There is a sequence A(n) where n is a positive integer such that A(n+1

Hi

This is combination of geometric and arithmetic sequences.

n>0, min n = 1.

\(a_2 = 10 + \frac{1}{2}a_1\)

\(a_3 = 10 + \frac{1}{2}a_2 = 10 + \frac{1}{2}(10 + \frac{1}{2}a_1) = 10 + 5 + \frac{1}{4}a_1\)

\(a_4 = 10 + \frac{1}{2}a_3 = 10 + \frac{1}{2}(10 + 5 + \frac{1}{4}a_1) = 10 + 5 + \frac{5}{2} + \frac{1}{8}a_1\)

\(a_5 = 10 + \frac{1}{2}a_4 = 10 + 5 + \frac{5}{2} + \frac{5}{4} + \frac{1}{16}a_1\)

...

\(a_n = 10 + 5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{n-3}} + \frac{1}{2}^{n-1}a_1\)

When n=1000 our fraction \(\frac{1}{2^{999}}\) is close to 0.

\(5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{997}}\) We can apply logic of infinite geometric series where |r|<1 because our n is quite big (1000).

\(S = \frac{5}{1-1/2} = 5*2 = 10\)

\(a_{1000} ≈ 10 + 10 + 0 = 20\)

Answer C

Hope this helps

Regards