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As we have learned

Sum of n terms of a GP -

S_{n}= \left\{\begin{matrix} a\frac{\left ( r^{n}-1 \right )}{r-1}, &if \: r\neq 1 \\ n\, a, & if \, r= 1 \end{matrix}\right.

 

- wherein

a\rightarrow first term

r\rightarrow common ratio

n\rightarrow number of terms    

 

S_{20}= 1 +\left ( \frac{4-1}{2} \right )+\left ( \frac{8-1}{4} \right )+\left ( \frac{16-1}{8} \right )+......

=1+2+2.......20times....\left ( 1/2+1/4+1/8+......20times \right ) 

 

39-1/2(\left ( \frac{1-(1/2)^{19}}{1-1/2} \right ))= 39-1+1/(2)^{19}

38+\frac{1}{2^{19}}

 

 

 


Option 1)

38+\frac{1}{2^{19}}

This is correct

Option 2)

38+\frac{1}{2^{20}}

This is incorrect

Option 3)

39+\frac{1}{2^{20}}

This is incorrect

Option 4)

39+\frac{1}{2^{19}}

This is incorrect

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As we learnt in 

Common ratio of a GP (r) -

The ratio of two consecutive terms of a GP

- wherein

eg: in 2, 4, 8, 16, - - - - - - -

r = 2

and in 100, 10, 1, 1/10 - - - - - - -

r = 1/10

 


 {A}, {B},{C} are in  G.P where {A}, {B},{C}  are Terms of an A.P

 

Let first term is a and common difference is d, and common ratio be r then

A = a+d

B = a+4d

C= a+8d

\therefore \: \frac{B}{A}=\frac{C}{B}=\frac{r}{1}

\frac{a+4d}{a+d} =\frac{a+8d}{a+4d}=\frac{r}{1}

\therefore \frac{a+4d+a+d}{a+4d-a-d}=\frac{r+1}{r-1}

\Rightarrow \frac{2a+5d}{3d}=\frac{r+1}{r-1}-----(i)

    \frac{a+8d+a+4d}{a+8d-a-4d} =\frac{r+1}{r-1}

\Rightarrow \frac{2a+12d}{4d}=\frac{r+1}{r-1}------(ii)

from (i) and (ii)

\frac{2a+5d}{2a+12d}=\frac{3}{4}

8a+20d=6a+36d

2a=16d

a=8d

\Rightarrow \frac{r+1}{r-1}=\frac{2\times8d+5d}{3d}=\frac{16d+5d}{3d}=\frac{21}{3}=7

\therefore r+1=7r-7

8=6r

r=\frac{4}{3}


Option 1)

\frac{8}{5}

Incorrect option

Option 2)

\frac{4}{3}

Correct option

Option 3)

1

Incorrect option

Option 4)

\frac{7}{4}

Incorrect option

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