b_(n+2) ∙ b_(n+3) ∙ b_(n+4) = 3^(3n+3) geometric series b


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Question Number 146943 by mathdanisur last updated on 16/Jul/21

$b_{n + 2} \cdot b_{n + 3} \cdot b_{n + 4} = 3^{3 n + 3}$ $g e o m e t r i c s e r i e s b_{8} = ?$
	Answered by Olaf_Thorendsen last updated on 16/Jul/21

	$Let b_{0} = \frac{1}{9} and q = 3$ $b_{n + 2} = \frac{3^{n + 2}}{9} = 3^{n}$ $b_{n + 3} = \frac{3^{n + 3}}{9} = 3^{n + 1}$ $b_{n + 4} = \frac{3^{n + 4}}{9} = 3^{n + 2}$ $b_{n + 2} b_{n + 3} b_{n + 4} = 3^{n} . 3^{n + 1} . 3^{n + 2} = 3^{3 n + 3}$ $b_{8} = b_{0} q^{8} = \frac{3^{8}}{9} = 3^{6} = 729$
	Commented by mathdanisur last updated on 16/Jul/21

	$c o o l S e r t h a n k y o u, b u t b_{0} = \frac{1}{9} h o w ?$
	Answered by Rasheed.Sindhi last updated on 17/Jul/21

	$b_{n + 2} \cdot b_{n + 3} \cdot b_{n + 4} = 3^{3 n + 3}$ $\underline{g e o m e t r i c s e r i e s b_{8} = ?}$ $n = - 1$ $b_{1} . b_{2} . b_{3} = 3^{3 (- 1) + 3} = 1 . . . . . . . . . . (i)$ $n = 0$ $b_{2} . b_{3} . b_{4} = 3^{3 (0) + 3} = 27 . . . . . . . . . . . (i i)$ $(i i) / (i) :$ $\frac{b_{4}}{b_{1}} = 27$ $\frac{b_{1} r^{3}}{b_{1}} = 27 \Rightarrow r = 3 (b_{n} = b_{1} r^{n - 1})$ $(i) : b_{1} . b_{2} . b_{3} = 1$ $b_{1} . 3 b_{1} . 9 b_{1} = 1$ $b_{1}^{3} = \frac{1}{27} \Rightarrow b_{1} = \frac{1}{3}$ $b_{8} = b_{1} r^{7} = \frac{1}{3} . 3^{7} = 3^{6} = 729$
	Commented by mathdanisur last updated on 17/Jul/21

	$c o o l S e r t h a n k y o u$
	Answered by Rasheed.Sindhi last updated on 17/Jul/21
	$Let b is first term and r is common ratio. determinant (((b_n =br^(n−1) ))) b_(n+2) ∙ b_(n+3) ∙ b_(n+4) = 3^(3n+3) br^(n+2-1) .br^(n+3-1) .br^(n+4-1) =3^(3n+3) br^(n+1) .br^(n+2) .br^(n+3) =3^(3n+3) b^3 r^(3n+6) =3^(3n+3) br^(n+2) =3^(n+1) ⇒ { ((n=-2 : b=(1/3))),((n=−1 : ((1/3))r=1⇒r=3)) :} b_8 =br^(8−1) =((1/3))(3)^7 =3^6 =729$
	$L e t b i s f i r s t t e r m a n d r i s c o m m o n$ $r a t i o .$ $\begin{matrix} b_{n} = {b r}^{n - 1} \end{matrix}$ $b_{n + 2} \cdot b_{n + 3} \cdot b_{n + 4} = 3^{3 n + 3}$ ${b r}^{n + 2 - 1} . {b r}^{n + 3 - 1} . {b r}^{n + 4 - 1} = 3^{3 n + 3}$ ${b r}^{n + 1} . {b r}^{n + 2} . {b r}^{n + 3} = 3^{3 n + 3}$ $b^{3} r^{3 n + 6} = 3^{3 n + 3}$ ${b r}^{n + 2} = 3^{n + 1} \Rightarrow {\begin{cases} n = - 2 : b = \frac{1}{3} \\ n = - 1 : (\frac{1}{3}) r = 1 \Rightarrow r = 3 \end{cases}$ $b_{8} = {b r}^{8 - 1} = (\frac{1}{3}) {(3)}^{7} = 3^{6} = 729$
	Commented by mathdanisur last updated on 17/Jul/21

	$c o o l S e r t h a n k y o u$
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