Consider two G.Ps.

Question

Consider two G.Ps. 2, 22, 23, ..... and 4, 42, 43, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is $${(2)^{{{225} \over 8}}}$$, then $$\sum\limits_{k = 1}^n {k(n - k)} $$ is equal to :

Accepted Answer

Given G.P's 2, 22, 23, .... 60 terms

4, 42, ....  n terms

Now, G.M $$ = {2^{{{225} \over 8}}}$$

$${\left( {{{2.2}^2}...\,{{4.4}^2}...} \right)^{{1 \over {60 + n}}}} = {2^{{{225} \over 8}}}$$

$$\left( {{2^{{{{n^2} + n + 1830} \over {60 + n}}}}} \right) = {2^{{{225} \over 8}}}$$

$$ \Rightarrow {{{n^2} + n + 1830} \over {60 + n}} = {{225} \over 8}$$

$$ \Rightarrow 8{n^2} - 217n + 1140 = 0$$

$$n = {{57} \over 8},\,20,\,$$ so $n = 20$

$$\therefore$$ $$\sum\limits_{k = 1}^{20} {k(20 - k) = 20 \times {{20 \times 21} \over 2} - {{20 \times 21 \times 41} \over 6}} $$

$$ = {{20 \times 21} \over 2}\left[ {20 - {{41} \over 3}} \right] = 1330$$

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