Divide : $(51 m^{3} p^{2} - 34 m^{2} p^{3})$ by $17 m p$

m^{2} p

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m p^{2}

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3 m^{2} p - 2 m p^{2}

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m p

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Solution

The correct option is C $3 m^{2} p - 2 m p^{2}$
$\frac{51 m^{3} p^{2} - 34 m^{2} p^{3}}{17 m p}$

$= \frac{17 m p (3 m^{2} p - 2 m p^{2})}{17 m p}$

$= 3 m^{2} p - 2 m p^{2}$

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Similar questions

p^{11} + p^{9} + p^{7} + p^{5} + p^{3} + p^{2}

S_{n} = n^{2} p

S_{m} = m^{2} p, m \neq n

S_{p} = p^{3}

S_{n} = n^{2} p

S_{m} = m^{2} p

m \neq n

S_{n} = p^{3}

If $S_{n} = n^{2} p$ and $S_{m_{3}} = m^{2} p, m \neq n$ , in an A.P., prove that $S_{p} = p^{3}$ .

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