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: Understanding the problem:
We are given an arithmetic progression (AP) with the following conditions:
- The sum of the first three terms is 30.
- The sum of the last three terms is 36.
- The first term \( a = 9 \).
We are asked to find the number of terms in the AP.

Step 2: Sum of the first three terms:
The sum of the first three terms of an AP is given by the formula: \[ S_3 = \frac{n}{2} \left( 2a + (n-1)d \right) \] where \( n \) is the number of terms, \( a \) is the first term, and \( d \) is the common difference.
For the sum of the first three terms: \[ S_3 = a + (a + d) + (a + 2d) \] Substituting \( a = 9 \), we get: \[ S_3 = 9 + (9 + d) + (9 + 2d) = 30 \] Simplifying: \[ 27 + 3d = 30 \] \[ 3d = 3 \] \[ d = 1 \] Thus, the common difference \( d = 1 \).

Step 3: Sum of the last three terms:
Let the total number of terms in the AP be \( n \). The last three terms are \( a + (n-3)d \), \( a + (n-2)d \), and \( a + (n-1)d \). The sum of these last three terms is: \[ S_{\text{last 3}} = (a + (n-3)d) + (a + (n-2)d) + (a + (n-1)d) \] Substituting \( a = 9 \) and \( d = 1 \), we get: \[ S_{\text{last 3}} = (9 + (n-3)) + (9 + (n-2)) + (9 + (n-1)) \] Simplifying: \[ S_{\text{last 3}} = 3 \times 9 + (n-3 + n-2 + n-1) = 27 + 3n - 6 = 36 \] \[ 3n + 21 = 36 \] \[ 3n = 15 \] \[ n = 5 \] Thus, the number of terms in the AP is \( n = 5 \).

Conclusion:
The number of terms in the AP is \( 5 \).