In an AP, given a₃ = 15, S₁₀ = 125, find d and a₁₀.

To solve the problem involving an arithmetic progression (AP), we need to find the common difference \(d\) and the 10th term \(a_{10}\) given the equations \(a_3 = 15\) and \(S_{10} = 125\).

Understanding the Terms

In an arithmetic progression, the \(n\)th term can be expressed as:

a_n = a + (n - 1)d

where \(a\) is the first term and \(d\) is the common difference. The sum of the first \(n\) terms is given by:

S_n = \frac{n}{2} \times (2a + (n - 1)d)

Step 1: Using the First Equation

From the first equation \(a_3 = 15\):

a + 2d = 15 (1)

Step 2: Using the Second Equation

From the second equation \(S_{10} = 125\):

S_{10} = \frac{10}{2} \times (2a + 9d) = 125

This simplifies to:

5(2a + 9d) = 125

Dividing both sides by 5 gives:

2a + 9d = 25 (2)

Step 3: Solving the System of Equations

Now we have two equations:

• (1) \(a + 2d = 15\)
• (2) \(2a + 9d = 25\)

From equation (1), we can express \(a\) in terms of \(d\):

a = 15 - 2d

Substituting this into equation (2):

2(15 - 2d) + 9d = 25

This simplifies to:

30 - 4d + 9d = 25

Combining like terms gives:

5d = 5

Thus, we find:

d = 1

Step 4: Finding the First Term

Now, substitute \(d\) back into equation (1) to find \(a\):

a + 2(1) = 15

This leads to:

a = 13

Step 5: Calculating the 10th Term

Now that we have both \(a\) and \(d\), we can find the 10th term \(a_{10}\):

a_{10} = a + 9d = 13 + 9(1) = 22

Final Results

The common difference \(d\) is 1, and the 10th term \(a_{10}\) is 22.