JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
Consider two G.Ps. $2,2^2,2^3,\ldots$ and $4,4^2,4^3,\ldots$ 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 :
Sequence and Series · Geometric Progression · Standard Sums (Σ formulas)
Solution
Formulas used. (1) G.M. of $N$ numbers $=(\text{product})^{1/N}$(2) $\sum_{k=1}^{n}k(n-k)=\frac{(n-1)n(n+1)}{6}$Step 1 - Product of all terms. $$2^{1+2+\cdots+60}\cdot4^{1+2+\cdots+n}=2^{1830}\cdot2^{n(n+1)}=2^{1830+n^2+n}.$$ Step 2 - The G.M. $$\frac{1830+n^2+n}{60+n}=\frac{225}{8}\ \Longrightarrow\ 8n^2-217n+1140=0 .$$ $$n=\frac{217\pm103}{16}=20\ \text{or}\ \frac{57}{8}.$$ $n$ is a natural number, so $n=20$. Step 3 - The required sum. $\sum k(n-k)=n\sum k-\sum k^2=\dfrac{n(n+1)}{6}\left[3n-(2n+1)\right]=\dfrac{(n-1)n(n+1)}{6}$: $$\frac{19\cdot20\cdot21}{6}=1330 .$$ Answer: (C)