Geometric progression where first term a = 128, common ratio r = ⅔

Length of nth piece = p cm

= arⁿ⁻¹ cm

= 128 (⅔)ⁿ⁻¹ cm


So,

128 (⅔)ⁿ⁻¹ = p

(2⁷)(2ⁿ⁻¹) / (3ⁿ⁻¹) = p

(2⁶⁺ⁿ)(3¹⁻ⁿ) = p


ln p = ln (2⁶⁺ⁿ)(3¹⁻ⁿ)

ln p = ln 2⁶⁺ⁿ + ln 3¹⁻ⁿ

ln p = (6 + n)ln2 + (1 - n)ln3

ln p = (n + 6)ln2 + (-n + 1)ln3


A = 1, B = 6, C = -1, D = 1

Since |r| < 1, the series converges.

So there is a sum to infinity.


Total length of string means we cut an infinite number of pieces and we see what the total length is.


Sum to infinity

= a/(1 - r)

= 128/(1 - ⅔)

= 128/(⅓)

= 128 × 3

= 384


So the maximum length we can ever cut in total is 384cm, never more than that.

How many pieces must be cut off?

Sum of n pieces = Sn

We want Sn > 380


a(1 - rⁿ)/(1 - r) > 380

128(1 - (⅔)ⁿ)/(1 - ⅔) > 380

384(1 - (⅔)ⁿ) > 380

1 - (⅔)ⁿ > 380/384

1 - 380/384 > (⅔)ⁿ

1/96 > (⅔)ⁿ


Take ln on both sides,


ln(1/96) > ln(⅔)ⁿ

ln(1/96) > n ln(⅔)

ln(1/96) / ln(⅔) < n

(Flip the sign because we are dividing by ln(⅔) which is a negative value)

11.2570677 < n


Since n > 11.2570677 , minimum integer value of n = 12


So 12 pieces must be cut before it exceeds 380 cm