Test Index

KEAM 2012 Math Paper

© examsnet.com

Question : 42 of 120

Marks: +1, -0

If

3^{rd},7^{th}

12^{th}

terms of an A.P. are three consecutive terms of a G.P., then the common ratio of the G.P. is

$\frac{5}{4}$
$\frac{9}{4}$
$\frac{2}{9}$
$\frac{1}{2}$
$\frac{12}{7}$

Solution:

Let first term and common difference of an AP are

a

and

d.

\therefore \;\; T_{3}=a+2 d

T_{7}=a+6 d

and

\;\; T_{12}=a+11 d

But

T_{3}, T_{7}

and

T_{12}

are consecutive terms of an GP.

\therefore (a+6 d)^{2}=(a+2 d)(a+11 d)

a^{2}+36 d^{2}+12 a d=a^{2}+13 a d+22 d^{2}

a d=14 d^{2}

\Rightarrow \;\; a=14 d

\therefore

Terms are

14 d+2 d, 14 d+6 d, 14 d+11 d

ie.,

16 d, \;\; 20 d, \;\; 25 d

\therefore

Common ratio

=\; \frac{20 d}{16 d}=\; \frac{5}{4}

© examsnet.com

Go to Question:

Prev Question

Next Question