Mathematics, 28.10.2019 11:31 smuqtadir124
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.
Answers: 1
![\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 $6r$}\\](/tpl/images/0349/3287/b7c81.png)
![\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 \frac{ar^5}{ar^2} $ = $\frac{768}{96}$\\ $\implies r^3 = 8$$\therefore r = 2$\\\textup{Now substituting $r=2$ in $(2)$, we get $a = 24.$ }](/tpl/images/0349/3287/02eab.png)
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Solution: a g. p has a first term of a, a common ratio of r and its 6th term is 768. another g. p h...
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