Question

if the geometric progression 384,192,96,and 3/128,3/64,3/32.......,have their nterm equal. find the value of "n"​

Answer 1

Here is the solution:

Dont forget to give BRAINLIEST

Answer 2

Answer:

The value of n is 8.

Step-by-step explanation:

For the first geometric progression we have :

first term,  $a_{1}$ = 384

second term, $a_{2}$ = 192 = $a_{1}$ r  = 384r , therefore the common ratio, r = $\frac{192}{384} = 0.5$

third term,$a_{3}$ = 96 = $a_{2}$ r  = 192r, so we can check the common ratio,r = $\frac{96}{192}$ = 0.5

So the nth term for the first geometric progression is given by

 = $a_{1}$ × $r^{n-1}$ = 384 × $(0.5)^{n-1}$ .... (i)

For the second geometric progression we have :

first term,  $a_{1}$ = $\frac{3}{128}$

second term, $a_{2}$ = $\frac{3}{64}$ = $a_{1}$ r  = $\frac{3}{128}$ r , therefore the common ratio, r = $\frac{\frac{3}{64} }{\frac{3}{128} } = \frac{128}{64} = 2$

third term,$a_{3}$ = $\frac{3}{32}$ = $a_{2}$ r  = $\frac{3}{64}$ r, so we can check the common ratio,r = $\frac{\frac{3}{32} }{\frac{3}{64} } = \frac{64}{32} = 2$

So the nth term for the second geometric progression is given by

 = $a_{1}$ × $r^{n-1}$ =  $\frac{3}{128}$ × $(2)^{n-1}$ .... (ii)

Since both progressions have their nth term equal, we can equate equations (i) and (ii) and so we can write

384 × $(0.5)^{n-1}$  =   $\frac{3}{128}$ × $(2)^{n-1}$     ⇒    $\frac{384 \times 128}{3} = 16384 = \frac{2^{n-1}}{0.5^{n-1}} = 4^{n - 1}$

Therefore 16384 = $4^{n-1}$      ⇒   $4^{7}$ = $4^{n - 1}$          ⇒    7 = n -1       ∴ n = 8