Let \alpha and \beta   be the roots of equation

x^{2}-6x-2=0.\; if a_{n}=\alpha ^{n}-\beta ^{n},\; for\: n\geq 1,\; then\; the \: value\; of\; \frac{a_{10}-2a_{8}}{2a_{9}}

is equal to:

  • Option 1)

    6

  • Option 2)

    -6

  • Option 3)

    3

  • Option 4)

    -3

Answers (6)

x^2-6x-2=0\\ \alpha+\beta=6\\ \alpha.\beta=-2\\ Given, \frac{a_1_0-2a_8}{2a_9}, where a_n=\alpha^n-\beta^n\\ \frac{a_1_0-2a_8}{2a_9}=\frac{\alpha^1^0-\beta^1^0-2(\alpha^8-\beta^8)}{2(\alpha^9-\beta^9)}\\ Replace -2 with \alpha\beta\\ \frac{\alpha^1^0-\beta^1^0+(\alpha\beta)(\alpha^8-\beta^8)}{2(\alpha^9-\beta^9)}\\=\frac{\alpha^1^0-\beta^1^0+\alpha^9\beta-\alpha\beta^1^0}{2(\alpha^9-\beta^9)}\\ \frac{(\alpha+\beta)(\alpha^9-\beta^9)}{2(\alpha^9-\beta^9)}\\ =\frac{(\alpha+\beta)}{2}=\frac{6}{2}=3


Option 1)

6

Option 2)

-6

Option 3)

3

Option 4)

-3

probably option 3) 3

S surya

3

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