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2015-03-31T21:42:04+05:30

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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

3 4 3
see 2000 is neither divisible by 3 so i took the nearest number before 2000 which has the form 3*2^(n-1) ther i have done the apporoximatio
u can also do like this
we get 2^{n-1} < 666.66667 approximately as its a inequation we can tell the nearest small number which is divisible by 2 is 512 so n - 1 = 8 , n = 9
thnx
welcome
2015-04-01T17:23:52+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.

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