Sequences and Series question (nth of geometric sequence) (1 Viewer)

BlueGas · May 13, 2015

I need help for questions f, h, i and j.

In the following geometric sequences find the term indicated:

BLIT2014 · May 13, 2015

Not 100% sure but

F) ( 1/square root 2 ) * square root 2 ^8-1=8

InteGrand · May 13, 2015

BlueGas said:
I need help for questions f, h, i and j.

In the following geometric sequences find the term indicated:

$Note that for a geometric sequence with starting term $a$ and common ratio $r$, the $n$th term is given by: $T_n = ar^{n-1}$.$

$(f) Here, $a=\frac{1}{\sqrt{2}},r=\sqrt{2}$, so $T_8 = ar^7 = \frac{1}{\sqrt{2}}\times (\sqrt{2})^7 = (\sqrt{2})^6 = 8$.$

$(h) Here, $a=1,r=-\sqrt{3}$, so $T_6 = ar^5 = 1\times (-\sqrt{3})^5 = -9\sqrt{3}$.$

$(i) Here, $a=m^{-4},r=m^{2}$, so $T_{10} = ar^9 = m^{-4}\times (m^{2})^9 = m^{14}$.$

$(j) Here, $a=\frac{m}{n},r=mn$, so $T_{6} = ar^5 =\frac{m}{n}\times (mn)^5 = m^6 n^4$.$

BlueGas · May 13, 2015

InteGrand said:
$Note that for a geometric sequence with starting term $a$ and common ratio $r$, the $n$th term is given by: $T_n = ar^{n-1}$.$

$(f) Here, $a=\frac{1}{\sqrt{2}},r=\sqrt{2}$, so $T_8 = ar^7 = \frac{1}{\sqrt{2}}\times (\sqrt{2})^7 = (\sqrt{2})^6 = 8$.$

$(h) Here, $a=1,r=-\sqrt{3}$, so $T_6 = ar^5 = 1\times (-\sqrt{3})^5 = -9\sqrt{3}$.$

$(i) Here, $a=m^{-4},r=m^{2}$, so $T_{10} = ar^9 = m^{-4}\times (m^{2})^9 = m^{14}$.$

$(j) Here, $a=\frac{m}{n},r=mn$, so $T_{6} = ar^5 =\frac{m}{n}\times (mn)^5 = m^6 n^4$.$

How'd you figure out that r = m^2 for question i)?

If you did m^-4 x m^-2, wouldn't you get m^8 or m^-6?

InteGrand · May 13, 2015

BlueGas said:
How'd you figure out that r = m^2 for question i)?

If you did m^-4 x m^-2, wouldn't you get m^8 or m^-6?

$No, we don't multiply two consecutive terms of a geometric progression to find the common ratio $r$, but rather we take the ratio of one term to its preceding term. In the case of (i), we could take the ratio of the third term to the second term say, and we get $r=\frac{1}{m^{-2}}=m^2$. (We could do this with any two consecutive terms in the sequence and we'd get the same value for $r$, which is why it's called the \emph{common} ratio.)$

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Sequences and Series question (nth of geometric sequence) (1 Viewer)

BlueGas

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BLIT2014

The pessimistic optimist.

InteGrand

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BlueGas

Well-Known Member

InteGrand

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