Question

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

Answer 1

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

Answer 2

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}`

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