Question

The 5th term of an AP is 20 and the 12th term is 41. Find the first term.

• A. 7
• B. 8
• C. 9
• D. 10

Hint:

The \(n\)-th term of an arithmetic progression is \( a_n = a + (n-1)d \). Use the difference of terms to find the common difference and then substitute back to find the first term.

Answer 1

The Correct Option is B

To solve the problem, we will use the formula for the nth term of an arithmetic progression (AP):

\(a_n = a + (n-1)d\)

where \(a\) is the first term, \(d\) is the common difference, and \(a_n\) is the nth term.

Given:

• The 5th term (\(a_5\)) is 20: \(a + 4d = 20\)
• The 12th term (\(a_{12}\)) is 41: \(a + 11d = 41\)

We have two equations:

1. \(a + 4d = 20\)
2. \(a + 11d = 41\)

Subtract the first equation from the second to eliminate \(a\):

\((a + 11d) - (a + 4d) = 41 - 20\)

\(7d = 21\)

Solve for \(d\):

\(d = 3\)

Substitute \(d = 3\) back into the first equation to find \(a\):

\(a + 4(3) = 20\)

\(a + 12 = 20\)

Solving for \(a\):

\(a = 20 - 12\)

\(a = 8\)

Therefore, the first term of the AP is 8. The correct answer is: 8

Answer 2

Step 1: Let the first term be \( a \) and common difference be \( d \). The \(n\)-th term of an AP is given by: \[ a_n = a + (n-1)d \] Given, \[ a_5 = a + 4d = 20 \quad \cdots (1) \] \[ a_{12} = a + 11d = 41 \quad \cdots (2) \] Step 2: Subtract equation (1) from (2): \[ (a + 11d) - (a + 4d) = 41 - 20 \implies 7d = 21 \implies d = 3 \] Step 3: Substitute \( d = 3 \) into equation (1): \[ a + 4 \times 3 = 20 \implies a + 12 = 20 \implies a = 8 \] Hence, the first term is \( {8} \).