Question

If the third and sixth terms of a geometric progression are 12 and 96, then find
the number of terms in the progression, which are less than 2000.

Answer 1

so let the first term be a 
and common ratio be r 

so sixth term is 

ar⁵ = 96 ..........i 
ar² = 12  .... ii

dividing ii from i

r = 2 

and a = 3

so according to question 

$ar^{n-1} \leq 2000$ 

putting a and r 
 and simplifying 

$3*2^{n-1} \leq 2000$

⇒ $3*2^{n-1} \leq 2^4*5^3$
 
≈ $3*2^{n-1}<1536$

⇒ n = 10 ANSWER

Answer 2

let the first term = a    and   the common ratio = r

3rd term = a r² = 12 
6th term = a r⁵ = 96       
  (a r⁵) / (a r²) = r³ = 96/ 12 = 8    =>  r = 2
  Since  a r² = 12    =>  a = 12 / r² = 3
So the series is  :  3,  6, 12, 24, 48, 96, 192, 384, 768, 1536.  (we can simply count the number of terms here.).  Or, we can find it mathematically as below.

nth term = a $r^n$
Let us find the highest n such that  nth term is less than 2000.
  .     3 * $2^n$ < 2000
  =>     $2^n$ < 2000/3 = 666.66..
 =>      $2^n$ < 666.66..
  We have $2^9$ = 512    and    $2^{10}$  = 1024.
  So the highest term  less than 2000 is for  n = 9.

Hence, there are n+1 terms less than 2000:  so 10 terms.
=================================
 We can also find it by  :
             Log $2^n$ < log (2000/3)
             n log 2 < log (2000/3)
             n < [ log (2000/3) ] / Log 2
             n  < 9.28
         So n = 9.
Hence there are 10 terms in the GP that are less than 2000.