Let α and β are two positive roots of x^2-2ax+ab=0 where 0<b<a, then the value of S_n=1+2b/a+3ba^2+.....+nba^(n-1), ∀n∈N cannot exceed

Correct answer: (D) α+β^(α-β4). S_n<S_(∞) ⇒S_n<1+2b/a+3b^a2+.... upto ∞ ⇒S_n<1/(1-ba)+(1×ba1-ba^2=11-ba^2=aa-b^2⇒S_n<4a24a2-4ab^2Now, α+β=2a and αβ=ab⇒S_n<α+β^2)/(α+β^2-4αβ)2=α+β^2α-β^22 ⇒S_n<α+β^(α-β4)

Q25 / 118 · NTA Abhyas (JEE Main) 2020 · Mathematics · Sequences and Series

Let $α$ and $β$ are two positive roots of $x^{2} - 2 a x + a b = 0$ where $0 < b < a,$ then the value of $S_{n} = 1 + 2 (\frac{b}{a}) + 3 {(\frac{b}{a})}^{2} + . . . . . + (n) {(\frac{b}{a})}^{n - 1}, \forall n \in N$
cannot exceed

A
$\frac{α}{β}$
B
$|\frac{α + β}{α - β}|$
C
$|\frac{β}{α}|$
Correct answer
D
${(\frac{α + β}{α - β})}^{4}$

Solution

$S_{n} < S_{\infty}$
$\Rightarrow S_{n} < 1 + 2 (\frac{b}{a}) + 3 {(\frac{b}{a})}^{2} + . . . .$ upto $\infty$

$\Rightarrow S_{n} < \frac{1}{1 - \frac{b}{a}} + \frac{1 \times \frac{b}{a}}{{(1 - \frac{b}{a})}^{2}} = \frac{1}{{(1 - \frac{b}{a})}^{2}} = {(\frac{a}{a - b})}^{2}$
$\Rightarrow S_{n} < {(\frac{4 a^{2}}{4 a^{2} - 4 a b})}^{2}$
Now, $α + β = 2 a$ and $α β = a b$
$\Rightarrow S_{n} < {(\frac{{(α + β)}^{2}}{{(α + β)}^{2} - 4 α β})}^{2} = {(\frac{{(α + β)}^{2}}{{(α - β)}^{2}})}^{2}$
$\Rightarrow S_{n} < {(\frac{α + β}{α - β})}^{4}$

Hence, the correct answer is option: 'D'.

Let α and β are two positive roots of x2-2ax+ab=0 where 0<b<a, then the value of Sn=1+2ba+3ba2+.....+nban-1, ∀n∈N cannot exceed

Solution

More Sequences and Series Questions

Let $α$ and $β$ are two positive roots of $x^{2} - 2 a x + a b = 0$ where $0 < b < a,$ then the value of $S_{n} = 1 + 2 (\frac{b}{a}) + 3 {(\frac{b}{a})}^{2} + . . . . . + (n) {(\frac{b}{a})}^{n - 1}, \forall n \in N$
cannot exceed