NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.Misc.Q.18
If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0,
where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17: 15.
It is given that a and b are the roots of x2 – 3x + p = 0
So, a + b = 3 and ab = p ………………………………… (1)
Also, c and d are the roots of x2 – 12x + q = 0 So, c + d = 12 and cd = q … (2)
It is given that a, b, c, d is in G.P.
Let a = x, b = xr, c = xr2, d = xr3
From (1) and (2), we get
x + xr = 3
x (1 + r) = 3
xr2 + xr3 =12
xr2(1 + r) = 12
On dividing, we obtain
xr2(1 + r) ÷ x (1 + r) = 12 ÷ 3
r2 = 4
r = ±2
When r = 2, x = 3 ÷ (1 + 2) = 3 ÷ 3 = 1
When r = -2, x = 3 ÷ (1 - 2) = 3 ÷ (-1) = -3
Case 1:
When r = 2 and x =1, ab = x2 r = 2, cd = x2r5 = 32
So, (q + p) ÷ (q – p) = (32 + 2) ÷ (32 – 2) = 34 ÷ 30 = 17 ÷ 15
Hence, (q + p): (q – p) = 17: 15
Case 2:
When r = -2 and x = -3, ab = x2 r = -18, cd = x2r5 = -288
So, (q + p) ÷ (q – p) = (-288 - 18) ÷ (-288 + 18) = -306 ÷ (-270) = 17 ÷ 15
Hence, (q + p): (q – p) = 17: 15
Thus, in both the cases, we obtain (q + p): (q – p) = 17: 15