If a 2 b 3 c 4 - abc 3 a 3 bc -1 2 - a m b n c p then m-n-p has the value equal to

The correct option is B 54 #160; ( a^2b^3c )^4( abc )^3( a^3bc^ 1)^2=√(a^mb^nc^p) #160; ⇒a^8+3 6b^12+3 2c^4+3+2=a^m/2b^n/2c^p/2 #160; ...

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Square Root of a Number

If a 2 b 3...

Question

If $\frac{{(a^{2} b^{3} c)}^{4} . {(a b c)}^{3}}{{(a^{3} {b c}^{- 1})}^{2}} = \sqrt{a^{m} b^{n} c^{p}}$ then $m + n + p$ has the value equal to

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Solution

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\frac{(a^{2} b^{3} c)^{4} . (a b c)^{3}}{(a^{3} b c^{- 1})^{2}} = \sqrt{a^{m} b^{n} c^{p}}

m + n + p

a^{2} + \frac{1}{a^{2}} = 34

a^{3} + \frac{1}{a^{3}}

a_{1}, a_{2}, a_{3}, . . . . . . . . . ., a_{n}

A . P .

(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + . . . . . . . + (n^{2} - a_{n}) = \frac{(n - 1) n (n + 1)}{3}

(\frac{a_{5} + a_{3} - a_{2}}{6})

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

A_{i} = \frac{x - a_{i}}{| x - a_{i} |}, i = 1, 2, . . . . ., n,

a_{1} < a_{2} < a_{3} < . . . . . < a_{n} .

1 \leq m \leq n, m \in N,

R = lim x \to a_{m}^{+} (A_{1} A_{2} . . . . A_{n})

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