if the GP is 384 ,192,96,.......and 3by128,3by64,3by32,.....have their nth term equal find the value of n
Answers
If the G.P is 384,192,96,......and 3/128,3/64,3/32,....have their nth term equal find the value of n.
G.P. one =>
384, 192, 96, ...
first term, a = 384
common ratio, r
r = \frac{192}{384} = \frac{12}{24} = \frac{1}{2}r=
384
192
=
24
12
=
2
1
nth term, tn
\begin{lgathered}t_{n} = a {r}^{n - 1} = 384 ({ \frac{1}{2} )}^{n - 1} = 384 {( \frac{ {1}^{n - 1} }{ {2}^{n - 1} } }^{}) \\ = 384 \times \frac{1}{ {2}^{n - 1} } \\ t_{n} = \frac{384}{ {2}^{n - 1} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)\end{lgathered}
t
n
=ar
n−1
=384(
2
1
)
n−1
=384(
2
n−1
1
n−1
)
=384×
2
n−1
1
t
n
=
2
n−1
384
...(i)
G.P. two =>
3/128, 3/64, 3/32, ...
first term, a = 3/128
common ratio, r'
\begin{lgathered}r_{o} = \frac{ \frac{3}{64} }{ \frac{3}{128} } = \frac{3}{64} \times \frac{128}{3} = \frac{128}{64} \\ r_{o} = 2\end{lgathered}
r
o
=
128
3
64
3
=
64
3
×
3
128
=
64
128
r
o
=2
nth term, tn
t_{n} = a { r_{o}}^{n - 1} = ( \frac{3}{128} ) \times {2}^{n - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)t
n
=ar
o
n−1
=(
128
3
)×2
n−1
...(i)
According to given condition;
their nth terms are equal.
\begin{lgathered}\frac{384}{ {2}^{n - 1} } = \frac{3}{128} \times {2}^{n - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (from \: i \: and \: ii) \\ \frac{384 \times 128}{3} = ( {2}^{n - 1} )( {2}^{n - 1} ) \\ 128 \times 128 = {2}^{n - 1 + n - 1} \\ {128}^{2} = {2}^{2n - 2} \\ {( {2}^{7} )}^{2} = {( {2}^{n - 1} )}^{2} \\ {2}^{7} = {2}^{n - 1} \\ n - 1 = 7 \\ n = 8\end{lgathered}
2
n−1
384
=
128
3
×2
n−1
...(fromiandii)
3
384×128
=(2
n−1
)(2
n−1
)
128×128=2
n−1+n−1
128
2
=2
2n−2
(2
7
)
2
=(2
n−1
)
2
2
7
=2
n−1
n−1=7
n=8