Question

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

Answer 1

Answer:

13

Step-by-step explanation:

the sum to n terms of a geometric sequence is

$S_{n}$ = $\frac{a(r^{n}-1) }{r-1}$

where a is the first term and r the common ratio

here a = 6 , r = $\frac{12}{6}$ = 2 and $S_{n}$ = 49146 , then

$\frac{6(2^{n}-1) }{2-1}$ = 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 = $\frac{log8192}{log2}$ = 13

Answer 2

Given G.P. :-

• $\sf 6,12,24,......$
• $\sf first \: term,a = 6$
• $\sf Sum,S_n = 49146$

Solution:-

$\red{ \underline { \boxed{ \sf{Common \: Ratio,r = \frac{any\:observed\:term}{previous\:term}}}}}$

$\longrightarrow$$\sf Common \: Ratio,r = \frac{12}{6}$

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

Now,

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

where

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

Putting Values -

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

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

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

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

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

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

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

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.