Q: The spool for a cassette tape has the circumference of 70 mm. The thickness of the tape makes each turn is 0.075mm longer than the pervious one. How many times will the spool turn playing a tape that is 54 metres long?

I have attempted to answer it

using the sum formula
$S_{n} = \frac{n}{2}(2a+(n-1)d)$
a = 70 d = common difference = 0.075

$S_{n} = \frac{n}{2}(140+(n-1)0.075) = 54$

$108000= 140000n + 75n^2-75n$

in the end I get n = -1866.4382or0.771523
which is clearly wrong

I cant seem to find the answer when I try to use the nth term

I just see this as

$\displaystyle \ell = \sum C_k = \sum^n_{k=0} (70+0.075k) = 54000$

$70(n+1) + 0.075 \dfrac{n(n+1)}{2} = 54000$

$0.0375 n^2 + (70+0.0375)n -53930= 0$

the positive solution is

$n=586.1~turns$

Originally Posted by romsek
the positive solution is
$n=586.1~turns$
Yer out by ~1: 587.0936477344342039884...

a = 1st term = 70
d = difference = .075
t = total = 54000

dn^2 + (2a - d)n - 2t = 0

To the corner for ~1 minute...
(enough time to eat 12 Easter eggs)

Originally Posted by DenisB
Yer out by ~1: 587.0936477344342039884...
$586 \to 53989.3 ~mm$

$587 \to 54103.4~mm$

$n$ = number of turns ... (?)