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

Answer 1

The value of the ratio of $\dfrac{a}{c}$  is 4

Step-by-step explanation:

Given as :

For An Arithmetic Progression

The 7th terms = a

The 11th terms = b

The 13th terms = c

Let The first term of A.P = x

Let The common difference = d

According to question

∵ nth terms of an A.P = $t_n$  = x + (n - 1) d

So, for n = 7

$t_7$  = x + (7 - 1) d

i.e   a  = x + 6 d                  .........1

Similarly

So, for n = 11

$t_1_1$  = x + (11 - 1) d

i.e   b  = x + 10 d                 .........2

Again

So, for n = 13

$t_1_3$  = x + (13 - 1) d

i.e   c  = x + 12 d                 .........3

Again

For Geometric Progression'

The three consecutive terms are a , b , c

So, For G.P

  b² = a c                       ..........4

From eq 1 , eq 2 , eq 3 , eq 4

( x + 10 d )²  =  ( x + 6 d )  ( x + 12 d )

Now, solving the equation

x² + 100 d² + 20 x d = x² + 12 x d + 6 x d + 72 d²

Or, ( x² - x² ) + ( 100 d² - 72 d² ) + ( 20 x d - 18 x d ) = 0

Or, 0 + 28 d² + 2 x d = 0

Or,   2 d ( 14 d + x ) = 0

Or,    d = 0  ,   14 d + x = 0

∴            x = - 14 d

Now, A/Q

The value of  $\dfrac{a}{c}$  =  $\dfrac{x+6d}{x+12d}$

put the value of x

So, $\dfrac{a}{c}$  =  $\dfrac{-14d+6d}{-14d+12d}$

Or, $\dfrac{a}{c}$ = $\dfrac{-8d}{-2d}$

∴    $\dfrac{a}{c}$  = 4

So, The ratio of $\dfrac{a}{c}$ = 4

Hence, The value of the ratio of $\dfrac{a}{c}$  is 4   Answer