & T_2-T_1=3-2=1 \\ \because & T_3-T_2=5-3=2 \\ & T_2-T_1 \neq T_3-T_2 \end{matrix} $

Hence, it is not an AP.

Again,

$ \frac{T_2}{T_1}=3 / 2 \Rightarrow \frac{T_3}{T_2}=5 / 3 $

$ \because \quad \frac{T_2}{T_1} \neq \frac{T_3}{T_2} $

It is not a GP.

Hence, it is a sequence.

(iii) $13,8,3,-2,-7$

$\because \quad T_2-T_1=T_3-T_2$

Hence, it is an AP.

$ \begin{aligned} & T_2-T_1=8-13=-5 \\ & T_3-T_2=3-8=-5 \\ & T_2-T_1=T_3-T_2 \end{aligned} $

Solution

(i) $1^{2}+2^{2}+3^{2}+\cdots+n^{2}$

Consider the identity, $(k+1)^{3}-k^{3}=3 k^{2}+3 k+1$

On putting $k=1,2,3, \ldots,(n-1), n$ successively, we get

$ \begin{aligned} & 2^{3}-1^{3}=3 \cdot 1^{2}+3 \cdot 1+1 \\ & 3^{3}-2^{3}=3 \cdot 2^{2}+3 \cdot 2+1 \\ & 4^{3}-3^{3}=3 \cdot 3^{2}+3 \cdot 3+1 \end{aligned} $

$ \begin{aligned} & n^{3}-(n-1)^{3}=3 \cdot(n-1)^{2}+3 \cdot(n-1)+1 \\ & (n+1)^{3}-n^{3}=3 \cdot n^{2}+3 \cdot n+1 \end{aligned} $

Adding columnwise, we get

$n^{3}+3 n^{2}+3 n=3 \quad \sum_{r=1}^{n} r^{2}+3 \frac{n(n+1)}{2}+n \quad \because \sum_{r=1}^{n} r^{2}=\frac{n(n+1)}{2} \\ $

$\Rightarrow \quad 3 \sum_{r=1}^{n} r^{2}=n^{3}+3 n^{2}+3 n-\frac{3 n(n+1)}{2}+n \\ $

$\Rightarrow \quad \sum_{r=1}^{n} r^{2}=\frac{2 n^{3}+3 n^{2}+n}{2}=\frac{n(n+1)(2 n+1)}{2} \\ $

$ \Rightarrow \quad \sum_{r=1}^{n} r^{2}=\frac{n(n+1)(2 n+1)}{6} $

Hence, $\sum_{r=1}^{n} r^{2}=1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$ (ii) $1^{3}+2^{3}+3^{3}+\cdots+n^{3}$

Consider the identity $(k+1)^{4}-k^{4}=4 k^{3}+6 k^{2}+4 k+1$

On putting $k=1,2,3, \cdots(n-1), n$ successively, we get

$ \begin{aligned} & 2^{4}-1^{4}=4 \cdot 1^{3}+6 \cdot 1^{2}+4 \cdot 1+1 \\ & 3^{4}-2^{4}=4 \cdot 2^{3}+6 \cdot 2^{2}+4 \cdot 2+1 \\ & 4^{4}-3^{4}=4 \cdot 3^{3}+6 \cdot 3^{2}+4 \cdot 3+1 \end{aligned} $

$ \begin{aligned} & n^{4}-(n-1)^{4}=4(n-1)^{3}+6(n-1)^{2}+4(n-1)+1 \\ & (n+1)^{4}-n^{4}=4 \cdot n^{3}+6 \cdot n^{2}+4 \cdot n+1 \end{aligned} $

Adding columnwise, we get

$(n+1)^{4}-1^{4}=4 \cdot(1^{3}+2^{3}+\cdots+n^{3})+6(1^{2}+2^{2}+3^{3}+\cdots+n^{2}) \\ $

$+4(1+2+3+\cdots+n)+(1+1+\cdots+1) n \text { terms } \\ $

$\Rightarrow n^{4}+4 n^{3}+6 n^{2}+4 n=4 \sum_{r=1}^{n} r^{3}+6 \sum_{r=1}^{n} r^{2}+4 \sum_{r=1}^{n} r+n \\ $

$\Rightarrow n^{4}+4 n^{3}+6 n^{2}+4 n=4 \sum_{r=1}^{n} r^{3}+6 \frac{n(n+1)(2 n+1)}{6}+4 \frac{n(n+1)}{2}+n \\ $

$\Rightarrow \sum_{r=1}^{n} r^{3}=\frac{n^{2}(n+1)^{2}}{4} \\ $

$\Rightarrow \sum_{r=1}^{n} r^{3}=\frac{n(n+1)}{2}=\sum_{r=1}^{n} r^{2} \\ $

$\text { Hence, } \sum_{r=1}^{n} r^{3}=1^{3}+2^{3}+\cdots+n^{3}=\frac{n(n+1)^{2}}{2}=\sum_{r=1}^{n} r^{2} $

(iii)

$ \begin{aligned} 2+4+6+\cdots+2 n & =2[1+2+3+\cdots+n] \\ & =2 \times \frac{n(n+1)}{2}=n(n+1) \end{aligned} $

(iv) Let

$ S_{n}=1+2+3+\cdots+n $

Clearly, it is an arithmetic series with first term, $a=1$,

common difference,

$d=1$

and

last term $=n$

$ S_{n}=\frac{n}{2}(1+n)=\frac{n(n+1)}{2} $

Hence, $\quad 1+2+3+\cdots+n=\frac{n(n+1)}{2}$.