Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P . If these are also the three consecutive terms of a G.P., then (a/c) is equal to:

Question

Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P . If these are also the three consecutive terms of a G.P., then (a/c) is equal to: (A) (1/2) (B) 4 (C) 2 (D) (7/13)

Tardigrade Admin · Accepted Answer

a = A + 6d b = A + 10d c = A + 12d a,b,c are in G.P. ⇒ (A + 10d)2 = (A 6d)(a + 12d) ⇒ (A/d) = - 14 (a/c) = (A + 6d/A + 12d) = ( 6 + (A/d)/12 + (A)d) = (6 - 14/12 -14) = 4

Q. Let $a, b$ and $c$ be the $7^{t h}, 1 1^{t h}$ and $1 3^{t h}$ terms respectively of a non-constant $A . P$ . If these are also the three consecutive terms of a $G . P$ ., then $\frac{a}{c}$ is equal to:

$\frac{1}{2}$

$4$

$2$

$\frac{7}{13}$

$a = A + 6 d$
$b = A + 10 d$
$c = A + 12 d$
a,b,c are in G.P.
$\Rightarrow (A + 10 d)^{2} = (A 6 d) (a + 12 d)$
$\Rightarrow \frac{A}{d} = - 14$
$\frac{a}{c} = \frac{A + 6 d}{A + 12 d} = \frac{6 + \frac{A}{d}}{12 + \frac{A}{d}} = \frac{6 - 14}{12 - 14} = 4$

Q. Let a,b and c be the 7th,11th and 13th terms respectively of a non-constant A.P . If these are also the three consecutive terms of a G.P., then ca​ is equal to:

21​

4

2

137​

a=A+6d b=A+10d c=A+12d a,b,c are in G.P. ⇒(A+10d)2=(A6d)(a+12d) ⇒dA​=−14 ca​=A+12dA+6d​=12+dA​6+dA​​=12−146−14​=4

Q. Let $a, b$ and $c$ be the $7^{t h}, 1 1^{t h}$ and $1 3^{t h}$ terms respectively of a non-constant $A . P$ . If these are also the three consecutive terms of a $G . P$ ., then $\frac{a}{c}$ is equal to:

$\frac{1}{2}$

$4$

$2$

$\frac{7}{13}$

$a = A + 6 d$
$b = A + 10 d$
$c = A + 12 d$
a,b,c are in G.P.
$\Rightarrow (A + 10 d)^{2} = (A 6 d) (a + 12 d)$
$\Rightarrow \frac{A}{d} = - 14$
$\frac{a}{c} = \frac{A + 6 d}{A + 12 d} = \frac{6 + \frac{A}{d}}{12 + \frac{A}{d}} = \frac{6 - 14}{12 - 14} = 4$