Key Concepts and Formulas

1. Sum of an infinite G.P.: $S_\infty = \frac{a}{1-r}$, where $|r| < 1$.
2. Sum of the first $n$ terms of a G.P.: $S_n = \frac{a(1-r^n)}{1-r}$.
3. Sum of the first $N$ terms of an A.P.: $S_N = \frac{N}{2}[2A_1 + (N-1)D]$.
4. $N^{th}$ term of an A.P.: $A_N = A_1 + (N-1)D$.

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Step-by-Step Solution

Part 1: Analyzing the Geometric Progression (G.P.)

Step 1: Set up equations from the given G.P. information. We are given that the first term of the G.P. is $a$ and the common ratio is $r$. The sum of the infinite G.P. is 5. Using the formula for the sum of an infinite G.P., we get: $\frac{a}{1-r} = 5 \quad (*)$ This implies that $|r| < 1$ for the sum to converge.

The sum of its first five terms is $\frac{98}{25}$. Using the formula for the sum of the first $n$ terms of a G.P.: $S_5 = \frac{a(1-r^5)}{1-r} = \frac{98}{25} \quad (**)$

Step 2: Combine the two G.P. equations to find $r^5$. We can rewrite the equation $(**)$ by separating the term $\frac{a}{1-r}$: $S_5 = \left(\frac{a}{1-r}\right)(1-r^5) = \frac{98}{25}$ Substitute the value of $\frac{a}{1-r}$ from equation $(*)$: $5(1-r^5) = \frac{98}{25}$ Now, solve for $1-r^5$: $1-r^5 = \frac{98}{25 \times 5} = \frac{98}{125}$ Solve for $r^5$: $r^5 = 1 - \frac{98}{125} = \frac{125 - 98}{125} = \frac{27}{125}$ Thus, we have found that $r^5 = \frac{27}{125}$. We don't need to find the individual values of $a$ and $r$.

Part 2: Analyzing the Arithmetic Progression (A.P.)

Step 3: Identify the first term and common difference of the A.P. The A.P. has the following properties:

• First term, $A_1 = 10ar$.
• Common difference, $D = 10ar^2$.
• We need to find the sum of the first 21 terms, so $N = 21$.

Step 4: Calculate the sum of the first 21 terms of the A.P. Using the formula $S_N = \frac{N}{2}[2A_1 + (N-1)D]$: $S_{21} = \frac{21}{2}[2(10ar) + (21-1)(10ar^2)]$ $S_{21} = \frac{21}{2}[20ar + 20(10ar^2)]$ $S_{21} = \frac{21}{2}[20ar + 200ar^2]$ Factor out 20 from the terms inside the bracket: $S_{21} = \frac{21}{2} \times 20[ar + 10ar^2]$ $S_{21} = 21 \times 10[ar + 10ar^2]$ $S_{21} = 21[10ar + 100ar^2]$

Step 5: Calculate the $11^{th}$ term of the A.P. The question asks for the sum in terms of $a_{11}$, which represents the $11^{th}$ term of this A.P. Using the formula $A_N = A_1 + (N-1)D$: $a_{11} = A_1 + (11-1)D$ Substitute the values of $A_1$ and $D$: $a_{11} = 10ar + (10)(10ar^2)$ $a_{11} = 10ar + 100ar^2$

Step 6: Relate the sum of the first 21 terms to the $11^{th}$ term. From Step 4, we have $S_{21} = 21[10ar + 100ar^2]$. From Step 5, we have $a_{11} = 10ar + 100ar^2$. By comparing these two expressions, we can see that: $S_{21} = 21 \times (10ar + 100ar^2)$ $S_{21} = 21 \times a_{11}$ The value of $r^5$ was not explicitly needed to solve this part of the problem, which is a common feature in such problems where algebraic relationships are key.

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Common Mistakes and Tips

• Distinguish between G.P. and A.P. terms: Be careful not to confuse the first term $a$ of the G.P. with the terms of the A.P. ($a_n$).
• Use substitutions strategically: In G.P. problems involving infinite sums and finite sums, substituting the infinite sum into the finite sum formula often simplifies the problem significantly.
• Focus on the target expression: The options are given in terms of specific A.P. terms ($a_{11}$ or $a_{16}$). Calculate these terms once you have the A.P.'s first term and common difference.

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Summary

The problem involves two parts: first, analyzing a Geometric Progression to find information about its common ratio, and second, using this information to analyze an Arithmetic Progression. We used the given sum of the infinite G.P. and the sum of its first five terms to deduce the value of $r^5$. Then, we defined the first term and common difference of the A.P. in terms of $a$ and $r$. By calculating the sum of the first 21 terms of the A.P. and its $11^{th}$ term, we found that the sum is exactly 21 times the $11^{th}$ term.

The final answer is $\boxed{\text{21 } a_{11}}$.