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 sum of 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 the same as the last one $a+d$

Therefore the four numbers are $a+d,a-d,a,a+d$

The first three of the above four are in G.P

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

But $d=6$ given

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

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

$\therefore a=2$

Putting for $a$ and $d$ the four numbers are

$8,-4,2,8$ which satisfies the given conditions.