Key Concepts and Formulas

• The $n$-th Term of a Geometric Progression (GP): If the first term of a GP is $A$ and the common ratio is $R$, then the $n$-th term is given by $T_n = A \cdot R^{n-1}$.
• Properties of Exponents: For any positive real number $x \neq 1$, if $x^M = x^N$, then $M=N$.

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Step-by-Step Solution

Step 1: Analyze the first Geometric Progression (GP1) and find its $11^{\text{th}}$ term. We are given that the first term of GP1 is $a$ and its third term is $b$. Let the common ratio of GP1 be $r_1$. Using the formula for the $n$-th term, $T_n = A \cdot R^{n-1}$: For the third term ($n=3$): $T_3 = a \cdot r_1^{3-1} = a \cdot r_1^2$ We are given $T_3 = b$, so: $b = a \cdot r_1^2$ From this, we can express $r_1^2$ in terms of $a$ and $b$: $r_1^2 = \frac{b}{a}$ Now, we need to find the $11^{\text{th}}$ term of GP1. Using the formula with $n=11$: $T_{11} = a \cdot r_1^{11-1} = a \cdot r_1^{10}$ We can rewrite $r_1^{10}$ as $(r_1^2)^5$. Substituting the expression for $r_1^2$: $T_{11} = a \cdot \left(\frac{b}{a}\right)^5$ This gives us the $11^{\text{th}}$ term of the first GP in terms of $a$ and $b$.

Step 2: Analyze the second Geometric Progression (GP2) and find its $p^{\text{th}}$ term. We are given that the first term of GP2 is $a$ and its fifth term is $b$. Let the common ratio of GP2 be $r_2$. Using the formula for the $n$-th term: For the fifth term ($n=5$): $T_5' = a \cdot r_2^{5-1} = a \cdot r_2^4$ We are given $T_5' = b$, so: $b = a \cdot r_2^4$ From this, we can express $r_2^4$ in terms of $a$ and $b$: $r_2^4 = \frac{b}{a}$ Since $a$ and $b$ are positive real numbers, $b/a$ is positive. This implies $r_2$ is a real number. We can also write $r_2$ as: $r_2 = \left(\frac{b}{a}\right)^{1/4}$ Now, we need to find the $p^{\text{th}}$ term of GP2. Using the formula with $n=p$: $T_p' = a \cdot r_2^{p-1}$ Substitute the expression for $r_2$: $T_p' = a \cdot \left(\left(\frac{b}{a}\right)^{1/4}\right)^{p-1}$ Using the exponent rule $(x^m)^n = x^{mn}$: $T_p' = a \cdot \left(\frac{b}{a}\right)^{\frac{p-1}{4}}$ This gives us the $p^{\text{th}}$ term of the second GP in terms of $a$, $b$, and $p$.

Step 3: Equate the $11^{\text{th}}$ term of GP1 and the $p^{\text{th}}$ term of GP2 and solve for $p$. The problem states that the $11^{\text{th}}$ term of the first GP is equal to the $p^{\text{th}}$ term of the second GP: $T_{11} = T_p'$ Substituting the expressions we found: $a \cdot \left(\frac{b}{a}\right)^5 = a \cdot \left(\frac{b}{a}\right)^{\frac{p-1}{4}}$ Since $a$ is a positive real number, $a \neq 0$. We can divide both sides by $a$: $\left(\frac{b}{a}\right)^5 = \left(\frac{b}{a}\right)^{\frac{p-1}{4}}$ We are given that $a$ and $b$ are distinct positive real numbers. This implies that $\frac{b}{a}$ is a positive real number and $\frac{b}{a} \neq 1$. For an equation of the form $x^M = x^N$, where $x > 0$ and $x \neq 1$, we can equate the exponents: $5 = \frac{p-1}{4}$ Now, we solve this linear equation for $p$. Multiply both sides by 4: $5 \times 4 = p-1$ $20 = p-1$ Add 1 to both sides: $20 + 1 = p$ $p = 21$

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Common Mistakes & Tips

• Distinguishing Common Ratios: Always use different variables for the common ratios of different GPs to avoid confusion. In this case, we used $r_1$ and $r_2$.
• Exponent Manipulation: Be careful when simplifying expressions involving exponents, especially fractional exponents. The rule $(x^m)^n = x^{mn}$ is frequently used.
• Justifying Exponent Equality: Remember that equating exponents in $x^M = x^N$ is valid only when the base $x$ is positive and not equal to 1. The condition that $a$ and $b$ are distinct positive real numbers ensures that $b/a > 0$ and $b/a \neq 1$.

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Summary

The problem involves two geometric progressions with the same first term $a$. For the first GP, the third term is $b$, allowing us to find its common ratio's square as $b/a$. We used this to express the $11^{\text{th}}$ term as $a(b/a)^5$. For the second GP, the fifth term is $b$, leading to its common ratio raised to the fourth power being $b/a$. We used this to express the $p^{\text{th}}$ term as $a(b/a)^{\frac{p-1}{4}}$. Equating these two terms and simplifying, we arrived at an equation where the exponents must be equal. Solving the resulting linear equation $5 = \frac{p-1}{4}$ yields $p=21$.

The final answer is \boxed{21} which corresponds to option (C).