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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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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Question : If $\frac{1}{p}+\frac{1}{q}=\frac{1}{p+q}$, the value of $\left (p^{3}-q^{3}\right )$ is:
Question : The value of $\frac{p^{2}- (q - r)^{2}}{(p + r)^{2} - (q)^{2}}$ + $\frac{q^{2} - (p - r)^{2}}{(p + q)^{2} - (r)^{2}}$ + $\frac{r^{2}- (p - q)^{2}}{(q + r)^{2} - (p)^{2}}$ is:
Question : If $p=9, q=\sqrt{17}$, then the value of $(p^2-q^2)^{-\frac{1}{3}}$ is equal to:
Question : The average of 5 consecutive integers starting with 'm' is n. What is the average of 6 consecutive integers starting with (m + 2)?
Question : The value of $ \frac{(p-q)^3+(q-r)^3+(r-p)^3}{12(p-q)(q-r)(r-p)}$, where $p \neq q \neq r$, is equal to:
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