Question

If the g. P 3/625 ,3/125,3/25,..and G.p1875,625,125,...have their nth terms equal, find n and nth term.

Answer 1

Answer:

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Answer 2

Given :

The geometric progression series $\dfrac{3}{625} ,\dfrac{3}{125}, \dfrac{3}{25} , ..........$

And

The geometric progression series is 1875 , 625 , 125

The nth term of both g.p series are same

To Find :

The number of term, and nth term

Solution :

For  geometric progression series

(i ) $\dfrac{3}{625} ,\dfrac{3}{125}, \dfrac{3}{25} , ..........$  are in g.p

So, common ratio = r = 5

For n term, g.p = $t_n$ = a $r^{n-1}$

So,        $t_n$ = $\dfrac{3}{625}$ × $5^{n-1}$

Again

For  geometric progression series

(ii )  1875 , 625 , 125 , ............. are in g.p

So, common ratio = r = 5

For n term, g.p = $T_n$ = a $r^{\dfrac{n}{2} -1}$

So,        $T_n$ = 1875 × $5^{\dfrac{n}{2} -1}$

Since nth term of both series are equal

∴       $t_n$  =  $T_n$

Or,    $\dfrac{3}{625}$ × $5^{n-1}$ =   1875 × $5^{\dfrac{n}{2} -1}$

Or,  $5^{n-1}$ = 625 × 625  × $5^{\dfrac{n}{2} -1}$

Or,  $\dfrac{5^{n} }{5}$  = 625 × 625  × $\dfrac{5^{\dfrac{n}{2} } }{5}$

Or,  $5^{n}$ =  625 × 625  × $5^{\dfrac{n}{2} }$

Or,  $\dfrac{5^{n} }{5\dfrac{n}{2} }$ =  625 × 625

or,  $5^{\dfrac{n}{2} }$  = $5^{8}$

∴     $\dfrac{n}{2}$  = 8

i.e   n = 16

So, The number of terms = n = 16

Again

For n= 16

  $t_1_6$ = $\dfrac{3}{625}$ × $5^{16-1}$

Or, $t_1_6$ = $\dfrac{3}{625}$ × $5^{15}$

Or, $t_1_6$ = $\dfrac{3}{625}$ × 3.05 × $10^{10}$

∴  , $t_1_6$ = 1464 × $10^{5}$

Hence,  The number of terms is 16  , And The 16th term is  1464 × $10^{5}$  Answer