Question

In set of $4$ number the first three number are in $GP$ and the last three are in $AP$ with common different is $6$
If first number is same as the fourth number. Find the numbers

Answer 1

Last three numbers are in AP.

Let last three numbers be $a-d,a,a+d$.

First and fourth number are same.

So, first number = $a+d$

Hence, series is $a+d,a-d,a,a+d$.

Now, first three numbers are in GP.

So, $(a-d{)}^2=(a+d)a$

Common difference, d=6

$(a-6{)}^2=(a+6)a$

$a^2+36-12a=a^2+6a$

$18a=36$

$a=2$

Hence, the numbers are $8,-4,2,8$.