Question

SOLUTION: a G.P has a first term of a, a common ratio of r and its 6th term is 768. another G.P has a first term of a, a common ratio of 6r and its 3rd term is 3456. evaluate a and r.​

Answer 1

`\textbf{The value of the first term $a = 24$ and the common difference $r = 2$}\\`

Explanation:

`\textup{The general form of the GP would be:}\\$$a, ar, ar^2, ar^3, . . .$$$\therefore $The $n^(th)$ term would be $ar^(n-1)$\\Also, it is stated that the sixth term of the GP is $768$\\  \[768 = ar^5  \tag{1}\] \\\textup{Now consider the second GP with the first term $a$ and with common difference $6r$}\\`

`\textup{Therefore, the GP would be:}\\$$a, a(6r), a{(6r)}^2, a{(6r)}^3,. . .$$\\\textup{Consequently the $n^(th)$ term would be:} $a{{(6r)}^(n-1)}$\\Now given its third term is $3456$\\$\implies 3456 = a{(6r)}^2\\$ \[96 = ar^2  \tag{2}\] $ \frac{\textcircled{1}}{\textcircled{2}} \implies  (ar^5)/(ar^2) $ = $(768)/(96)$\\  $\implies r^3 = 8$$\therefore r = 2$\\\textup{Now substituting $r=2$ in $(2)$, we get $a = 24.$  }`