Question : The value of $\frac{1}{(p-n)(n-q)}+\frac{1}{(n-q)(q-p)}+\frac{1}{(q-p)(p-n)}$ is:
Option 1: $1$
Option 2: $0$
Option 3: $p + q + n$
Option 4: $\frac{2n}{p+q}$
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Answer (1)
25th Jan, 2024
Correct Answer: $0$
Solution :
Given:
$\frac{1}{(p-n)(n-q)}+\frac{1}{(n-q)(q-p)}+\frac{1}{(q-p)(p-n)}$
= $\frac{q-p+p-n+n-q}{(p-n)(n-q)(q-p)}$
= $0$
Hence, the correct answer is 0.
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