Consider two G.P.s $2, 2^{2}, 2^{3} . . . . . .$ and $4, 4^{2}, 4^{3} . . . . . .$ of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)}^{\frac{225}{8}}$ then $\sum_{k = 1}^{n} k (n - k)$ is equal to

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1540

560

1330

2600

answer is C.

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Detailed Solution

$\begin{aligned} {((2^{1} 2^{2} \dots {.2}^{60}) (4^{1} \cdot 4^{2} \dots \dots 4^{n}))}^{\frac{1}{60 + n}} = 2^{\frac{225}{8}} \\ {(2^{30 \times 61} 4^{\frac{n (n + 1)}{2}})}^{\frac{1}{60 + n}} = 2^{\frac{225}{8}} \\ 2^{1830 + n^{2} + n} = 2^{\frac{(225) (60 + n)}{8}} \\ = 8 n^{2} - 217 n + 1140 = 0 \\ n = 20, \frac{57}{8} \\ \sum_{k = 1}^{n} nk - k^{2} = \frac{n^{2} (n + 1)}{2} - \frac{n (n + 1) (2 n + 1)}{6} \\ = 1330 \end{aligned}$