In any A.P. if sum of first six terms is 5 times the sum of next six terms then which term is zero?

Options

A. 10 th
B. 11 th
C. 12 th
D. 13 th

Explanation

To solve the problem, we need to find which term in the arithmetic progression (A.P.) is zero given that the sum of the first six terms is five times the sum of the next six terms.

### Step-by-Step Solution:

1. **Define the Terms of the A.P.:**
Let the first term of the A.P. be \( a \) and the common difference be \( d \). The \( n \)-th term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]

2. **Sum of the First Six Terms:**
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For the first six terms (\( n = 6 \)):
\[ S_6 = \frac{6}{2} \times (2a + 5d) = 3(2a + 5d) = 6a + 15d \]

3. **Sum of the Next Six Terms:**
The next six terms are the 7th to the 12th terms. We can calculate their sum:
\[ S_{7 \text{ to } 12} = T_7 + T_8 + T_9 + T_{10} + T_{11} + T_{12} \]
Where:
\[ T_7 = a + 6d, \quad T_8 = a + 7d, \quad T_9 = a + 8d, \quad T_{10} = a + 9d, \quad T_{11} = a + 10d, \quad T_{12} = a + 11d \]
Therefore:
\[ S_{7 \text{ to } 12} = (a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) + (a + 10d) + (a + 11d) = 6a + (6 + 7 + 8 + 9 + 10 + 11)d \]
The sum of the coefficients of \( d \) is:
\[ 6 + 7 + 8 + 9 + 10 + 11 = 51 \]
Thus:
\[ S_{7 \text{ to } 12} = 6a + 51d \]

4. **Set Up the Equation:**
According to the problem, the sum of the first six terms is five times the sum of the next six terms:
\[ 6a + 15d = 5(6a + 51d) \]

5. **Expand and Simplify:**
Expanding the right side:
\[ 6a + 15d = 30a + 255d \]
Rearranging gives:
\[ 6a + 15d - 30a - 255d = 0 \]
Simplifying this:
\[ -24a - 240d = 0 \]
Dividing through by -24:
\[ a + 10d = 0 \]

6. **Conclusion:**
The equation \( a + 10d = 0 \) implies that the first term \( a \) plus ten times the common difference \( d \) equals zero. This means:
\[ a = -10d \]
The term that is zero in the A.P. is the 11th term:
\[ T_{11} = a + 10d = 0 \]

### Final Answer:
The 11th term of the A.P. is zero.