sequences and series - How do I solve the following exercise?

Let $a_1,a_2,a_3$ 3 numbers in arithmetic progression and $b_1,b_2,b_3$ 3 numbers in geometric progression so that $b_1\neq b_2$ and $a_1+a_2+a_3=b_1+b_2+b_3$. Show that if $a_1b_1,a_2b_2,a_3b_3$ are in arithmetic progression then $a_1=b_2$ and $a_1+b_1=a_3+b_3$.
Any idea how I should do it?

Answer

It isn't true.

Let $a_1 = A-k; a_2 = A; a_3 = A+k$

$b_1 = \frac Br; b_2 = B; b_3 = B*r$

$a_1 + a_2 + a_3 = 3A = b_1 + b_2 + b_3= B(1+ r + \frac 1r)$

And $a_2b_2 - a_1b_1 = a_3b_3 - a_2b_2$

$AB - (A-K)\frac Br = (A+K)Br - AB$

$2A = \frac {A-K}r + (A+K)r$

$(A+k)r^2 - 2Ar +(A-K) = 0$

$r = \frac {2A \pm \sqrt{4A^2 - 4(A+k)(A-k)}}{2(A+k)}$

$= \frac {A \pm k}{(A+k)}$

But $r \ne 1$ so $r = \frac {A-k}{A+k}$.

So $3A = B(1 + \frac {A-k}{A+k} + \frac {A+k}{A-k})$

$b_2 = B = \frac {3A}{1 + \frac {A-k}{A+k} + \frac {A+k}{A-k}}$

$= \frac {3A(A-K)(A+k)}{(A-k)(A+k) +(A-k)^2 + (A+k)^2}$

$=\frac {3A(A-k)(A+k)}{3A^2 +k^2}$

$= \frac {3A^3 - 3Ak^2}{3A^2 + k^2}$

$=\frac {3A^3 - Ak^2 - 2Ak^2}{3A^2 + k^2} = A - \frac {Ak^2}{3A^2 +k^2}$

.... which doesn't seem to work... I see no reason why $\frac {Ak^2}{3A^2 +k^2}$ should equal $k$ so that $A - \frac {Ak^2}{3A^2 +k^2} = A - k = a_1$.

So let $A = 5; k = 1$ and $r = \frac 23$ and $B =\frac {15}{1 + \frac 23 + \frac 32} = \frac {90}{19}$.

Then $4,5,6$ are arithmetic progression.

$\frac {135}{19}, \frac {90}{19}, \frac {60}{19}$ are in geometric progression.

$4 + 5+ 6 = 15$ and $\frac {135+90 + 60}{19} = 15$.

And $4*\frac {135}{19}, 5*\frac {90}{19}, 6*\frac {60}{19}$ are in arithmetic progression??? Lets see.

They are in arithmetic progression if $5*90 - 4*135 = 6*60 - 5*90$

$\iff 10*90 = 6*60 + 4*135\iff 900 = 360 + 540 = 900$.

So they are.

... And No. That isn't a true result.

(Unless $4 = \frac {90}{19}$ which , last time I checked it wasn't.)