Question

Given a sequence of $4$ numbers, first three of which are in $G.P$ . and the last three are in $A.P$ . with common difference six. If first and last terms of this sequence are equal, then the last term is:

• A. 16
• B. 8
• C. 4
• D. 2

Answer 1

Answer: (B)

Let $a, b, c, d$ be four numbers of the sequence.
Now, according to the question $b^{2} = ac$ and $c - b = 6$ and $a - c = 6$
Also, given $a = d$
$\therefore b^{2}=ac \Rightarrow b^{2} =a \left[\frac{a+b}{2}\right]$
$\left(\because 2c=a+b\right)$
$\Rightarrow \, a^{2}-2b^{2}+ab=0$
Now, $C-b=6 $ and $a-c=6,$
gives $a - b = 12$
$\Rightarrow b=a-12$
$\therefore a^{2}-2b^{2}+ab=0$
$\Rightarrow a^{2}-2 \left(a-12\right)^{2}+a\left(a-12\right)=0$
$\Rightarrow a^{2}-2 a^{2}-288+48a+a^{2}-12a=0$
$\Rightarrow 36a =288 \Rightarrow a=8$
Hence, last term is $d = a = 8$.