This is a basic level question and it is homework for someone that I am trying to help out. It is indicated as a Fibonacci puzzle. But I am not able to fit the numbers into a general geometric formula. However, solving the question using very simple and basic algebra is not an issue.
"A fruit collector has x number of fruits on Monday. He has to throw (y+2)/2 each day through out the week. y is the beginning number of fruits for a day. On Sunday he is left with just 1 fruit. What was the total fruits he had when he began the week on Monday?"
I literally did 6 calculations to get it done, and it's a shame. I was not able to figure out a common radio at first degree, but a common ratio of second degree after finding the differences between terms. e.g.
T1 = 1, T2 = 4, T3 = 10, T4 = 22 $$4-1 = 3$$ $$10-4 = 6 = 3*2$$ $$22-10 = 12 = 6*2$$ $$46-22 = 24 = 12*2$$
I would appreciate if someone could point me out how to use $$T_n = ar^n-1$$, $$S_n = a(r^n-1)/(r-1)$$ in this scenario.
Using the given information, the number of fruit $n$ days from the end (where the last day is day $1$) satisfies the formula $$T_n=T_{n+1}-\frac{T_{n+1}+2}2=\frac{T_{n+1}}2-1\ ,$$ that is, $$T_{n+1}=2T_n+2\ .$$ A geometric sequence has the form $$T_{n+1}=rT_n\ ,$$ where $r$ is a fixed number, so this is not a geometric sequence and you cannot use the formulae you have quoted.
...at least, not directly. However you might notice that the formula can be written as $$T_{n+1}+2=2(T_n+2)\ ,$$ and so $T_n+2$ is a geometric sequence having ratio $r=2$ and initial term $T_1+2=3$. So a general formula will be $$T_n+2=3\times2^{n-1}$$ and therefore $$T_n=3\times2^{n-1}-2\ .$$