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

Answer:

13

Step-by-step explanation:

the sum to n terms of a geometric sequence is

[tex]S_{n}[/tex] = [tex]\frac{a(r^{n}-1) }{r-1}[/tex]

where a is the first term and r the common ratio

here a = 6 , r = [tex]\frac{12}{6}[/tex] = 2 and [tex]S_{n}[/tex] = 49146 , then

[tex]\frac{6(2^{n}-1) }{2-1}[/tex] = 49146

6([tex]2^{n}[/tex] - 1) = 49146 ( divide both sides by 6 )

[tex]2^{n}[/tex] - 1 = 8191 ( add 1 to both sides )

[tex]2^{n}[/tex] = 8192 ( take log of both sides )

log [tex]2^{n}[/tex] = log 8192 , that is

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

n = [tex]\frac{log8192}{log2}[/tex] = 13

[tex]\red{ \underline { \boxed{ \sf{Common \: Ratio,r = \frac{any\:observed\:term}{previous\:term}}}}}[/tex]

[tex]\longrightarrow[/tex][tex]\sf Common \: Ratio,r = \frac{12}{6} [/tex]

[tex]\quad \boxed{\sf {Common \:Ratio,r = 2 }}[/tex]

Now,

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

where

Putting Values -

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

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

[tex]\begin{gathered}\\\implies\quad \sf 8191= \frac{2^n-1}{1} \\\end{gathered} [/tex]

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

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

[tex]\begin{gathered}\\\implies\quad \sf 2^n =8192 \\\end{gathered} [/tex]

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

Comparing powers on both sides-

[tex]\begin{gathered}\\\implies\quad \boxed{\sf{ n = 13}}\\\end{gathered} [/tex]

[tex]\longrightarrow[/tex]Therefore, to get a sum of 49146 , 13 terms of this G.P. should be taken.

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