If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5} a n d a_{6}$ are consecutive odd natural numbers such that $a_{6} > a_{1}, a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = p^{3} a n d a_{5} = 39$ , then the value of $(a_{3} + a_{4} - 12 p)$ is equal to

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Solution

The correct option is A 1
Given : $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ and $a_{6}$ are consecutive odd natural numbers.
$a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = P^{3}$
$a_{r} = 39 (a_{4} = a_{5} - 2, a_{3} = a_{4} - 2, a_{2} = a_{3} - 2, a_{1} = a_{2} - 2, a_{6} + 2)$
$∴ a_{1} = 31$ $a_{2} = 33$ $a_{3} = 35$ $a_{4} = 37$ $a_{6} = 39$
$31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^{3}$
$P = 6$
$a_{3} + a_{4} - 12 P = 35 + 37 - 12 \times 6$
$= 72 - 72$

$a_{3} + a_{4} - 12 P = 0$

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Similar questions

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_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

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_{1}, a_{2}, . . . . . . . . . . . . ., a_{50}

G . P .

\frac{a_{1} - a_{3} + a_{5} - . . . . . + a_{49}}{a_{2} - a_{4} + a_{6} - . . . . . + a_{50}}

\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

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