Sequences and Series

Let $\alpha, \beta, \gamma$ be the roots of the equation

$ \begin{gathered} 3 x^3-26 x^2+52 x-24=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \alpha+\beta+\gamma=\frac{26}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{52}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \alpha \beta \gamma=8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered} $

$\alpha, \beta, \gamma$ are in GP $\Rightarrow \beta^2=\alpha \gamma$

From Eq. (iii), $\beta^3=8 \Rightarrow \beta=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

∴ From Eqs. (i) and (iv),

$ \begin{aligned} & \alpha+\gamma=\frac{20}{3} \quad \alpha \gamma=4 \Rightarrow \gamma=\frac{4}{\alpha} \\ & \therefore \alpha+\frac{4}{\alpha}=\frac{20}{3} \Rightarrow 3 \alpha^2-20 \alpha+12=0 \\ & \Rightarrow(\alpha-6)(3 \alpha-2)=0 \quad \alpha=6, \alpha=\frac{2}{3} \end{aligned} $

For $\alpha=6, \gamma=\frac{2}{3}$ and for $\alpha=\frac{2}{3}, \gamma=6$

$ \begin{aligned} & \therefore \quad \alpha=6, \beta=2 \text { and } \gamma=\frac{2}{3} \\ & \text { or } \quad \alpha=\frac{2}{3}, \beta=2 \text { and } \gamma=6 \end{aligned} $

$ \text { } \begin{aligned}Now, 3 \alpha+2 \beta+\gamma & =3 \times \frac{2}{3}+2 \times 2+6 \\ & =2+4+6=12 \end{aligned} $

Let the coordinate $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$.

According to question, $x$ - coordinate as well as $y$ coordinates are in GP.

Let $x_1=a, x_2=a r$ and $x_3=a r^2,y_1=b, y_2=b r$ and $y_3=b r^2$

Now, $A(a, b), B(a r, b r), C\left(a r^2, b r^2\right)$.

Slope of $A B=\frac{b(1-r)}{a(1-r)}=\frac{b}{a}$ and

Slope of $\mathrm{BC}=\frac{b r(1-r)}{\operatorname{ar}(1-r)}=\frac{b}{a}$

As slope of $A B=$ Slope of $B C$

$\therefore$ A, B and C are collinear.

$\therefore$ A, B and C lie on a straight line.

Given, $a_0=1, a_{n+1}=3 n^2+n+a_n$ .... (i)

From Eq.(i), we get

$\begin{array}{ll} & a_1=3(0)^2+0+a_0=1 \\ \Rightarrow \quad & a_1=1 \\ & a_2=3(1)^2+1+a_1=3+1+1=5 \\ \Rightarrow \quad & a_2=5 \end{array}$

$\begin{aligned} & a_3=3(2)^2+2+a_2=12+2+5=19 \\ & \Rightarrow \quad a_3=19 \\ \end{aligned}$

Now, check the option for $n=3$ which expression satisfy by $a_3=19$

Option (a), $a_n=n^3+n^2+1$

Put $n=3$, then

$\begin{array}{rlrl} a_3 & =(3)^3+(3)^2+1 \\ & =27+9+1=37 \\ \because \quad a_3 & \neq 37 \\ \therefore \quad a_n & \neq n^3+n^2+1 \\ & \text { Option (b), } a_n =n^3-n^2+1 \end{array}$

Option (b), $a_n=n^3-n^2+1$

Put $n=3$, then

$\begin{aligned} a_3 & =(3)^3-(3)^2+1 \\ & =27-9+1=19 \\ \therefore \quad &a_n =n^3-n^2+1 \end{aligned}$

$\begin{aligned} & \text { Given, } x^2-\sqrt{3} x+1=0 \\ & \Rightarrow \quad x=\frac{\sqrt{3} \pm \sqrt{3-4}}{2} \\ & x=\frac{\sqrt{3}}{2} \pm \frac{1}{2} i=\cos \frac{\pi}{6} \pm i \sin \frac{\pi}{6} \\ & \Rightarrow \quad x^{2 n}=\cos \frac{2 n \pi}{6} \pm i \sin \frac{2 n \pi}{6} \\ & \Rightarrow \quad x^{2 n}=\cos \frac{n \pi}{3} \pm i \sin \frac{n \pi}{3} \\ & \text { and } x^{-2 n}=\cos \frac{n \pi}{3} \pm i \sin \frac{n \pi}{3} \\ & \operatorname{Now},\left(x^n-\frac{1}{x^n}\right)^2=\left(x^{2 n}+\frac{1}{x^{2 n}}-2\right) \\ & =-2+2 \cos \frac{n \pi}{3} \\ & \Rightarrow \sum_\limits{n=1}^{24}\left(x^n-\frac{1}{x^n}\right)^2=\sum_\limits{n=1}^{24}\left(-2+2 \cos \frac{n \pi}{3}\right) \\ & =-48+2\left[\cos \frac{\pi}{3}+\cos \frac{2 \pi}{3}+\ldots+\cos \frac{24 \pi}{3}\right] \\ & =-48+2\left[\frac{\cos \left(\frac{2 \pi}{3}+\frac{23 \pi}{6}\right) \cdot \sin \frac{24 \pi}{6}}{\sin \frac{\pi}{6}}\right] \\ & =-48+2 \cdot 0=-48 \\ & \end{aligned}$

$\because$ $p,q,r,s$ are in AP.

Let $p = a,q = a + d,r = a + 2d,s = a + 3d$

Given, $p,q$ are roots of

${x^2} - 2x + A = 0$

Then, $p + q = 2$ and $pq = A$

$ \Rightarrow 2a + d = 2$ .... (i)

Also, r and s are roots of ${x^2} - 18x + B = 0$

Then, $r + s = 18$ and $rs = B$

$ \Rightarrow 2a + 5d = 18$ ..... (ii)

Solving Eqs. (i) and (ii), we get

$a = - 1,d = 4$

Now, $pq = A$ (product of roots)

or $A = a(a + d) = ( - 1)( - 1 + 4)$

$A = - 3$

and $rs = B$

or $B = (a + 2d)(a + 3d)$

$ = ( - 1 + 8)( - 1 + 12) = 77$

$\therefore$ $A = - 3,B = 77$

$f(x)=x^3+a x^2+b x+c$

Let roots of $f(x)$ are $\alpha, \beta, \gamma$ and since roots are in AP.

$\therefore \quad 2 \beta=\alpha+\gamma$

Now, Sum of roots

$\begin{aligned} \alpha+\beta+\gamma & =-a \\ \text { or }2 \beta+\beta & =-a \text { or } \beta=-\frac{a}{3} \end{aligned}$

It is given that, roots are integers.

$\therefore \beta$ is an integer when $a$ is multiple of 3 .

Option (a), (c) and (d) are multiple of 3.

$\Rightarrow \quad a \neq 1214$

Option (b) is correct.