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Q.
If the roots of cubic $x^3-6 x^2-24 x+c=0$ are the first 3 terms of an A.P. then sum to first 10 terms of the A.P. is
Sequences and Series
Solution:
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 $
$a^2-a d+a^2+a d+a^2-d^2=-24$
$3 a^2-d^2=-24 $
$d^2=36 \Rightarrow d=6 \text { or }-6$
Hence the AP's are
$-4, 2, 8, \ldots \ldots \ldots \ldots ; S_{10}=5[-8+9(6)]=(46)(5)=230 \Rightarrow(C)$
or $8, 2, -4, \ldots \ldots \ldots ; S_{10}=5[16+9(-6)]=(5)(-38)=-190 \Rightarrow(B)$.