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Question
If a and b are the roots of x2 − 3x + p = 0 and c, d are the roots x2 − 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q − p) = 17 : 15.
Solution
We have,
a +b = 3, ab = p, c + d =12 and cd = q
a, b, c and d form a G.P.
∴ First term = a, b = ar, c = ar2 and d = ar3
Then, we have
a + b = 3 and c + d = 12
\[\Rightarrow a + ar = 3 \]
\[ \Rightarrow a( 1 + r ) = 3 . . . \left( i \right)\]
\[\text { Similarly, } a r^2 (1 + r) = 12 . . . \left( ii \right)\]
\[ \Rightarrow \frac{a r^2 \left( 1 + r \right)}{a\left( 1 + r \right)} = \frac{12}{3}\]
\[ \Rightarrow r^2 = 4 \]
\[ \Rightarrow r = 2\]
\[ \therefore a \left( 1 + r \right) = 3 \]
\[ \Rightarrow a = 1\]
\[\text { Now }, p = ab \]
\[ \Rightarrow p = a \times ar = 2\]
\[\text { And, } q = cd \]
\[ \Rightarrow q = a r^2 \times a r^3 = 2^5 = 32\]
\[ \therefore \frac{q + p}{q - p} = \frac{32 + 2}{32 - 2} = \frac{34}{30} = \frac{17}{15}\]
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