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This problem involves a sequence of numbers in an Arithmetic Progression (A.P.).

Given:
- The sum of the number of shells in the 3rd and 11th cell is 68.
- The number of shells in the 11th cell is 24 more than that in the 3rd cell.

Let the first term of the A.P. be \( a \) and the common difference be \( d \).

1. The number of shells in the 3rd cell is given by \( a + 2d \).
2. The number of shells in the 11th cell is \( a + 10d \).

From the given information:
- The sum of the number of shells in the 3rd and 11th cells is 68:
  \[
  (a + 2d) + (a + 10d) = 68
  \]
  Simplifying this equation:
  \[
  2a + 12d = 68
  \]
  \[
  a + 6d = 34 \quad \text{(Equation 1)}
  \]

- The number of shells in the 11th cell is 24 more than in the 3rd cell:
  \[
  (a + 10d) = (a + 2d) + 24
  \]
  Simplifying this equation:
  \[
  a + 10d = a + 2d + 24
  \]
  \[
  8d = 24
  \]
  \[
  d = 3 \quad \text{(Equation 2)}
  \]

Substituting \( d = 3 \) into Equation 1:
\[
a + 6(3) = 34
\]
\[
a + 18 = 34
\]
\[
a = 16
\]

So, the first term \( a = 16 \) and the common difference \( d = 3 \).

Now, let's address the sub-questions:

### (i) Difference between the number of sea shells in the 19th and 20th cells
- The number of shells in the 19th cell is \( a + 18d = 16 + 18(3) = 16 + 54 = 70 \).
- The number of shells in the 20th cell is \( a + 19d = 16 + 19(3) = 16 + 57 = 73 \).
  
The difference is:
\[
73 - 70 = 3
\]

### (ii) Number of sea shells in the first cell
The number of sea shells in the 1st cell is \( a = 16 \).

### (iii) Total number of sea shells in the first 5 cells
- The number of shells in the 1st cell is \( a = 16 \).
- The number of shells in the 2nd cell is \( a + d = 16 + 3 = 19 \).
- The number of shells in the 3rd cell is \( a + 2d = 16 + 6 = 22 \).
- The number of shells in the 4th cell is \( a + 3d = 16 + 9 = 25 \).
- The number of shells in the 5th cell is \( a + 4d = 16 + 12 = 28 \).

Thus, the total number of shells in the first 5 cells is:
\[
16 + 19 + 22 + 25 + 28 = 110
\]

Would you like further details on any part of the solution? Here are 5 related questions to practice:

1. What is the number of sea shells in the 15th cell?
2. How many total sea shells are in the first 10 cells?
3. What is the difference between the number of sea shells in the 25th and 30th cells?
4. If the common difference were changed to 4, how would that affect the number of shells in the 1st and 5th cells?
5. Can you derive a general formula for the sum of the first \( n \) terms in this sequence?

**Tip:** In an A.P., the general term formula for the \(n\)-th term is \(a_n = a + (n-1)d\).

In a board game, the number of sea shells in various cells forms an A.P. If the number of sea shells in the 3rd and 11th cell together is 68 and the number of shells in the 11th cell is 24 more than that of the 3rd cell, then answer the following questions based on this data. (i) What is the difference between the number of sea shells in the 19th and 20th cells? (ii) How many sea shells are there in the first cell? (iii) Altogether, how many sea shells are there in the first 5 cells?

This problem involves solving an arithmetic progression (A.P.) related to a board game. Learn how to find the number of sea shells in specific cells, the difference between consecutive terms, and the sum of terms using A.P. concepts. Perfect for Grades 9-11.

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