I donts need this french thing, solve the maths for me

Okay, boyufavour, let's tackle these arithmetic progression problems.

19. The 10th term of an A.P. is 28 and the 15th term is 43. Find the first term and the common difference.

The formula for the $n$-th term of an A.P. is $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.

Step 1: Set up a system of equations using the given information.
For the 10th term ($n=10$, $T_{10}=28$):

$$a + (10-1)d = 28$$ $$a + 9d = 28 \quad (1)$$

For the 15th term ($n=15$, $T_{15}=43$):

$$a + (15-1)d = 43$$ $$a + 14d = 43 \quad (2)$$

Step 2: Solve the system of equations for $d$.
Subtract equation (1) from equation (2):

$$(a + 14d) - (a + 9d) = 43 - 28$$ $$5d = 15$$ $$d = \frac{15}{5}$$ $$d = 3$$

Step 3: Substitute the value of $d$ into equation (1) to find $a$.

$$a + 9(3) = 28$$ $$a + 27 = 28$$ $$a = 28 - 27$$ $$a = 1$$

The first term is 1 and the common difference is 3.

$$\boxed{\text{First term } a = 1, \text{ Common difference } d = 3}$$

20. The sum of the first 10 terms of an A.P. is 120 and the sum of the next 5 terms is 140. Find the first term and the common difference.

The formula for the sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2}(2a + (n-1)d)$.

Step 1: Set up equations based on the given sums.
For the sum of the first 10 terms ($n=10$, $S_{10}=120$):

$$S_{10} = \frac{10}{2}(2a + (10-1)d) = 120$$ $$5(2a + 9d) = 120$$ $$2a + 9d = \frac{120}{5}$$ $$2a + 9d = 24 \quad (1)$$

The sum of the next 5 terms means $T_{11} + T_{12} + T_{13} + T_{14} + T_{15} = 140$.
This is equivalent to $S_{15} - S_{10} = 140$.
Since $S_{10} = 120$, we have:

$$S_{15} - 120 = 140$$ $$S_{15} = 140 + 120$$ $$S_{15} = 260$$

Now, use the formula for $S_{15}$ ($n=15$, $S_{15}=260$):

$$S_{15} = \frac{15}{2}(2a + (15-1)d) = 260$$ $$\frac{15}{2}(2a + 14d) = 260$$ $$15(a + 7d) = 260$$ $$a + 7d = \frac{260}{15}$$ $$a + 7d = \frac{52}{3} \quad (2)$$

Step 2: Solve the system of linear equations for $a$ and $d$.
From equation (2), express $a$ in terms of $d$:

$$a = \frac{52}{3} - 7d$$

Substitute this expression for $a$ into equation (1):

$$2\left(\frac{52}{3} - 7d\right) + 9d = 24$$ $$\frac{104}{3} - 14d + 9d = 24$$ $$\frac{104}{3} - 5d = 24$$ $$-5d = 24 - \frac{104}{3}$$ $$-5d = \frac{72}{3} - \frac{104}{3}$$ $$-5d = \frac{72 - 104}{3}$$ $$-5d = \frac{-32}{3}$$ $$d = \frac{-32}{3 \times -5}$$ $$d = \frac{32}{15}$$

Step 3: Substitute the value of $d$ back into the expression for $a$.

$$a = \frac{52}{3} - 7\left(\frac{32}{15}\right)$$ $$a = \frac{52}{3} - \frac{224}{15}$$

To subtract, find a common denominator (15):

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