Question

Let $p$ and $q$ be the roots of the equation $x^2- 2x + A = 0$ and let $r$ and $s$ be the roots of the equation $x^2- 18x + B = 0.$ if $p < q < r < s$ are in arithmetic progression, then $A =$ ... and $B =$ ....

Answer 1

Given, $p + q = 2, pq = A$
and $ r + s = 18, rs = B$
and it is given that $p, q, r, s$ are in an AP.
Therefore, let $p = a - 3d, q = a - d, r = a + d$
and $ s = a + 3d$
Since, $ p < q < r < s$
We have, $ d > 0 $
Now, $ 2 = p + q = a - 3d + a - d = 2a - 4d$
$\Rightarrow a- 2d =1 $ ...(i)
Again, $ 18 = r + s = a + d + a + 3d $
$ 18 = 2a + 4d$
$\Rightarrow 9 = a + 2d $ ...(ii)
On substracting Eq. (i) from Eq. (ii). we get
$ 8 = 4d $
$\Rightarrow d = 2 $
On putting in Eq. (ii), we get $a =5$
$\therefore p = a - 3d = 5 - 6 = -1$
$ q = a - d = 5 - 2 = 3$
$ r = a + d = 5 + 2 = 7$
and $ s = a + 3d = 5 + 6 = 11$
Therefore, $A = pq = - 3 $ and $B = rs = 77$