Find the possible values of $a, b, c$

Question

How would you approach this problem?

Let

$$10,\; a,\; b,\; c,\; 90$$

be consecutive terms of a sequence. Find the possible values of $a,\; b,\; c$

a) if the sequence is arithmetic.
b) if the sequence is geometric.

[Hint: What kind of sequences do the terms $10,\; b,\; 90$ form in each case?]

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EDIT:

Here's my working for part a -

Since $u_1=10$,

$u_1+4d=90$

$\therefore\; 10+4d=90$

Hence, $\;d=20$

So,

$a=30$
$b=50$
$c=70$

How should I approach part b?

Answer 1

a) $10,b,90$ form an arithmetic sequence so that $b=\dfrac{10+90}2$. You find $a,c,$ similarly.

b) $10,b,90$ form a geometric sequence so that $b=\sqrt{10\cdot90}$. You find $a,c,$ similarly.

Answer 2

You should approach it the exact same way.

You figured if the sequence were arithmetic that there is a $d$ so that $u_{k+1} = u_k + d$ so

$90 =u_5 = d + u_4 = 2d + u_3=3d+u_2 = 4d+u_1$

And $u_1 = 10$ so $4d = 90-10$ or $d = 20$.

If the sequence is geometric then there is an $r$ so that $u_{k+1} = r*u_k$.

So $90 = u_5 = r*u_4 = r^2 *u_3=r^3*u_2 = r^4*u_1$

And $u_1 =10$ so $r^4 =\frac {90}{10}$ or $r = \sqrt 3$. (or $-\sqrt 3$)

.....

I guess the hint is that if you had a three terms arithmetic sequence $10, b ,90$ then $b$ is the arithmetic average $\frac {10+90}2= 50$. And $10, a, 50$ so $a= \frac {10+50}2$ and $50, c, 90$ so $c = \frac {50+90}2$ can be easily solved.

And if you have a geometric sequence $10, b, 90$ then $b$ is the geometric average of $\sqrt{10*90} = 30$. And so $10, a,30$ would have $a = \sqrt{10*30}$ and $30,c, 90$ would have $c =\sqrt{30*90}$.

Great! I love that hint!

(You can view my edit history to see my inconsistancy....)