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 $a^6 = 2 \Rightarrow a^5 = \frac{2}{a}$

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

Step 3: Final calculation

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

✅ Final Answer: $\boxed{16}$

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: $\boxed{4}$ solutions

Problem:

If one Arithmetic Mean (AM) $a$ and two Geometric Means $p$ and $q$ are inserted between any two positive numbers, find the value of: $$p^3 + q^3$$

Given:

• Let two positive numbers be $A$ and $B$.
• One AM: $a = \frac{A + B}{2}$
• Two GMs inserted: so the four terms in G.P. are: $$A, \ p = \sqrt[3]{A^2B}, \ q = \sqrt[3]{AB^2}, \ B$$

Now calculate:

$$pq = \sqrt[3]{A^2B} \cdot \sqrt[3]{AB^2} = \sqrt[3]{A^3B^3} = AB$$
$$p^3 = A^2B, \quad q^3 = AB^2$$
$$p^3 + q^3 = A^2B + AB^2 = AB(A + B)$$

Also,

$$2apq = 2 \cdot \frac{A + B}{2} \cdot AB = AB(A + B)$$

✅ Therefore,

$\boxed{p^3 + q^3 = 2apq}$

Short Solution:

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

Given: Harmonic mean is 4: $$\frac{2ab}{a + b} = 4 \quad \text{(1)}$$ Arithmetic mean $A = \frac{a + b}{2}$,
Geometric mean $G = \sqrt{ab}$

Given: $$2A + G^2 = 27$$ $$2 \cdot \frac{a + b}{2} + ab = 27 \Rightarrow a + b + ab = 27 \quad \text{(2)}$$ From (1): Multiply both sides by $a + b$: $$2ab = 4(a + b) \Rightarrow ab = 2(a + b) \quad \text{(3)}$$ Substitute (3) into (2): $$a + b + 2(a + b) = 27 \Rightarrow 3(a + b) = 27 \Rightarrow a + b = 9$$ Then from (3): $$ab = 2 \cdot 9 = 18$$ Now solve: $$x^2 - (a + b)x + ab = 0 \Rightarrow x^2 - 9x + 18 = 0$$ $$\Rightarrow x = 3, 6$$
Final Answer:
$$\boxed{3 \text{ and } 6}$$

Let the three terms in A.P. be a – d, a, a + d.

given that a – d + a + a + d = 21  

a = 7

then the three term in A.P. are 7 – d, 7, 7 + d

According to given condition 9 – d, 9, 21 + d are in G.P.

(9)² = (9 – d) (21 + d)

81 = 189 + 9d – 21d – d²

81 = 189 – 12d – d²

d² + 12d – 108 = 0

d(d + 18) – 6 (d + 18) = 0

(d – 6) (d + 18) = 0

We get, d = 6, –18

Putting d = 6 in the term 7 – d, 7, 7 + d we get 1, 7, 13.

? GP and Function Relation

Given: $f(x) = x^3 + 3x - 9$

The sum of infinite GP = max value of $f(x)$ on [−2, 3]

The difference between first two terms = $f'(0)$

---

Step 1: $f(x)$ is increasing ⇒ Max at $x = 3$

$f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27$

Step 2: $f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3$

⇒ $a(1 - r) = 3$

Step 3: Solve:
$a = 27(1 - r)$
$\Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3}$

✅ Final Answer: $r = \frac{2}{3}$