Since 64/3 = 21 + 1/3 > 21, I assume S is supposed to be the value of the infinite sum. So we have for some constants a and r (where |r | < 1),
[tex]S = \displaystyle \sum_{n=1}^\infty ar^{n-1} = \frac{64}3 \\\\ S_3 = \sum_{n=1}^3 ar^{n-1} = 21[/tex]
Consider the k-th partial sum of the series,
[tex]S_k = \displaystyle \sum_{n=1}^k ar^{n-1} = a \left(1 + r + r^2 + \cdots + r^{k-1}\right)[/tex]
Multiply both sides by r :
[tex]rS_k = a\left(r + r^2 + r^3 + \cdots + r^k\right)[/tex]
Subtract this from [tex]S_k[/tex]:
[tex](1 - r)S_k = a\left(1 - r^k\right) \implies S_k = a\dfrac{1-r^k}{1-r}[/tex]
Now as k goes to ∞, the r ᵏ term converges to 0, which leaves us with
[tex]S = \displaystyle \lim_{k\to\infty}S_k = \frac a{1-r} = \frac{64}3[/tex]
which we can solve for a :
[tex]\dfrac a{1-r} = \dfrac{64}3 \implies a = \dfrac{64(1-r)}3[/tex]
Meanwhile, the 3rd partial sum is given to be
[tex]\displaystyle S_3 = \sum_{k=1}^3 ar^{n-1} = a\left(1+r+r^2\right) = 21[/tex]
Substitute a into this equation and solve for r :
[tex]\dfrac{64(1-r)}3 \left(1+r+r^2\right) = 21 \\\\ \dfrac{64}3 (1 - r^3) = 21 \implies r^3 = \dfrac1{64} \implies r = \dfrac14[/tex]
Now solve for a :
[tex]a\left(1 + \dfrac14 + \dfrac1{4^2}\right) = 21 \implies a = 16[/tex]
It follows that
[tex]S_5 = a\left(1 + r + r^2 + r^3 + r^4\right) \\\\ S_5 = 16\left(1 + \dfrac14 + \dfrac1{16} + \dfrac1{64} + \dfrac1{256}\right) = \boxed{\frac{341}{16}} = 21 + \dfrac5{16} = 21.3125[/tex]
