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 $(d)=T_2-T_1=9-5=4$

• Option (a) 3: The common difference is calculated as ( T_2 - T_1 ). Given ( S_n = 3n + 2n^2 ), the first term ( T_1 = 5 ) and the second term ( T_2 = 9 ). Therefore, the common difference ( d = 9 - 5 = 4 ), not 3.

• Option (b) 2: The common difference is calculated as ( T_2 - T_1 ). Given ( S_n = 3n + 2n^2 ), the first term ( T_1 = 5 ) and the second term ( T_2 = 9 ). Therefore, the common difference ( d = 9 - 5 = 4 ), not 2.

• Option (c) 6: The common difference is calculated as ( T_2 - T_1 ). Given ( S_n = 3n + 2n^2 ), the first term ( T_1 = 5 ) and the second term ( T_2 = 9 ). Therefore, the common difference ( d = 9 - 5 = 4 ), not 6.

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} $

• Option (a) $4^{3}$ is incorrect because the product of the first 5 terms of the GP is calculated as $(a r^{2})^{5}$, which equals $4^{5}$, not $4^{3}$.

• Option (b) $4^{4}$ is incorrect because the product of the first 5 terms of the GP is calculated as $(a r^{2})^{5}$, which equals $4^{5}$, not $4^{4}$.

• Option (d) None of these is incorrect because the correct answer is indeed one of the given options, specifically $4^{5}$.

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} $

• Option (b) 22: This option is incorrect because, based on the derived equation ( a = -21d ), the 22nd term ( T_{22} ) is calculated as ( a + 21d ). Substituting ( a = -21d ) into this equation results in ( T_{22} = -21d + 21d = 0 ), not 22.

• Option (c) 198: This option is incorrect because the calculation of the 22nd term ( T_{22} ) using the derived equation ( a = -21d ) results in ( T_{22} = 0 ). There is no scenario in the given arithmetic progression where the 22nd term equals 198.

• Option (d) 220: This option is incorrect because, similar to the previous options, the calculation of the 22nd term ( T_{22} ) using ( a = -21d ) results in ( T_{22} = 0 ). The 22nd term cannot be 220 based on the given conditions and derived equations.

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 $

• Option (a) 3: If the common ratio of the GP were 3, then ( y = 3x ) and ( z = 9x ). Substituting these into the AP condition ( 4y = x + 3z ) would yield ( 4(3x) = x + 3(9x) ), which simplifies to ( 12x = 28x ), a contradiction.

• Option (c) 2: If the common ratio of the GP were 2, then ( y = 2x ) and ( z = 4x ). Substituting these into the AP condition ( 4y = x + 3z ) would yield ( 4(2x) = x + 3(4x) ), which simplifies to ( 8x = 13x ), a contradiction.

• Option (d) (\frac{1}{2}): If the common ratio of the GP were (\frac{1}{2}), then ( y = \frac{x}{2} ) and ( z = \frac{x}{4} ). Substituting these into the AP condition ( 4y = x + 3z ) would yield ( 4(\frac{x}{2}) = x + 3(\frac{x}{4}) ), which simplifies to ( 2x = \frac{5x}{4} ), a contradiction.

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} $

• Option (a) $\frac{q^{3}}{2}$: This option is incorrect because the formula for the sum of the first $q$ terms of an arithmetic progression (AP) is given by $S_q = \frac{q}{2} [2a + (q-1)d]$. When we substitute $a = q$ and $d = 2q$ into this formula, we get $S_q = q^3$, not $\frac{q^3}{2}$.

• Option (b) $m n q$: This option is incorrect because the sum of the first $q$ terms of the AP does not depend on the variables $m$ and $n$. The correct sum $S_q$ is derived solely from the given relationship $S_n = q n^2$ and the properties of the AP, leading to $S_q = q^3$.

• Option (d) $(m+n) q^{2}$: This option is incorrect because, similar to option (b), the sum of the first $q$ terms of the AP does not involve the variables $m$ and $n$. The correct sum $S_q$ is determined by the given relationship $S_n = q n^2$ and the properties of the AP, resulting in $S_q = q^3$.

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} $

• Option (a) 4: This option is incorrect because the ratio ( \frac{S_{3n}}{S_n} ) does not simplify to 4. The correct ratio, as derived, is 6, not 4.

• Option (c) 8: This option is incorrect because the ratio ( \frac{S_{3n}}{S_n} ) does not simplify to 8. The correct ratio, as derived, is 6, not 8.

• Option (d) 10: This option is incorrect because the ratio ( \frac{S_{3n}}{S_n} ) does not simplify to 10. The correct ratio, as derived, is 6, not 10.

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} $

• Option (a) 2: This option is incorrect because the minimum value of (4^x + 4^{1-x}) is 4, as shown by the application of the AM-GM inequality. The expression (4^x + 4^{1-x}) cannot be less than 4.

• Option (c) 1: This option is incorrect because the minimum value of (4^x + 4^{1-x}) is 4, not 1. The AM-GM inequality demonstrates that the expression is always at least 4.

• Option (d) 0: This option is incorrect because the minimum value of (4^x + 4^{1-x}) is 4, not 0. The AM-GM inequality shows that the expression cannot be zero or any value less than 4.

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} $

• Option (b) $\frac{n(n+1)}{2}$: This option represents the sum of the first ( n ) natural numbers, denoted as ( s_n ). However, the given problem involves the sum of the cubes of the first ( n ) natural numbers, which is a different series. Therefore, this option does not correctly represent the sum of the given series.

• Option (c) $\frac{n^{2}+3 n+2}{2}$: This option does not correspond to any known formula related to the sum of cubes or the sum of natural numbers. It appears to be an arbitrary expression that does not fit the context of the problem, which involves the sum of the cubes of the first ( n ) natural numbers.

• Option (d) None of these: This option is incorrect because the correct answer is provided in option (a). The correct sum of the series ( \sum_{r=1}^{n} \frac{S_{r}}{S_4} ) is indeed ( \frac{n(n+1)(n+2)}{6} ), as derived in the solution.

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} $

• Option (a) $49^{2}-1$: This option is incorrect because the calculation of the 50th term, $t_{50}$, involves adding 2 to the square of 49, not subtracting 1 from it. The correct formula derived from the series pattern is $t_{50} = 49^2 + 2$.

• Option (b) $49^{2}$: This option is incorrect because it does not account for the additional 2 that is added to the square of 49 in the correct formula. The correct term is $t_{50} = 49^2 + 2$, not just $49^2$.

• Option (c) $50^{2}+1$: This option is incorrect because it uses the wrong base for the square. The correct term is based on 49, not 50. The correct formula is $t_{50} = 49^2 + 2$, not $50^2 + 1$.

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} $

• Option (b) $6 cm$: This option is incorrect because $6 cm$ is the length of the middle edge (breadth) of the rectangular solid block, not the longest edge. The longest edge must be either $12 cm$ or $3 cm$ depending on the value of $r$.

• Option (c) $18 cm$: This option is incorrect because the calculations show that the possible lengths of the edges are $12 cm$, $6 cm$, and $3 cm$. There is no edge with a length of $18 cm$.

• Option (d) $3 cm$: This option is incorrect because $3 cm$ is the length of the shortest edge of the rectangular solid block, not the longest edge. The longest edge must be either $12 cm$ or $3 cm$ depending on the value of $r$.

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