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

Problem Statement:
An arithmetic progression (AP) has the following properties:- The sum of its first three terms is 30.
- The sum of its last three terms is 36.
- Its first term is \( a = 9 \).
Determine the total number of terms in this AP.

Methodology:
Utilize the standard formula for the sum of the first \( n \) terms of an AP: \( S_n = \frac{n}{2} \left( 2a + (n - 1) d \right) \), where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

Calculation of Common Difference (d):
Given that the sum of the first three terms (\( S_3 \)) is 30, we apply the sum formula:\[S_3 = \frac{3}{2} \left( 2a + (3 - 1) d \right) = 30\]Substitute the known value of \( a = 9 \):\[\frac{3}{2} \left( 2(9) + 2d \right) = 30\]Simplify and solve for \( d \):\[\frac{3}{2} \left( 18 + 2d \right) = 30\]\[3(18 + 2d) = 60\]\[54 + 6d = 60\]\[6d = 6\]\[d = 1\]The common difference is \( d = 1 \).

Calculation of Number of Terms (n):
The sum of the last three terms is 36. If \( n \) is the total number of terms, these are the terms \( a_{n-2}, a_{n-1}, a_n \). Their sum is:\[a_{n-2} + a_{n-1} + a_n = 36\]Using the formula \( a_k = a + (k-1)d \), we express the last three terms:\[a_{n-2} = a + (n-3)d, \quad a_{n-1} = a + (n-2)d, \quad a_n = a + (n-1)d\]Substitute these into the sum equation:\[(a + (n-3)d) + (a + (n-2)d) + (a + (n-1)d) = 36\]Combine terms:\[3a + (3n - 6)d = 36\]Substitute \( a = 9 \) and \( d = 1 \):\[3(9) + (3n - 6)(1) = 36\]\[27 + 3n - 6 = 36\]\[3n + 21 = 36\]\[3n = 15\]\[n = 5\]

Result:
The arithmetic progression has \( \boxed{5} \) terms.