A series in which some terms are in AP and other terms in GP In an increasing sequence of four positive integer, the first three terms are in AP, the last three terms are in GP and the fourth term exceed the first term by $30$, then the common difference of AP lying in the intervsl $[1,9]$ is:
MY ATTEMPT:
Let the series is $W,X,Y,Z$. 
The first three terms are $a-d,a,a+d$ and the last three terms are $b/r$,$b$,$br$. 
So we get $b=a+d$ and $$r=\frac{30 + a-d}{a+d}$$
Then I replaced $b$ and $r$ in terms of $a$ and $d$. 
I obtained a quadratic equation. After this I could not do anything as I used all the data given.

Let $a,b,c,d$ be the four positive integers.
The first three are in AP and the last three are in GP so we have
$$b-a=x,c-b=x,\dfrac{d}{c}=\dfrac{c}{b}\rightarrow bd=c^2$$
furthermore The 4th is the 1st plus 30, that is $d=a+30$
Let substitute the last information into the system
$$b-a=x;\;c-b=x;\;b(a+30)=c^2$$
we get 
$$c=b+x;\;a=b+x\rightarrow b(b+x+30)=(b+x)^2\rightarrow b^2+bx+30=b^2+2bx+x^2\rightarrow $$
$$b=\frac{x^2}{30-3x}$$
as the numbers are positive  integers the only value for $x$ is $x=9$ and the four terms are $a= 18,\;b= 27,\;c= 36,\;d= 48$
