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If a and b are the roots of are 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.

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Question

If a and b are the roots of are 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.

Sum
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Solution

It is given that a and b are the roots of x– 3x + p = 0

∴ a + b = 3 and ab = p … (1)

Also, c and d are the roots of  x2 – 12x + q = 0

∴ c + d = 12 and cd = q … (2)

It is given that a, b, c, d are in G.P.

Let a = x, b = xr, c = xr2, d = xr3

From (1) and (2), we obtain

x + xr = 3

⇒ x (1 + r) = 3

xr2 + xr3 =12

⇒ xr(1 + r) = 12

On dividing, we obtain

`(x^2 (1 + r))/(x (1 + r)) = (12)/(3)`

= r2 = 4

= r = ±2

When r = 2, `x = 3/(1 + 2) = 3/2 = 1`

When r = -2, `x = 3/(1 - 2) = 3/(-1) = -3`

Case I:

When r = 2 and x = 1

ab = x2 r = 2

cd = x2 r5 = 32

∴ `(q + p)/(q - p) = (32 + 2)/(32 - 2) = 34/30 = 17/15`

i.e. (q + p) : (q - p) = 17 :15

Case II:

When r = -2, x = -3

ab = x2 r = -18

cd =  x2 r5 = -288

∴ `(q + p)/(q - p) = (-288 - 18)/(-288 + 18) = (-306)/(-270) = 17/15`

i.e., (q + p) : (q - p) = 17 : 15

Thus, in both the cases, we obtain (q+p) : (q − p) = 17 : 15

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Chapter 8: Sequences and Series - Miscellaneous Exercise [Page 148]

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NCERT Mathematics [English] Class 11
Chapter 8 Sequences and Series
Miscellaneous Exercise | Q 9. | Page 148

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