The following sequence is a combination of an arithmetic sequence a geometric sequence: $$ 5 ; 5 ; 15 ; 10 ; 25 ; 20 ; 35 ; 40 ; \ldots $$ 4.1 Write down the next 2 terms. 4.2 Calculate $$ T_{42}-\mathrm{I}_{41} $$. 4.3 Determine the sum of the first 50 terms. 4.4 Prove that ALL the terms of this infinite sequence will be div

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### 4.1 Write down the next 2 terms.

The given sequence is:
$$ 5, 5, 15, 10, 25, 20, 35, 40, \ldots $$

We can observe the pattern:
- The odd-indexed terms (1st, 3rd, 5th, 7th, ...) form an arithmetic sequence: $$ 5, 15, 25, 35, \ldots $$ with a common difference of $$ 10 $$.
- The even-indexed terms (2nd, 4th, 6th, 8th, ...) form another arithmetic sequence: $$ 5, 10, 20, 40, \ldots $$ with a common difference of $$ 5 $$ and then multiplied by $$ 2 $$.

Continuing the pattern:
- The next odd term after $$ 35 $$ is $$ 35 + 10 = 45 $$.
- The next even term after $$ 40 $$ is $$ 40 + 5 = 45 $$.

Thus, the next two terms are:
$$ 45, 45 $$

### 4.2 Calculate $$ T_{42} - I_{41} $$.

To find $$ T_{42} $$ and $$ I_{41} $$:
- $$ T_n $$ refers to the $$ n $$-th term of the odd-indexed sequence.
- $$ I_n $$ refers to the $$ n $$-th term of the even-indexed sequence.

From the patterns:
- The $$ n $$-th term of the odd-indexed sequence can be expressed as:
$$ T_n = 5 + (n-1) \cdot 10 = 10n - 5 $$
- The $$ n $$-th term of the even-indexed sequence can be expressed as:
$$ I_n = 5 + (n-1) \cdot 5 = 5n $$

Now, we need to calculate $$ T_{42} $$ and $$ I_{41} $$:
- $$ T_{42} $$ corresponds to the 21st term of the odd sequence (since odd terms are at positions 1, 3, 5, ...):
$$ T_{21} = 10 \cdot 21 - 5 = 210 - 5 = 205 $$
- $$ I_{41} $$ corresponds to the 41st term of the even sequence:
$$ I_{41} = 5 \cdot 41 = 205 $$

Now, we calculate:
$$ T_{42} - I_{41} = 205 - 205 = 0 $$

### 4.3 Determine the sum of the first 50 terms.

The first 50 terms consist of 25 odd terms and 25 even terms.

1. **Sum of the first 25 odd terms**:
   $$
   S_{odd} = \sum_{n=1}^{25} T_n = \sum_{n=1}^{25} (10n - 5) = 10 \sum_{n=1}^{25} n - 5 \cdot 25
   $$
   Using the formula for the sum of the first $$ n $$ natural numbers:
   $$
   \sum_{n=1}^{k} n = \frac{k(k+1)}{2}
   $$
   For $$ k = 25 $$:
   $$
   \sum_{n=1}^{25} n = \frac{25 \cdot 26}{2} = 325
   $$
   Thus,
   $$
   S_{odd} = 10 \cdot 325 - 125 = 3250 - 125 = 3125
   $$

2. **Sum of the first 25 even terms**:
   $$
   S_{even} = \sum_{n=1}^{25} I_n = \sum_{n=1}^{25} (5n) = 5 \sum_{n=1}^{25} n = 5 \cdot 325 = 1625
   $$

Now, the total sum of the first 50 terms:
$$
S_{50} = S_{odd} + S_{even} = 3125 + 1625 = 4750
$$

### 4.4 Prove that ALL the terms of this infinite sequence will be divisible by 5.

From the sequences:
- The odd terms are $$ 5, 15, 25, 35, \ldots $$ which can be expressed as $$ 5n $$ for $$ n = 1, 3, 5, \ldots $$. Clearly, all these terms are divisible by $$ 5 $$.
- The even terms are $$ 5, 10, 20, 40, \ldots $$ which can also be expressed as $$ 5m $$ for $$ m = 1, 2, 4, 8, \ldots $$. All these terms are also divisible by $$ 5 $$.

Since both sequences consist entirely of terms that are multiples of $$ 5 $$, we can conclude that all terms in the infinite sequence are divisible by $$ 5 $$.

### Summary of Answers:
- **4.1**: Next two terms are $$ 45, 45 $$.
- **4.2**: $$ T_{42} - I_{41} = 0 $$.
- **4.3**: Sum of the first 50 terms is $$ 4750 $$.
- **4.4**: All terms are divisible by $$ 5 $$.
- **4.1**: The next two terms are 45 and 45. - **4.2**: $$ T_{42} - I_{41} = 0 $$. - **4.3**: The sum of the first 50 terms is 4750. - **4.4**: All terms in the sequence are divisible by 5.

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