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

Answer: (C)

1330

Answer 2

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): $1330$