Find the sum of the geometric series 2 + 4 + 8 + 16 + ext... + 16384 and determine the number of terms.

$\frac{4}{2} = 2$
$\frac{8}{4} = 2$
$\frac{16}{8} = 2$
Kwa kuwa uwiano ni sawa (constant ratio), hii ni geometric series.
The type of series is a $\boxed{\text{Geometric Series}}$.

First term, $a = 2$.
Common ratio, $r = 2$.
Last term, $T_n = 16384$.

$T_n = a \cdot r^{n-1}$.

Tunajua $2^{10} = 1024$.
$2^{11} = 2048$
$2^{12} = 4096$
$2^{13} = 8192$
$2^{14} = 16384$
Hivyo, $n = 14$.
The number of terms in the series is $\boxed{14}$.

$S_n = \frac{a(r^n - 1)}{r - 1}$ (kwa sababu $r > 1$).

The sum of terms in the series is $\boxed{32766}$.

B) Given that the sum of the first n terms of the sequence $3 + 7 + 11 + 15 + \dots$ is 820, find the last term of the sequence.

Angalia tofauti kati ya namba zinazofuatana:
$7 - 3 = 4$
$11 - 7 = 4$
$15 - 11 = 4$
Kwa kuwa tofauti ni sawa (common difference), hii ni arithmetic sequence.
First term, $a = 3$.
Common difference, $d = 4$.
Sum of n terms, $S_n = 820$.

$S_n = \frac{n}{2}(2a + (n-1)d)$.

Hapa, $a=2$, $b=1$, $c=-820$.

Kuna majibu mawili:
$n_1 = \frac{-1 + 81}{4} = \frac{80}{4} = 20$
$n_2 = \frac{-1 - 81}{4} = \frac{-82}{4} = -20.5$
Kwa kuwa idadi ya terms haiwezi kuwa hasi au sehemu, $n = 20$.

$T_n = a + (n-1)d$.

The last term of the sequence is $\boxed{79}$.