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How many terms of the geometric progression 6,12,24,......must be taken to get a sum of 49146​

User Lilie
by
7.2k points

2 Answers

9 votes

Answer:

13

Explanation:

the sum to n terms of a geometric sequence is


S_(n) =
(a(r^(n)-1) )/(r-1)

where a is the first term and r the common ratio

here a = 6 , r =
(12)/(6) = 2 and
S_(n) = 49146 , then


(6(2^(n)-1) )/(2-1) = 49146

6(
2^(n) - 1) = 49146 ( divide both sides by 6 )


2^(n) - 1 = 8191 ( add 1 to both sides )


2^(n) = 8192 ( take log of both sides )

log
2^(n) = log 8192 , that is

n log 2 = log 8192 ( divide both sides by log 2 )

n =
(log8192)/(log2) = 13

User YaoFeng
by
8.3k points
14 votes

Given G.P. :-


  • \sf 6,12,24,......

  • \sf first \: term,a = 6

  • \sf Sum,S_n = 49146

Solution:-


\red{ \underline { \boxed{ \sf{Common \: Ratio,r = (any\:observed\:term)/(previous\:term)}}}}


\longrightarrow
\sf Common \: Ratio,r = (12)/(6)


\quad \boxed{\sf {Common \:Ratio,r = 2 }}

Now,


\green{ \underline { \boxed{ \sf{Sum\: of \: G.P. \: to\: n\:term,S_n = (a(r^n-1))/(r-1)}}}}

where

  • a = first term
  • r = common Ratio
  • n= number of terms

Putting Values -


\begin{gathered}\\\implies\quad \sf 49146 = (6(2^n-1))/(2-1) \\\end{gathered}


\begin{gathered}\\\implies\quad \sf \frac{\cancel{49146}}{\cancel{6}} = (2^n-1)/(1) \\\end{gathered}


\begin{gathered}\\\implies\quad \sf 8191= (2^n-1)/(1) \\\end{gathered}


\begin{gathered}\\\implies\quad \sf 2^n-1 =8191 \\\end{gathered}


\begin{gathered}\\\implies\quad \sf 2^n =8191+1 \\\end{gathered}


\begin{gathered}\\\implies\quad \sf 2^n =8192 \\\end{gathered}


\begin{gathered}\\\implies\quad \sf 2^n =2^(13)\quad(8192= 2^(13)) \\\end{gathered}

Comparing powers on both sides-


\begin{gathered}\\\implies\quad \boxed{\sf{ n = 13}}\\\end{gathered}


\longrightarrowTherefore, to get a sum of 49146 , 13 terms of this G.P. should be taken.

User Bejado
by
8.0k points

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