# Geometric progression question. Year less?

Here is the math question.

A 100m cliff erodes by 2/7 of its height each year.

(a) What will the height of the cliff be after 10 years?

This is how I worked out the question.

100*(5/7)^10


Which is 3.46m to 2d.p.

However my teacher said that it is wrong (to the whole class and also insulted me a bit >_>) and that I didn't follow this formula.

Tn = ar^(n-1)


And that I should have done

100(5/7)^9


Which is instead 4.84m to 2d.p.

This doesn't make sense because if you used the formula and the question was

What will the height of the cliff be after 1 year?

 100(5/7)^(1-1)
=100(5/7)^0
=100


That doesn't make sense at all!

Am I right or is the teacher right?

ADDITIONALLY my teacher said his answer is an interpretation of the question. Is his answer a valid interpretation of this question? Or is it just incorrect mathematics?

BTW Also my teacher said my answer has no common sense and that I won't be able to do the HSC well if I keep reading questions wrong.

• Your argument looks OK. $\frac{5}{7}$ th is left after each year, so after $10$ years it should be $(\frac{5}{7})^{10}$ th left of the original. So you are $\color{blue}{right}$. And you have checked it out for the base case. +1 for laying it out so neatly and checking it. – Shailesh May 5 '16 at 1:07
• With the way you have phrased the question, I agree with your answer. It is always good to take formulae and check them against extreme cases (such as after no years elapse or after "infinitely many" years pass) to make sure they make sense. It is possible that the exact phrasing as it was originally given was slightly different (or intended to be different), which just highlights the importance of precise wording in mathematics. (for example, it could have been worded as "in 2001 the height was... what is the height in 2010?") – JMoravitz May 5 '16 at 1:09
Well, we agree that $T_{n+1} = (5/7) T_n,$ so $T_n = T_0 (5/7)^n$. Clearly, after $0$ years, the height of the tree is $100$, so you have $T_n = 100 (5/7)^n$.
Hence, after one year, we have $$T_1 = 100 (5/7)$$ and after 10 years, we have $$T_1 = 100 (5/7)^{10}$$ as you have claimed.