Key Concepts and Formulas

• Geometric Mean (G.M.): For $N$ positive numbers $a_1, a_2, \ldots, a_N$, their Geometric Mean is $(a_1 \cdot a_2 \cdot \ldots \cdot a_N)^{1/N}$.
• Sum of First $m$ Natural Numbers: $\sum_{k=1}^m k = \frac{m(m+1)}{2}$.
• Sum of Squares of First $m$ Natural Numbers: $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$.
• Summation of $k(n-k)$: $\sum_{k=1}^n k(n-k) = n\sum_{k=1}^n k - \sum_{k=1}^n k^2$.

Step-by-Step Solution

Step 1: Calculate the product of terms for the first G.P. The first G.P. is $2, 2^2, 2^3, \ldots$ with 60 terms. The terms are $2^1, 2^2, \ldots, 2^{60}$. The product $P_1$ is $2^1 \cdot 2^2 \cdot \ldots \cdot 2^{60} = 2^{(1+2+\ldots+60)}$. Using the sum of the first $m$ natural numbers formula with $m=60$: $1+2+\ldots+60 = \frac{60(60+1)}{2} = \frac{60 \cdot 61}{2} = 30 \cdot 61 = 1830$ So, $P_1 = 2^{1830}$.

Step 2: Calculate the product of terms for the second G.P. The second G.P. is $4, 4^2, 4^3, \ldots$ with $n$ terms. The terms are $4^1, 4^2, \ldots, 4^n$. The product $P_2$ is $4^1 \cdot 4^2 \cdot \ldots \cdot 4^n = 4^{(1+2+\ldots+n)}$. Using the sum of the first $m$ natural numbers formula with $m=n$: $1+2+\ldots+n = \frac{n(n+1)}{2}$ So, $P_2 = 4^{\frac{n(n+1)}{2}}$. To express this with base 2, we use $4=2^2$: $P_2 = (2^2)^{\frac{n(n+1)}{2}} = 2^{2 \cdot \frac{n(n+1)}{2}} = 2^{n(n+1)}$.

Step 3: Determine the total product and the geometric mean. The total number of terms is $N = 60 + n$. The total product $P$ of all terms is $P_1 \cdot P_2$: $P = 2^{1830} \cdot 2^{n(n+1)} = 2^{(1830 + n(n+1))} = 2^{(n^2+n+1830)}$. The Geometric Mean (G.M.) of these $N$ terms is $P^{1/N}$: $\text{G.M.} = \left(2^{(n^2+n+1830)}\right)^{\frac{1}{60+n}} = 2^{\frac{n^2+n+1830}{60+n}}$.

Step 4: Solve for the unknown number of terms, $n$. We are given that the G.M. is $2^{\frac{225}{8}}$. Equating the exponents: $\frac{n^2+n+1830}{60+n} = \frac{225}{8}$. Cross-multiplying gives: $8(n^2+n+1830) = 225(60+n)$ $8n^2 + 8n + 14640 = 13500 + 225n$ Rearranging into a quadratic equation: $8n^2 + (8-225)n + (14640-13500) = 0$ $8n^2 - 217n + 1140 = 0$ Using the quadratic formula $n = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$: $n = \frac{217 \pm \sqrt{(-217)^2 - 4(8)(1140)}}{2(8)}$ $n = \frac{217 \pm \sqrt{47089 - 36480}}{16}$ $n = \frac{217 \pm \sqrt{10609}}{16}$ Since $\sqrt{10609} = 103$: $n = \frac{217 \pm 103}{16}$ This yields two possible values for $n$: $n_1 = \frac{217+103}{16} = \frac{320}{16} = 20$ $n_2 = \frac{217-103}{16} = \frac{114}{16} = \frac{57}{8}$ Since $n$ must be a positive integer (number of terms), we take $n=20$.

Step 5: Evaluate the required summation. We need to find $\sum_{k=1}^n k(n-k)$ with $n=20$. $\sum_{k=1}^{20} k(20-k) = \sum_{k=1}^{20} (20k - k^2)$ Using the linearity of summation and the standard formulas: $= 20 \sum_{k=1}^{20} k - \sum_{k=1}^{20} k^2$ $= 20 \left(\frac{20(20+1)}{2}\right) - \left(\frac{20(20+1)(2 \cdot 20+1)}{6}\right)$ $= 20 \left(\frac{20 \cdot 21}{2}\right) - \left(\frac{20 \cdot 21 \cdot 41}{6}\right)$ $= 20 (210) - (10 \cdot 7 \cdot 41)$ $= 4200 - 2870$ $= 1330$

Common Mistakes & Tips

• Exponent Manipulation: Ensure accurate application of exponent rules, especially when changing bases (e.g., from 4 to 2).
• Quadratic Formula: Double-check the calculations when using the quadratic formula, particularly the discriminant.
• Integer Solutions: Remember that the number of terms ($n$) must be a positive integer; discard any non-integer or negative solutions.
• Summation Formulas: Have the formulas for $\sum k$ and $\sum k^2$ memorized for quick application.

Summary The problem involved calculating the geometric mean of terms from two geometric progressions. By expressing all terms as powers of 2, we found the total product and geometric mean. Equating this with the given geometric mean allowed us to solve for the number of terms, $n=20$. Finally, we evaluated the required summation $\sum_{k=1}^n k(n-k)$ using standard summation formulas for $n=20$, which resulted in 1330.

The final answer is $\boxed{1330}$.