Phrases Previous Year Questions (PYQs) – Progressions – Page 1 of 3

Given:

$$x=1+2^{1/6}+4^{1/6}+8^{1/6}+{16}^{1/6}+{32}^{1/6}$$

Step 1: Write in powers of $a=2^{1/6}$

$$x=1+a+a^2+a^3+a^4+a^5=1+\frac{a(a^5-1)}{a-1}$$

Step 2: Use identity $a^6=2\Rightarrow a^5=\frac{2}{a}$

$$x=1+\frac{2-a}{a-1}=\frac{1}{a-1}\Rightarrow 1+\frac{1}{x}=a\Rightarrow {(1+\frac{1}{x})}^{24}=a^{24}$$

Step 3: Final calculation

$$a=2^{1/6}\Rightarrow a^{24}=(2^{1/6}{)}^{24}=2^4=\boxed{{\displaystyle 16}}$$

✅ Final Answer: $\boxed{{\displaystyle 16}}$

Step 1: Recognize the series

The exponent is an infinite geometric series: $$1+|\sin x|+|\sin x{|}^2+|\sin x{|}^3+\cdots$$

This is a geometric series with first term $a=1$, common ratio $r=|\sin x|\in [0,1]$, so: $$\text{Sum}=\frac{1}{1-|\sin x|}$$

Step 2: Rewrite the equation

$$5^{\frac{1}{1-|\sin x|}}=25=5^2$$

Equating exponents: $$\frac{1}{1-|\sin x|}=2\Rightarrow 1-|\sin x|=\frac{1}{2}\Rightarrow |\sin x|=\frac{1}{2}$$

Step 3: Solve for $x\in (-\pi ,\pi )$

We want all $x\in (-\pi ,\pi )$ such that $|\sin x|=\frac{1}{2}$

So $\sin x=\pm \frac{1}{2}$. Within $(-\pi ,\pi )$, the values of $x$ satisfying this are:

• $x=\frac{\pi }{6}$
• $x=\frac{5\pi }{6}$
• $x=-\frac{\pi }{6}$
• $x=-\frac{5\pi }{6}$

✅ Final Answer: $\boxed{{\displaystyle 4}}$ solutions