Question

The sum of the 6th and 15th elements of an arithmetic progression is equal to the sum of 7th, 10th and 12th elements of the same progression. Which element of the series should necessarily be equal to zero?

• A. 10th
• B. 8th
• C. 1st
• D. None of these

Answer 1

Answer: (B)

8th

Answer 2

The correct option is (B): 8th
Explanation:Let the arithmetic progression (AP) be represented as $A_1, A_2, A_3, \ldots$ with the first term $a$ and common difference $d$. The $n$th term of an AP is given by $A_n = a + (n-1)d$.
According to the problem, we need to equate the sums of specific terms:
$A_6 + A_{15} = A_7 + A_{10} + A_{12}$
This can be expressed as:
$(a + 5d) + (a + 14d) = (a + 6d) + (a + 9d) + (a + 11d)$
Simplifying both sides:
$a + 19d = 3a + 26d$
Rearranging the equation gives:
$a + 19d - 3a - 26d = 0$
This leads to:
$-2a - 7d = 0 \quad \Rightarrow \quad 2a + 7d = 0$
From this, we find:
$a = -\frac{7}{2}d$
Now, to determine the 8th term:
$A_8 = a + 7d = -\frac{7}{2}d + 7d = \frac{7}{2}d$
For $A_8$ to equal zero, $d$ must be set to zero. Therefore, the 8th term $A_8$ is necessarily zero.
Thus, the answer is Option B: 8th term.