JEE mathematics

If α and β are the roots of the equation x² - 6x - 2 = 0, and if an = αⁿ - βⁿ for n ≥ 1, find the value of (a₁₀ - — JEE Mathematics

SSitaKhadka16 Asked 24d ago 727 views 1 answer

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 6x - 2 = 0$, and if $a_n = \alpha^n - \beta^n$ for $n \ge 1$, find the value of $\frac{a_{10} - 2a_8}{2a_9}$.

1 Answer

AAaravMehta91 ✓ Accepted · 23d ago ▲ 23

Since $\alpha$ and $\beta$ are the roots of $x^2 - 6x - 2 = 0$, they satisfy the equation:

$$\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^2 - 2 = 6\alpha$$

$$\beta^2 - 6\beta - 2 = 0 \implies \beta^2 - 2 = 6\beta$$

Multiply the first equation by $\alpha^8$ and the second equation by $\beta^8$:

$$\alpha^{10} - 2\alpha^8 = 6\alpha^9$$

$$\beta^{10} - 2\beta^8 = 6\beta^9$$

Subtracting the second equation from the first:

$$(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8) = 6(\alpha^9 - \beta^9)$$

By definition, $a_n = \alpha^n - \beta^n$, so:

$$a_{10} - 2a_8 = 6a_9$$

Dividing both sides by $2a_9$:

$$\frac{a_{10} - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$$

Answer: 3

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Discussion (5)

C
Saved me before my mock test. Much clearer than my coaching notes.
CamilleDubois28 · 23d ago
AS
For revision — the key formula used here comes up almost every year.
Arjun Sharma · 21d ago
L
Does this approach generalise to the JEE Advanced version of this question?
LiamAnderson39 · 21d ago
NR
Adding for context: NCERT covers the base concept in the same chapter.
Nikhil Rao · 19d ago
VA
Good follow-up questions — remember to always state your assumptions in the JEE subjective section.
Vidaara Admin · Vidaara Team · 17d ago
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