Question

If $a<b<c<d$ are in A.P, then values of ${\alpha }_1$ & ${\alpha }_2$ respectively are:

• A. $-12,308$
• B. $12,308$
• C. $-12,6$
• D. None of these

Answer 1

The correct option is A $-12,308$

From the above equations,
$a+b=4$
$c+d=36$
Now,
$a+c=2b$
and $b+d=2c$
Therefore,
$a+b+c=3b$
$4+c=3b$
$3b-c=4$ ...(i)
Similarly,
$b+c+d=3c$
$3c-b=36...\left(ii\right)$
Hence, $b=6$
and $c=14$
Hence, $a=-2$ and $d=22$
Hence,
${\alpha }_1=ab=-12$
${\alpha }_2=cd=22\left(14\right)=308$