Q

I have a doubt, kindly clarify. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is :

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

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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)

Incorrect option

Option 2)

Correct option

Option 3)

Incorrect option

Option 4)

Incorrect option

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