Question

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?

Answer 1

Hint: Use the nth term formula and proceed as instructed in question.

Complete step-by-step answer:
It has been given
$
  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 \\
$

Note: Must remember all the general terms, sum formula, quadratic roots in order to solve such similar problems. Similar questions can be asked for Harmonic Progression and AGP etc.