Solve these ones

The sum of the first term is equal to the first term itself.

$$S_1 = 2(1)^2 + 3(1)$$ $$S_1 = 2(1) + 3$$ $$S_1 = 2 + 3$$ $$T_1 = 5$$

$$S_2 = 2(2)^2 + 3(2)$$ $$S_2 = 2(4) + 6$$ $$S_2 = 8 + 6$$ $$S_2 = 14$$

The second term is the sum of the first two terms minus the first term.

$$T_2 = S_2 - S_1$$ $$T_2 = 14 - 5$$ $$T_2 = 9$$

The common difference is the difference between any term and its preceding term.

$$d = T_2 - T_1$$ $$d = 9 - 5$$ $$d = 4$$

The first term is $\boxed{\textbf{5}}$ and the common difference is $\boxed{\textbf{4}}$.

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: Geometric Progression

The $n^{th}$ term of a geometric progression is given by $T_n = 3 \times 2^{n-1}$. Find the first term and the common ratio.

Substitute $n=1$ into the formula for the $n^{th}$ term.

$$T_1 = 3 \times 2^{1-1}$$ $$T_1 = 3 \times 2^0$$ $$T_1 = 3 \times 1$$ $$T_1 = 3$$

Substitute $n=2$ into the formula for the $n^{th}$ term.

$$T_2 = 3 \times 2^{2-1}$$ $$T_2 = 3 \times 2^1$$ $$T_2 = 3 \times 2$$ $$T_2 = 6$$

The common ratio is the ratio of any term to its preceding term.

$$r = \frac{T_2}{T_1}$$ $$r = \frac{6}{3}$$ $$r = 2$$

The first term is $\boxed{\textbf{3}}$ and the common ratio is $\boxed{\textbf{2}}$.

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