Q49 / 90 · JEE Main 2019 (09 Jan Shift 2) · Mathematics

Let $a, b$ and $c$ be the $7^{t h}, 11^{t h}$ and $13^{t h}$ 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:

A
$2$
B
$\frac{7}{13}$
C
$\frac{1}{2}$
Correct answer
D
$4$

Solution

Let the first terms of A.P. be

A

and common difference be

d,

then

a = A + 6 d

b = A + 10 d

c = A + 12 d

Given

a, b, c

are in G. P.

\Rightarrow b^{2} = a c

\Rightarrow {(A + 10 d)}^{2} = (A + 6 d) (A + 12 d)

\Rightarrow A^{2} + 100 d^{2} + 20 A d = A^{2} + 18 A d + 72 d^{2}

\Rightarrow A = - 14 d

Now

\frac{a}{c} = \frac{A + 6 d}{A + 12 d} = \frac{- 14 d + 6 d}{- 14 d + 12 d} = \frac{- 8 d}{- 2 d} = 4

Hence, the correct answer is option: 'D'.

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