Arithmetic /geometric progression

Question

Given a sequence $2, x , y , 9$ where the first three terms form an arithmetic progression and the last three terms form a geometric progression, determine the value(s) of $x$ and $y$. Need assistance with this question please.

Answer 1

Hint:

• In an arithmetic sequence $\dots,a-d,a,a+d,\dots$, one has $(a-d)+(a+d) = 2a$. In words, sum of surrounding terms is two times the sum of surrounded term.
• In a geometric sequence $\dots,ar^{-1},a,ar\dots$, one has $ar^{-1} \cdot ar = a^2$.

\begin{cases} 2+y&=2x & \text{(arithmetic progression)} \\ 9x&=y^2 & \text{(geometric progression)} \end{cases}

Answer 2

For A.P. $T_1=a=2.............$ $T_2=a+(2-1)d$ $=2+d.$ equation($1$).............. $T_3=a+(3-1)d$ $=2+2d.$ equation ($2$).............

For G.P. $x/y=y/9$........ $ y2= 9x$........ $4+8d+4d×d =9(2+d)$. .......

$4d×d-d-14=0$........ $d=-7/4$....... Or... $d=2$...... . Put in equation $1$ and $2$......

And you get ...... $x=4$ and $y=6$..... And..... $x=1/$4 and $y=-3/2$.....

Answer 3

First from the arithmetic progression we have $x-2=y-x$ which gives us $y=2x-2$. Second from the geometric progression we have $\frac{y}{x}=\frac{9}{y}$. We plug in $y=2x-2$ into this equation and get $$\frac{2x-2}{x}=\frac{9}{2x-2}.$$ Can you solve this to find $x$ and hence $y$?