Question

In a set of four numbers the first three are in $G.P.$ and the last three in $A.P.$ with common difference $6$. If the first number is the same as the fourth, find the four numbers.

Answer 1

Last three of the four numbers are in $A.P.$ and hence they may be chosen as $a-d,a,a+d$
Also the first number is same as the last one i.e. $a+d$.
Therefore the four numbers are $a+d,a-d,a,a+d$. The first three of the above four are in $G.P.$
$\therefore (a-d{)}^2=a(a+d)$. But $d=6$ given
$\therefore (a-6{)}^2=a(a+6)$
or $a^2-12a+36=a^2+6a$
or $18a=36$ $\therefore a=2$.
Putting for $a$ and $d$ the four numbers are $8,-4,2,8,$ which satisfy the given conditions.