If for an $A . P .$ $a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .$
$a_{1} + a_{3} + a_{5} = - 12$ and $a_{1} a_{2} a_{3} = 8$
then the value of $a_{2} + a_{4} + a_{6}$ equals

- 12

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- 16

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- 18

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- 21

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Solution

The correct option is D $- 21$
$a_{1} + a_{3} + a_{5} = - 12$
$\Rightarrow a + a + 2 d + a + 4 d = - 12$
$\Rightarrow 3 a + 6 d = - 12$
$\Rightarrow a + 2 d = - 4 (1)$
$a_{1} . a_{2} . a_{3} = 8$
$\Rightarrow a \times (a + d) (a + 2 d) = 8$
$\Rightarrow a (a + d) \times - 4 = 8 (f r o m (1))$
$\Rightarrow a (a + d) = - 2$
$\Rightarrow a^{2} + a d = - 2$
$\Rightarrow a^{2} + a (\frac{- 4 - a}{2}) = - 2 (f r o m (1))$
$\Rightarrow 2 a^{2} - 4 a - a^{2} = - 4$
$\Rightarrow a^{2} - 4 a + 4 = 0$
$\Rightarrow {(a - 2)}^{2} = 0$
$\Rightarrow a - 2 = 0$
$\Rightarrow a = 2$
$a + 2 d = - 4 f r o m (1)$
$\Rightarrow 2 + 2 d = - 4$
$\Rightarrow 2 d = - 6$
$\Rightarrow d = - 3$
Now , $a_{2} + a_{4} + a_{6}$
$= a + d + a + 3 d + a + 5 d$
$= 3 a + 9 d$
$= 3 \times 2 + 9 \times - 3$
$= 6 - 27$
$= - 21$

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Similar questions

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 . 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}

\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}) =

a_{1}, a_{2}, a_{3}, \dots \dots

a_{3} + a_{5} + a_{8} = 11

a_{4} + a_{2} = - 2

a_{1} + a_{6} + a_{7}

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