Question

If the Geometric progression 384, 192, 96... and 3/128 3/64 3/32 ......, have their n term equal. Find the value 64 32 of 'n'.​

Answer 1

G.P. one =>

384, 192, 96, ...

first term, a = 384

common ratio, r

$r = \frac{192}{384} = \frac{12}{24} = \frac{1}{2}$

nth term, tn

$t_{n} = a {r}^{n - 1} = 384 ({ \frac{1}{2} )}^{n - 1} = 384 {( \frac{ {1}^{n - 1} }{ {2}^{n - 1} } }^{}) \\ = 384 \times \frac{1}{ {2}^{n - 1} } \\ t_{n} = \frac{384}{ {2}^{n - 1} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)$

G.P. two =>

3/128, 3/64, 3/32, ...

first term, a = 3/128

common ratio, r'

$r_{o} = \frac{ \frac{3}{64} }{ \frac{3}{128} } = \frac{3}{64} \times \frac{128}{3} = \frac{128}{64} \\ r_{o} = 2$

nth term, tn

$t_{n} = a { r_{o}}^{n - 1} = ( \frac{3}{128} ) \times {2}^{n - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)$

According to given condition;

their nth terms are equal.

$\frac{384}{ {2}^{n - 1} } = \frac{3}{128} \times {2}^{n - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (from \: i \: and \: ii) \\ \frac{384 \times 128}{3} = ( {2}^{n - 1} )( {2}^{n - 1} ) \\ 128 \times 128 = {2}^{n - 1 + n - 1} \\ {128}^{2} = {2}^{2n - 2} \\ {( {2}^{7} )}^{2} = {( {2}^{n - 1} )}^{2} \\ {2}^{7} = {2}^{n - 1} \\ n - 1 = 7 \\ n = 8$

Answer 2

The no. of terms for first geometric progression is 9. And the no. of terms in the second series is 8.

Step-by-step explanation:

First we need to find the r which is common ratio.

                    r = $\frac{\text{second term}}{\text{first term}}$

                     = $\frac{192}{384}$

                    = $\frac{1}{2}$

So, the total no. of terms are 9 in this GP.

  For second GP,

                     r = $\frac{\text{second term}}{\text{first term}}$

                       = $\frac{64}{128}$

                       = $\frac{1}{2}$

The total no. of terms are 8.

Basically, to find the no. of terms, write the whole series and just count it. Write it till the lowest common factor.