(i) The 3rd and 19th terms of an A.P. are 13 and 17 respectively. Find its 10th term. (ii) The 5th and 8th terms of an A.P. are 56 and 95 respectively. Find its 25th term.

Answer

Let's solve the problem step by step.

### Part (i)

**Given:**
- The 3rd term \( A_3 = 13 \)
- The 19th term \( A_{19} = 17 \)

**Step 1: Write the formulas for the terms.**
The formula for the nth term of an Arithmetic Progression (A.P.) is given by:
\[ A_n = A + (n - 1)D \]
where \( A \) is the first term and \( D \) is the common difference.

For the 3rd term:
\[ A_3 = A + (3 - 1)D = A + 2D = 13 \quad \text{(Equation 1)} \]

For the 19th term:
\[ A_{19} = A + (19 - 1)D = A + 18D = 17 \quad \text{(Equation 2)} \]

**Step 2: Set up the equations.**
From Equation 1:
\[ A + 2D = 13 \]
From Equation 2:
\[ A + 18D = 17 \]

**Step 3: Subtract Equation 1 from Equation 2.**
\[ (A + 18D) - (A + 2D) = 17 - 13 \]
This simplifies to:
\[ 16D = 4 \]
Thus:
\[ D = \frac{4}{16} = \frac{1}{4} \]

**Step 4: Substitute \( D \) back into Equation 1 to find \( A \).**
Substituting \( D \) into Equation 1:
\[ A + 2\left(\frac{1}{4}\right) = 13 \]
This simplifies to:
\[ A + \frac{1}{2} = 13 \]
So:
\[ A = 13 - \frac{1}{2} = \frac{26}{2} - \frac{1}{2} = \frac{25}{2} \]

**Step 5: Find the 10th term \( A_{10} \).**
Using the formula for the 10th term:
\[ A_{10} = A + (10 - 1)D = A + 9D \]
Substituting the values of \( A \) and \( D \):
\[ A_{10} = \frac{25}{2} + 9\left(\frac{1}{4}\right) = \frac{25}{2} + \frac{9}{4} \]
To add these fractions, convert \( \frac{25}{2} \) to have a common denominator:
\[ \frac{25}{2} = \frac{50}{4} \]
So:
\[ A_{10} = \frac{50}{4} + \frac{9}{4} = \frac{59}{4} = 14.75 \]

### Final Answer for Part (i):
The 10th term is \( 14.75 \).

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### Part (ii)

**Given:**
- The 5th term \( A_5 = 56 \)
- The 8th term \( A_8 = 95 \)

**Step 1: Write the formulas for the terms.**
For the 5th term:
\[ A_5 = A + (5 - 1)D = A + 4D = 56 \quad \text{(Equation 1)} \]

For the 8th term:
\[ A_8 = A + (8 - 1)D = A + 7D = 95 \quad \text{(Equation 2)} \]

**Step 2: Set up the equations.**
From Equation 1:
\[ A + 4D = 56 \]
From Equation 2:
\[ A + 7D = 95 \]

**Step 3: Subtract Equation 1 from Equation 2.**
\[ (A + 7D) - (A + 4D) = 95 - 56 \]
This simplifies to:
\[ 3D = 39 \]
Thus:
\[ D = \frac{39}{3} = 13 \]

**Step 4: Substitute \( D \) back into Equation 1 to find \( A \).**
Substituting \( D \) into Equation 1:
\[ A + 4(13) = 56 \]
This simplifies to:
\[ A + 52 = 56 \]
So:
\[ A = 56 - 52 = 4 \]

**Step 5: Find the 25th term \( A_{25} \).**
Using the formula for the 25th term:
\[ A_{25} = A + (25 - 1)D = A + 24D \]
Substituting the values of \( A \) and \( D \):
\[ A_{25} = 4 + 24(13) = 4 + 312 = 316 \]

### Final Answer for Part (ii):
The 25th term is \( 316 \).

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