Key Concepts and Formulas

• Arithmetic Progression (AP): The $n^{th}$ term is given by $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
• Geometric Progression (GP): The $n^{th}$ term is given by $T_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.
• Relationship between terms in an AP: $T_m = T_k + (m-k)d$ for any terms $T_m$ and $T_k$.

Step-by-Step Solution

Step 1: Define the terms of the GP and establish relationships. Let the first three terms of the GP be $2, p, q$. Let the common ratio of the GP be $r$. Then, $p = 2r$ and $q = 2r^2$. We are given that $q \neq 2$.

Step 2: Relate the GP terms to the AP terms. Let the AP have a first term $A$ and a common difference $D$. We are given that: The $7^{th}$ term of the AP is $2$: $T_7 = A + (7-1)D = A + 6D = 2$. The $8^{th}$ term of the AP is $p$: $T_8 = A + (8-1)D = A + 7D = p = 2r$. The $13^{th}$ term of the AP is $q$: $T_{13} = A + (13-1)D = A + 12D = q = 2r^2$.

Step 3: Find the common difference of the AP in terms of $r$. We can find the common difference $D$ of the AP using the relationship between consecutive terms: $D = T_8 - T_7 = (A + 7D) - (A + 6D) = 2r - 2$. So, $D = 2(r-1)$.

Step 4: Express the $13^{th}$ term of the AP using $T_7$ and $D$. We know that $T_{13} = T_7 + (13-7)D = T_7 + 6D$. Substitute the given values: $2r^2 = 2 + 6(2(r-1))$ $2r^2 = 2 + 12(r-1)$ $2r^2 = 2 + 12r - 12$ $2r^2 = 12r - 10$

Step 5: Solve the quadratic equation for the common ratio $r$. Divide the equation by 2: $r^2 = 6r - 5$ Rearrange into a standard quadratic form: $r^2 - 6r + 5 = 0$ Factor the quadratic equation: $(r-1)(r-5) = 0$ The possible values for $r$ are $r=1$ or $r=5$.

Step 6: Use the condition $q \neq 2$ to select the correct value of $r$. We are given that $q \neq 2$. Since $q = 2r^2$: If $r=1$, then $q = 2(1)^2 = 2$. This contradicts the given condition. Therefore, we must have $r=5$.

Step 7: Calculate the common difference $D$ of the AP. Using $D = 2(r-1)$ and $r=5$: $D = 2(5-1) = 2(4) = 8$.

Step 8: Calculate the first term $A$ of the AP. We know $T_7 = A + 6D = 2$. Substitute $D=8$: $A + 6(8) = 2$ $A + 48 = 2$ $A = 2 - 48 = -46$.

Step 9: Calculate the $5^{th}$ term of the GP. The $5^{th}$ term of the GP is given by $2r^{5-1} = 2r^4$. Substitute $r=5$: $5^{th}$ term of GP $= 2(5^4) = 2(625) = 1250$.

Step 10: Find the value of $n$ such that the $n^{th}$ term of the AP is equal to the $5^{th}$ term of the GP. We need to find $n$ such that $T_n = A + (n-1)D = 1250$. Substitute $A=-46$ and $D=8$: $-46 + (n-1)8 = 1250$ $(n-1)8 = 1250 + 46$ $(n-1)8 = 1296$ $n-1 = \frac{1296}{8}$ $n-1 = 162$ $n = 162 + 1 = 163$.

Common Mistakes & Tips

• Forgetting the condition $q \neq 2$: This condition is crucial for eliminating the extraneous solution for $r$. Always check all given constraints.
• Algebraic errors in solving the quadratic equation: Double-check your factorization or the quadratic formula application.
• Confusing the first term of the GP with the first term of the AP: Clearly distinguish between the first term of the GP (which is 2) and the first term of the AP (which we calculated as $A$).

Summary The problem involves linking an arithmetic progression and a geometric progression. We used the given terms to establish relationships between their respective first terms and common ratios/differences. By solving the resulting equations, we found the common ratio of the GP and the common difference and first term of the AP. Finally, we calculated the $5^{th}$ term of the GP and determined which term in the AP it corresponds to by solving for $n$.

The final answer is $\boxed{163}$.