Question

Consider two G.Ps. \(2, 2^2, 2^3, ….\) 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

• A. 560
• B. 1540
• C. 1330
• D. 2600

Answer 1

The Correct Option is C

Given G.P′s \(2, 2^2, 2^3, .… 60\) terms
\(4, 4^2, … n\) terms
Now, GM = \(2^{\frac {225}{8}}\)

\((2.2^2⋯4.4^2⋯)^{\frac {1}{60+n}}=2^{\frac {225}{8}}\)

\((2^{\frac {n^2+n+1830}{60+n})}=2^{\frac {225}{8}}\)

\({\frac {n^2+n+1830}{60+n}}={\frac {225}{8}}\)

\(⇒8n^2–217n+1140=0\)
\(n=\frac {57}{8}, 20\)
So, \(n=20\)
∴\(\sum_{k=1}^{n}k(n−k)\)\(=20×\frac {20×21}{2}−\frac {20×21×41}{6}\)

\(=\frac {20×21}{2}[20−\frac {41}{3}]\)
\(=1330\)

So, the correct option is (C): \(1330\)