Question

Question 18 If a and b are the roots of are roots of , where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15.

Class X1 - Maths -Sequences and Series Page 199

Answer 1

a and b are roots of x² - 3x + P = 0
so, sum of roots = - ( -3)/1 = 3
a + b = 3 --------(1)

and product of roots = P
ab = P ------------(2)

again, c and d are roots of x² - 12x + q = 0
sum of roots = - (-12)/1 = 12
c + d = 12 -----------(3)

product of roots = q
cd = q --------------(4)

A/C to question,
a , b , c , d are in GP
Let b = ar , c = ar² , d = ar³
put these values in equations (1) and (3) , then divide equation (1) by equation (3)
(a + ar)/(ar² + ar³) = 3/12
a(1 + r)/ar²(1 + r ) = 1/4
1/r² = 1/4
r² = (2)²
r = ±2

when, r = 2 by equation (1)
a + 2a = 3
a = 1 , b = ar = 1 × 2 = 2, c = ar² = 1 × 2² = 4
d = ar³ = 1 × 2³ = 8
from equation (3), P = ab = 1 × 2 = 2
from equation (4), q = cd = 4 × 8 = 32
so, ( q + p)/(q - p) = (32 + 2)/(32 - 2)
= 34/30 = 17/15
Hence, (q + p ) : ( q - p ) = 17 : 15

when, r = -2 by equation (1)
a - 2a = 3 , a = -3
a = -3 , b = ar = -3 × -2 = 6, c = ar² = -3 × 4 = -12
d = ar³ = -3 × -8 = 24
from equation (3), P = ab = -3 × 6 = -18
from equation (4), q = cd = -12 × 24 = -288
so, (q + p)/(q - p) = ( -18 - 288)/(-288 + 18)
= -306/-270
= 17/15
Hence, (q + p) : ( q - p) = 17 : 15