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यदि समीकरण, $x ^{2}+5(\sqrt{2}) x +10=0$, के $\alpha$ तथा $\beta$, $\alpha>\beta$ दो मूल है तथा $P_{n}=\alpha^{n}-\beta^{n}$,( प्रत्येक धन पूर्णांक $n$ के लिए) है, तो $\left(\frac{ P _{17} P _{20}+5 \sqrt{2} P _{17} P _{19}}{ P _{18} P _{19}+5 \sqrt{2} P _{18}^{2}}\right)$ का मान है ............. |
$4$
$3$
$2$
$1$
Solution
$x^{2}+5 \sqrt{2} x+10=0$
$\& p_{n}=\alpha^{n}-\beta^{n} \text { (Given) }$
$\text { Now } \frac{P_{17} P_{20}+5 \sqrt{2} p_{11} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}=\frac{P_{17}\left(P_{20} 5 \sqrt{2} P_{19}\right)}{P_{18}\left(P_{19}+5 \sqrt{2 P}_{18}\right)}$
$\frac{P_{17}\left(\alpha^{20}-\beta^{20}+5 \sqrt{2}\left(\alpha^{19}-\beta^{19}\right)\right)}{P_{18}\left(\alpha^{19}-\beta^{19}+5 \sqrt{2}\left(\alpha^{18}-\beta^{18}\right)\right)}$
$\frac{P_{17}\left(\alpha^{19}(\alpha+5 \sqrt{2})-\beta^{19}(\beta+5 \sqrt{2})\right)}{P_{18}\left(\alpha^{18}(\alpha+5 \sqrt{2})-\beta^{18}(\beta+5 \sqrt{2})\right)}$
$\text { Since } \alpha+5 \sqrt{2}=-10 / \alpha \ldots \ldots \ldots \ldots . .(1)$
$\&\, \beta+5 \sqrt{2}=-10 / \beta \ldots \ldots \ldots \ldots(2)$
Now put there values in above expression
$=-\frac{10 P_{1} P_{18}}{-10 P_{18} P_{17}}=1$
