Key Concepts and Formulas

• Sum of a Geometric Progression (G.P.): For a G.P. with first term $a$, common ratio $r$, and $n$ terms, the sum $S_n$ is given by $S_n = \frac{a(1-r^n)}{1-r}$ (for $r \neq 1$). If $r=1$, $S_n = na$.
• Identifying Sub-G.P.s: Terms at specific positions in a G.P. (e.g., odd-indexed terms) can themselves form a new G.P. The first term, common ratio, and number of terms of this new G.P. need to be determined.

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Step-by-Step Solution

Step 1: Define the G.P. and its Sum Let the given Geometric Progression have the first term $a$ and the common ratio $r$. The problem states there are 64 terms ($n=64$). The terms are $a, ar, ar^2, \ldots, ar^{63}$. The sum of all terms ($S_{all}$) is given by the G.P. sum formula: $S_{all} = \frac{a(1-r^{64})}{1-r}$

• Reasoning: This step establishes the mathematical representation for the total sum of the G.P. as per the problem statement, using the standard formula. We assume $r \neq 1$, as we will see later that the $r=1$ case does not satisfy the condition.

Step 2: Identify and Define the G.P. of Odd Terms The "odd terms" refer to the terms at the 1st, 3rd, 5th, ..., 63rd positions. These terms are $a, ar^2, ar^4, \ldots, ar^{62}$. This sequence itself forms a G.P. with:

• First term ($a'$): $a$
• Common ratio ($r'$): $\frac{ar^2}{a} = r^2$
• Number of terms ($n'$): The original G.P. has 64 terms. The odd-positioned terms are $T_1, T_3, \ldots, T_{63}$. The indices are $1, 3, 5, \ldots, 63$. This is an arithmetic progression of indices with first term 1, common difference 2. If the last term is $1 + (n'-1)2 = 63$, then $(n'-1)2 = 62$, so $n'-1 = 31$, which gives $n'=32$. Thus, there are 32 odd terms. The sum of these odd terms ($S_{odd}$) is: $S_{odd} = \frac{a'(1-(r')^{n'})}{1-r'} = \frac{a(1-(r^2)^{32})}{1-r^2} = \frac{a(1-r^{64})}{1-r^2}$
• Reasoning: This step identifies that the specified subset of terms (odd-positioned) also forms a G.P. By correctly identifying its first term, common ratio, and number of terms, we can apply the G.P. sum formula to find its sum.

Step 3: Set Up the Equation Based on the Problem Statement The problem states that "the sum of all the terms is 7 times the sum of the odd terms of the G.P." This can be written as: $S_{all} = 7 \times S_{odd}$ Substituting the expressions from Step 1 and Step 2: $\frac{a(1-r^{64})}{1-r} = 7 \times \frac{a(1-r^{64})}{1-r^2}$

• Reasoning: This step translates the word problem's condition into a precise mathematical equation by equating the sum of all terms with seven times the sum of the odd terms.

Step 4: Simplify the Equation We need to solve for $r$. We can make the following assumptions for simplification:

• $a \neq 0$ (otherwise, all terms are 0, and the condition $0=0$ is trivial).
• $r^{64} \neq 1$ (otherwise, $1-r^{64}=0$, leading to $0=0$, which doesn't help find a specific $r$).
• $r \neq 1$ (if $r=1$, $S_{all} = 64a$ and $S_{odd} = 32a$. The condition becomes $64a = 7 \times 32a \Rightarrow 2=7$, which is false). With these assumptions, we can cancel $a(1-r^{64})$ from both sides of the equation: $\frac{1}{1-r} = \frac{7}{1-r^2}$ Using the difference of squares factorization, $1-r^2 = (1-r)(1+r)$: $\frac{1}{1-r} = \frac{7}{(1-r)(1+r)}$ Since $r \neq 1$, $1-r \neq 0$. We can cancel $(1-r)$ from the denominator on both sides: $1 = \frac{7}{1+r}$
• Reasoning: This step involves algebraic manipulation to simplify the equation. By cancelling common non-zero factors and using algebraic identities, we reduce the equation to a simpler form, making it easier to solve for $r$.

Step 5: Solve for the Common Ratio ($r$) From the simplified equation $1 = \frac{7}{1+r}$, we can solve for $r$: Multiply both sides by $(1+r)$: $1 \times (1+r) = 7$ $1+r = 7$ Subtract 1 from both sides: $r = 7 - 1$ $r = 6$

• Reasoning: This final step involves basic algebraic operations to isolate the common ratio $r$, thus providing the solution to the problem.

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Common Mistakes & Tips

• Miscounting Terms: When forming the G.P. of odd terms, ensure the number of terms is correctly identified (e.g., 32 for 64 total terms).
• Incorrect Common Ratio for Sub-G.P.: The common ratio of the G.P. formed by odd terms is $r^2$, not $r$.
• Algebraic Errors: Be careful with factorizations (like difference of squares) and cancellations. Ensure that terms being cancelled are non-zero.

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Summary

The problem requires us to analyze a Geometric Progression and the sum of its terms, specifically comparing the total sum with the sum of terms at odd positions. By representing both sums using the G.P. sum formula and setting up an equation based on the given condition, we formed an algebraic expression. Through careful simplification, including the use of the difference of squares identity and cancellation of common factors (under valid assumptions), we arrived at a linear equation for the common ratio, which yielded the solution $r=6$.

The final answer is $\boxed{6}$.