Question

If the roots of cubic $x^3-6x^2-24x+c=0$ are the first 3 terms of an A.P. then sum to first 10 terms of the A.P. is

• A. - 250
• B. - 190
• C. 230
• D. 290

Answer 1

Answer: (B), (C)

Let the roots are $a-d,a,a+d$
$\therefore a-d+a+a+d=6\Rightarrow a=2$
again the product of roots taken 2 at a time
$a(a-d)+a(a+d)+(a+d)(a-d)=-24$
$a^2-ad+a^2+ad+a^2-d^2=-24$
$3a^2-d^2=-24$
$d^2=36\Rightarrow d=6\text{ or }-6$
Hence the AP's are
$-4,2,8,\dots \dots \dots \dots ;S_{10}=5[-8+9(6)]=(46)(5)=230\Rightarrow (C)$
or $8,2,-4,\dots \dots \dots ;S_{10}=5[16+9(-6)]=(5)(-38)=-190\Rightarrow (B)$.