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,

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

Given: The sequence is in H.P. with \(a_m = n\) and \(a_n = m\) \((m \ne n)\).

In an H.P., reciprocals of terms form an A.P. So, \(\tfrac{1}{a_m} = \tfrac{1}{n}\) and \(\tfrac{1}{a_n} = \tfrac{1}{m}\).

This A.P. leads to common difference \(d = \tfrac{1}{mn}\) and first term \(A = \tfrac{1}{mn}\).

Thus, the reciprocal of the \((m+n)\)th term is:

\(b_{m+n} = \tfrac{m+n}{mn}\)

Hence, the \((m+n)\)th term of H.P. is:

\(a_{m+n} = \dfrac{mn}{m+n}\)

Short Solution:

Solution:

Even numbers from 100 to 998: count $= \frac{998-100}{2}+1=450$.

Exclude evens divisible by $3$ or $5$ using inclusion–exclusion:

• Multiples of $6$ in $[100,998]$: $\lfloor 998/6 \rfloor-\lfloor 99/6 \rfloor = 166-16=150$.
• Multiples of $10$ in $[100,998]$: $\lfloor 998/10 \rfloor-\lfloor 99/10 \rfloor = 99-9=90$.
• Multiples of $30$ in $[100,998]$: $\lfloor 998/30 \rfloor-\lfloor 99/30 \rfloor = 33-3=30$.

Forbidden $=150+90-30=210$ ⇒ Allowed $=450-210={240}$.