Question

The sum of the first three terms of an AP is 30 and the sum of the last three terms is 36. If the first term is 9, then the number of terms is :

• A. 10
• B. 5
• C. 6
• D. 13

Answer 1

The Correct Option is B

Step 1: Understand the given problem:
We are given an arithmetic progression (AP) where: - The sum of the first three terms is 30, - The sum of the last three terms is 36, - The first term \( a = 9 \). We need to find the number of terms in the AP.

Step 2: Use the formula for the sum of the first \( n \) terms of an AP:
The sum of the first \( n \) terms of an AP is given by the formula: \[ S_n = \frac{n}{2} \left( 2a + (n - 1) d \right) \] where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

Step 3: Sum of the first three terms:
The sum of the first three terms is 30, so we can use the formula for the sum of the first three terms: \[ S_3 = \frac{3}{2} \left( 2a + (3 - 1) d \right) = 30 \] Substitute \( a = 9 \) into the equation: \[ \frac{3}{2} \left( 2(9) + 2d \right) = 30 \] Simplify: \[ \frac{3}{2} \left( 18 + 2d \right) = 30 \] Multiply both sides by 2: \[ 3(18 + 2d) = 60 \] Simplify: \[ 54 + 6d = 60 \] \[ 6d = 6 \] \[ d = 1 \] Thus, the common difference \( d = 1 \).

Step 4: Sum of the last three terms:
The sum of the last three terms is 36. Let the total number of terms be \( n \), so the last three terms are the terms \( a_{n-2}, a_{n-1}, a_n \). The sum of these terms is: \[ a_{n-2} + a_{n-1} + a_n = 36 \] The general formula for the \( n \)-th term of an AP is \( a_n = a + (n-1)d \). Therefore: \[ a_{n-2} = a + (n-3)d, \quad a_{n-1} = a + (n-2)d, \quad a_n = a + (n-1)d \] Substitute these into the equation for the sum of the last three terms: \[ (a + (n-3)d) + (a + (n-2)d) + (a + (n-1)d) = 36 \] Simplify: \[ 3a + (3n - 6)d = 36 \] Substitute \( a = 9 \) and \( d = 1 \) into this equation: \[ 3(9) + (3n - 6)(1) = 36 \] \[ 27 + 3n - 6 = 36 \] \[ 3n + 21 = 36 \] \[ 3n = 15 \] \[ n = 5 \]

Step 5: Conclusion:
The number of terms in the AP is \( \boxed{5} \).