Question

The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

• A. 1st
• B. 9th
• C. 12th
• D. None of the above

Hint:

Write every term as the first term plus a multiple of the common difference, then simplify the given equation. See which term the leftover expression matches.

Answer 1

The Correct Option is C

Step 1: Understand the question.
In an arithmetic progression (AP) each term is the previous term plus a fixed number \(d\), called the common difference. If the first term is \(a\), the \(n\)th term is \(T_n = a + (n-1)d\).
We are told one relation between some terms, and we must find which term must be zero.

Step 2: Write each term.
\[ T_3 = a + 2d, \quad T_{15} = a + 14d \] \[ T_6 = a + 5d, \quad T_{11} = a + 10d, \quad T_{13} = a + 12d \]

Step 3: Use the given condition.
The sum of the 3rd and 15th terms equals the sum of the 6th, 11th and 13th terms: \[ (a + 2d) + (a + 14d) = (a + 5d) + (a + 10d) + (a + 12d) \]
Left side: \(2a + 16d\). Right side: \(3a + 27d\).
\[ 2a + 16d = 3a + 27d \implies 0 = a + 11d \]

Step 4: Read the result.
The 12th term is \(T_{12} = a + (12 - 1)d = a + 11d\). We just found \(a + 11d = 0\), so \[ T_{12} = 0 \]

Step 5: Why the other options fail.
Option 1: the 1st term is \(a\). From \(a = -11d\), it is zero only if \(d = 0\), so it is not necessarily zero. For instance \(a = -11, d = 1\) satisfies the condition and has first term \(-11\).
Option 2: the 9th term is \(a + 8d = -11d + 8d = -3d\), which is not zero unless \(d = 0\).
Option 4 (None of the above) is wrong because the 12th term is definitely zero.

Final Answer:
The 12th term must be zero, which is option 3. \[ \boxed{T_{12} = 0\ \text{(Option 3)}} \]