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 $a-b=12$
$\Rightarrow b=a-12$
$\therefore a^2-2b^2+ab=0$
$\Rightarrow a^2-2{\left(a-12\right)}^2+a\left(a-12\right)=0$
$\Rightarrow a^2-2a^2-288+48a+a^2-12a=0$
$\Rightarrow 36a=288\Rightarrow a=8$
Hence, last term is $d=a=8$.