Let's try building out the first few integers in our sequence to see if we notice a pattern:
A(n+1) = 10 + 0.5A(n)
A(1) = A(0+1) --> Note: We can't solve for this because we need the value for A(0), which doesn't exist.
A(2)= A(1+2) $$=10+\frac{1}{2}A\left(1\right)$$
A(3) = A(2+1) $$=10+\frac{1}{2}A\left(2\right)=10+\frac{1}{2}\left(10+\frac{1}{2}A\left(1\right)\right)=10+5+\frac{1}{4}A\left(1\right)=15+\frac{1}{4}A\left(1\right)$$
A(4) = A(3+1) $$=10+\frac{1}{2}A\left(3\right)=10+\frac{1}{2}\left(15+\frac{1}{4}A\left(1\right)\right)=15+2.5+\frac{1}{8}A\left(1\right)=17.5+\frac{1}{8}A\left(1\right)$$
A(5) = A(4+1) $ $$=10+\frac{1}{2}A\left(4\right)=10+\frac{1}{2}\left(17.5+\frac{1}{8}A\left(1\right)\right)=17.5+1.25+\frac{1}{16}A\left(1\right)=18.75+\frac{1}{16}A\left(1\right)$$
At this point (or maybe a couple numbers back), we can probably recognize a pattern — for each number we increase, we add 10/(2^n-1) and reduce A(1) by 1/2. Knowing this, we can extend our pattern further:
$$A\left(6\right)=18.75\ +\ 0.625\ +\ \frac{1}{32}A\left(1\right)=19.375+\frac{1}{32}A\left(1\right)$$ $$A\left(7\right)=19.375\ +\ 0.3125\ +\ \frac{1}{64}A\left(1\right)=19.6875+\frac{1}{64}A\left(1\right)$$ $$A\left(8\right)=19.6875\ +\ 0.15625\ +\ \frac{1}{128}A\left(1\right)=19.84375+\frac{1}{124}A\left(1\right)$$
Now we should notice that 1) Our fraction of A(1) is becoming very small very fast, and 2) our number is increasing slower and slower as it approaches 20. Knowing this, we can extrapolate that by the time we reach A(1000), A(1) will be such a small fraction that it will be negligible, and our number will still be close to 20. So A(1000) will be closest to 20, or answer choice C.
