Question

Let p, q be the roots of the equation x2 - 2x + A = 0 andr, s be the roots of the
equation x2 - 18x + B = 0. If p <q <r<s are in AP, then the value of (A + B) is​

Answer 1

x² - 2x + A = 0 ___________ ( i )

It's roots are p & q

→ (p + q) = 2

And, → pq = A

x² - 18x + B = 0 ____________ ( ii )

It's roots are r & s

→ (r + s) = 18

And, → rs = B

Also given that p, q, r & s are in AP

So,

let p = a, q = (a + d), r = (a + 2d) &

s = (a + 3d)

Now,

→ p + q = 2

→ a + a + d = 2

→ 2a + d = 2 ______________ ( iii )

And,

→ r + s = 18

→ a + 2d + a + 3d = 18

→ 2a + 5d = 18 _____________ ( iv )

On solving ( iii) & ( iv ) we get;

a = -1 & d = 4

So, p = a = -1

q = a + d = -1 + 4 = 3

r = a + 2d = -1 + 2(4) = 7

s = a + 3d = -1 + 3(4) = 11

Now,

>> pq = A

>> (-1)(3) = A

>> A = -3

And,

>> rs = B

>> (7)(11) = B

>> B = 77

Hence,

(A + B) = -3 + 77 = 74 __________(Ans.)