Question

If $\sqrt{\frac{3+\sqrt{5}}{2}},sgn(x^2-3x+4)$ and $2\{\left[\frac{2}{3+x^2}\right]+sin\frac{\pi }{10}\},x\underset{–}{\in }R$ are the first three consecutive terms of GP and sum of its infinite terms is $\frac{\lambda }{(\sqrt{5}-1{)}^3}$, then find the value of $\lambda$.
[Note:$\left[k\right],\left\{k\right\}$ and $sgn\left(k\right)$ denote greatest integer function, fractional part function and signum function of $k$ respectively.]

Answer 1

Let common ratio of the GP be $r$ and first term be $a$. Since, sum of infinite terms exists, $|r|<1,r\ne 0$.

$\therefore a=\sqrt{\frac{3+\sqrt{5}}{2}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\sqrt{\frac{5+1+2\sqrt{5}}{4}}=\sqrt{\frac{(1+\sqrt{5}{)}^2}{4}}=\frac{\sqrt{5}+1}{2}$

Also, $x^2-3x+4>0\forall x\in R$ as its roots are imaginary.

Hence, the signum function can only take 1 as value.

$r=\frac{1}{a}$

Now, $\frac{2}{x^2+3}\le \frac{2}{3}⟹\left[\frac{2}{x^2+3}\right]=\left[\frac{2}{3}\right]=0$

$\therefore \{\left[\frac{2}{x^2+3}\right]+sin\frac{\pi }{10}\}=\{+sin\frac{\pi }{10}\}=sin\frac{\pi }{10}=\frac{\sqrt{5}-1}{4}$ (as $0<sin\frac{\pi }{10}<1$)

$\therefore a_3=\frac{\sqrt{5}-1}{2}$ (consider $sin5\times \frac{\pi }{10}$ as $sin\frac{\pi }{2}$ and solve)

This again shows that $r=\frac{1}{a}$.

Sum of GP = $\frac{a}{1-r}=\frac{a}{1-\frac{1}{a}}=\frac{a^2}{a-1}$

$=\frac{\frac{(1+\sqrt{5}{)}^2}{4}}{\frac{\sqrt{5}-1}{2}}\times \frac{(\sqrt{5}-1{)}^2}{(\sqrt{5}-1{)}^2}$

$=\frac{(5-1{)}^2}{2(\sqrt{5}-1{)}^3}$

$=\frac{8}{(\sqrt{5}-1{)}^3}$

$⟹\lambda =8$