Question

Out of the four number given, first three are in GP and last three are in AP whose common difference is 6. If the first and last numbers are same,then first term will be

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is D

8

Let the first three terms be

a, ar, a$r^2$
$\downarrow$ $\downarrow$ $\downarrow$
(1) (2) (3)

Let the last three terms be

ar, ar + d , ar + 2d
$\downarrow$ $\downarrow$ $\downarrow$
(2) (3) (4)

First and last term are equal.

$\Rightarrow$ ar + 2d = a ---------(A)

Equating 3rd term (or equating the $3^{rd}$ from G.P and $2^{nd}$ term from the A.P),

ar + d = a$r^2$ ----------------(B)

d = 6 (given)

Substituting in (A) and (B),

ar + 12 = a ---------(1)

ar + 6 = a$r^2$---------(A)

(1) - (2) $\Rightarrow$ 6 = a(1 - $r^2$)

From (2), 6 = a$r^2$ - ar

$\Rightarrow$ a(1 - $r^2$) = a($r^2$ - r)

$\Rightarrow$ (1 - $r^2$) (1 - r) = r(r - 1)

$\Rightarrow$ r=1 or 1 + r = -r

$\Rightarrow$ r=1 or r = $\frac{-1}{2}$

r ≠ 1 (since the terms are different)

Taking r = $\frac{-1}{2}$

a $\times$ $\frac{-1}{2}$ + 12 = a

12 = $\frac{3a}{2}$

$\Rightarrow$ a = 8