3. An athlete runs 20 Km on a certain Monday. Thereafter, he increases the distance by $$ 10 \% $$ every day: Calculate the number of kilometres he ran: a) on the following Saturday. b) altogether over the 6 days. 4. Consider the following quadratic sequence 6 ; $$ x ; 26 ; 45 $$; $$ y $$ Calculate the values of $$ x $$ and $$ y $$.

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1. **Problem 3**

An athlete runs 20 km on Monday. Each subsequent day he increases his distance by $$10\%$$.  
   The distance run on the $$n$$th day is given by  
   $$
   d_n = 20 \times (1.1)^{n-1}.
   $$

**(a) Kilometres on the following Saturday**

Monday is day 1, so Saturday is day 6.  
   Calculate the distance on Saturday:
   $$
   d_6 = 20 \times (1.1)^{6-1} = 20 \times (1.1)^5.
   $$
   Evaluating $$(1.1)^5$$ (approximately):
   $$
   (1.1)^5 \approx 1.61051,
   $$
   hence,
   $$
   d_6 \approx 20 \times 1.61051 \approx 32.21 \text{ km}.
   $$

**(b) Total kilometres over the 6 days**

The total distance is the sum:
   $$
   S = \sum_{n=1}^{6} d_n = 20 \sum_{k=0}^{5} (1.1)^k.
   $$
   This is a finite geometric series with first term $$a = 1$$, common ratio $$r = 1.1$$ and number of terms $$6$$. The sum of a geometric series is:
   $$
   S_N = a \frac{r^{N} - 1}{r - 1}.
   $$
   Therefore,
   $$
   S = 20 \times \frac{(1.1)^6 - 1}{1.1 - 1}.
   $$
   Since $$1.1 - 1 = 0.1$$ and calculating $$(1.1)^6 \approx 1.771561$$, we get:
   $$
   S \approx 20 \times \frac{1.771561 - 1}{0.1} = 20 \times \frac{0.771561}{0.1} = 20 \times 7.71561 \approx 154.31 \text{ km}.
   $$

2. **Problem 4**

The quadratic sequence is given as:  
   $$
   6; \quad x; \quad 26; \quad 45; \quad y.
   $$
   In a quadratic sequence, the second differences are constant.

**Step 1: Identify the first differences**

Let the terms be:
   $$
   a_1 = 6,\quad a_2 = x,\quad a_3 = 26,\quad a_4 = 45,\quad a_5 = y.
   $$
   First differences:
   $$
   d_1 = a_2 - a_1 = x - 6,
   $$
   $$
   d_2 = a_3 - a_2 = 26 - x,
   $$
   $$
   d_3 = a_4 - a_3 = 45 - 26 = 19,
   $$
   $$
   d_4 = a_5 - a_4 = y - 45.
   $$

**Step 2: Compute the second differences**

Second differences:
   $$
   \Delta_1 = d_2 - d_1 = (26 - x) - (x - 6) = 26 - x - x + 6 = 32 - 2x,
   $$
   $$
   \Delta_2 = d_3 - d_2 = 19 - (26 - x) = 19 - 26 + x = x - 7,
   $$
   $$
   \Delta_3 = d_4 - d_3 = (y - 45) - 19 = y - 64.
   $$

Since the sequence is quadratic, the second differences are equal. Thus, we set:
   $$
   \Delta_1 = \Delta_2.
   $$
   This gives:
   $$
   32 - 2x = x - 7.
   $$
   Solve for $$x$$:
   $$
   32 + 7 = x + 2x,
   $$
   $$
   39 = 3x,
   $$
   $$
   x = 13.
   $$

Now using $$\Delta_2 = \Delta_3$$:
   $$
   x - 7 = y - 64.
   $$
   Substitute $$x = 13$$:
   $$
   13 - 7 = y - 64,
   $$
   $$
   6 = y - 64,
   $$
   $$
   y = 70.
   $$

**Final Answers:**
   - $$ x = 13 $$
   - $$ y = 70 $$
**(a) On Saturday, the athlete ran approximately 32.21 km.** **(b) Over the 6 days, the athlete ran a total of approximately 154.31 km.** **For the quadratic sequence:** - $$ x = 13 $$ - $$ y = 70 $$

An athlete runs 20 Km on a certain Monday. Thereafter,
he increases the distance by every day:
Calculate the number of kilometres he ran:
a) on the following Saturday.
b) altogether over the 6 days.

Consider the following quadratic sequence
6 ; ;
Calculate the values of and .

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An athlete runs 20 Km on a certain Monday. Thereafter, he increases the distance by every day: Calculate the number of kilometres he ran: a) on the following Saturday. b) altogether over the 6 days. Consider the following quadratic sequence 6 ; ; Calculate the values of and .