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# In an increasing G.P. The sum of the first and the last term is 66, the product of the second and the last but one term is 128, and the sum of all the terms is 126. How are terms in progression? Verified
$a + a{r^{n - 1}} = 66 \\ {\text{And }} \\ ar.{a^{n - 2}} = {a^2}{r^{n - 1}} = 128 \\ \therefore {a^{n - 1}} = \dfrac{{128}}{a} \\ {\text{Putting in 1, we get }}a + \dfrac{{128}}{a} = 66 \\ \therefore {a^2} - 66a + 128 = 0 \\ {\text{splitting the middle terms we get}} \\ (a - 2)(a - 64) = 0 \\ \therefore a = 2,64 \\ {r^{n - 1}} = 32,\dfrac{1}{{32}} \\ {\text{We reject the second value as r > 1 ,}}\therefore {{\text{r}}^{n - 1}} = 32 \\ {\text{Sum = }}\dfrac{{a({r^n} - 1)}}{{r - 1}} = 126{\text{ }} \Rightarrow \dfrac{{2(32r - 1)}}{{r - 1}} = 126 \\ \because {r^{n - 1}} = 32 \\ \therefore 32r - 1 = 63r - 63 \\ \therefore r = 2{\text{ and }}{r^{n - 1}} = 32{\text{ gives}} \\ {{\text{2}}^{n - 1}} = {2^5} \Rightarrow n - 1 = 5 \Rightarrow n = 6 \\$