Question

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 to

• A. $-12$
• B. $-16$
• C. $-18$
• D. $-21$

Answer 1

Answer: (C)

$-18$

Given:$a_1+a_3+a_5=-12$

$a+a+2d+a+4d=-12$

$\Rightarrow 3a+6d=-12$

$\Rightarrow a+2d=-4$ .......$\left(1\right)$

$a_1{a}_2{a}_3=8$

$\Rightarrow a(a+d)(a+2d)=8$

$\Rightarrow a(a+d)\times -4=8$ from $\left(1\right)$

$\Rightarrow a(a+d)=-2$

$\Rightarrow (-4-2d)(-4-2d+d)=-2$

$\Rightarrow -2(2+d)(-d-4)=-2$

$\Rightarrow (d+2)(d+4)=-1$

$\Rightarrow d^2+6d+8+1=0$

$\Rightarrow d^2+6d+9=0$

$\Rightarrow d^2+2\times 3d+9=0$

$\Rightarrow {(d+3)}^2=0$

$\Rightarrow d=-3$

Put $d=-3$ in $\left(1\right)$

$a+2\times -3=-4$

$\Rightarrow a=-4+6=2$

$\therefore a=2,d=-3$

$a_2+a_4+a_6=a+d+a+3d+a+5d=3a+8d=3\times 2+8\times -3=6-24=-18$