JEE Advanced Maths – Arithmetic-Geometric Sequence (AGS) & Special Sequences PYQ with Solutions

Q 1 :

Let $\overset{r}{\overset{⏞}{75 . . . 57}}$ denote the $(r + 2)$ digit number where the first and the last digits are 7 and the remaining $r$ digits are 5. Consider the sum $S = 77 + 757 + 7557 + \dots + \overset{98}{\overset{⏞}{75 . . . 57}} .$ If $S = \frac{\overset{99}{\overset{⏞}{75 . . . 57}} + m}{n},$ where $m$ and $n$ are natural numbers less than 3000, then the value of $m + n$ is [2023]

(1219)

$Given S = 77 + 757 + 7557 + \dots + \overset{98}{\overset{⏞}{75 . . . 57}}$

$\frac{10 S = 770 + 7570 + \dots + 75 . . . . . . . . 570 + \overset{98}{\overset{⏞}{75 . . . 570}}}{- 9 S = 77 - \underset{98 times}{\underset{⏟}{13 - 13 - \dots - 13}} - \underset{98}{\underset{⏟}{75 \dots 570}}}$

$9 S = - 77 + 13 \times 98 + \underset{99}{\underset{⏟}{75 \dots 57}} + 13$

$S = \frac{\overset{99}{\overset{⏞}{75 . . . . . 57}} + 1210}{9}$

$\Rightarrow m = 1210, n = 9 \Rightarrow m + n = 1219 .$

Q 2 :

Let $α$ and $β$ be the roots of $x^{2} - x - 1 = 0,$ with $α > β .$ For all positive integers $n$ , define $a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1$

$b_{1} = 1$ and $b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2$

Then which of the following options is/are correct? [2019]

$\sum_{n = 1}^{\infty} \frac{α_{n}}{10^{n}} = \frac{10}{89}$
$b_{n} = α^{n} + β^{n} for all n \geq 1$
$a_{1} + a_{2} + a_{3} + \dots + a_{n} = a_{n + 2} - 1 for all n \geq 1$
$\sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \frac{8}{89}$

Select one or more options

(1, 2, 3)

$Since α, β are roots of x^{2} - x - 1 = 0 with α > β$

$∴$ $α = \frac{1 + \sqrt{5}}{2}, β = \frac{1 - \sqrt{5}}{2}$

$Also, α + β = 1, α β = - 1, α - β = \sqrt{5}$

$α^{2} - α - 1 = 0 and β^{2} - β - 1 = 0$

$a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1; b_{1} = 1 and b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2$

Let us now check the given options, one by one

$(1) \sum_{n = 1}^{\infty} \frac{a_{n}}{10^{n}} = \sum_{n = 1}^{\infty} \frac{α^{n} - β^{n}}{\sqrt{5} (10^{n})} = \frac{1}{\sqrt{5}} [\sum_{n = 1}^{\infty} {(\frac{α}{10})}^{n} - \sum_{n = 1}^{\infty} {(\frac{β}{10})}^{n}]$

$= \frac{1}{\sqrt{5}} [\frac{\frac{α}{10}}{1 - \frac{α}{10}} - \frac{\frac{β}{10}}{1 - \frac{β}{10}}]$ $= \frac{1}{\sqrt{5}} [\frac{α}{10 - α} - \frac{β}{10 - β}]$

$= \frac{1}{\sqrt{5}} [\frac{10 α - α β - 10 β + α β}{(10 - α) (10 - β)}]$

$= \frac{1}{5} [\frac{10 (α - β)}{100 - 10 (α + β) + α β}]$ $= \frac{1}{5} [\frac{10 \sqrt{5}}{100 - 10 - 1}] = \frac{10}{89}$

$Thus option (1) is correct.$

$(2) b_{n} = a_{n + 1} + a_{n - 1} = \frac{α^{n + 1} - β^{n + 1}}{α - β} + \frac{α^{n - 1} - β^{n - 1}}{α - β}$

$= \frac{(α^{n + 1} + α^{n - 1}) - (β^{n + 1} + β^{n - 1})}{α - β}$

$= \frac{α^{n - 1} (α^{2} + 1) - β^{n - 1} (β^{2} + 1)}{\sqrt{5}}$

$= \frac{1}{\sqrt{5}} [α^{n - 1} (α + 2) - β^{n - 1} (β + 2)]$ $[using α^{2} = α + 1, β^{2} = β + 1]$

$= \frac{1}{\sqrt{5}} [α^{n - 1} (\frac{1 + \sqrt{5}}{2} + 2) - β^{n - 1} (\frac{1 - \sqrt{5}}{2} + 2)]$

$= \frac{1}{\sqrt{5}} [α^{n - 1} (\frac{5 + \sqrt{5}}{2}) - β^{n - 1} (\frac{5 - \sqrt{5}}{2})]$

$= \frac{\sqrt{5}}{\sqrt{5}} [α^{n - 1} (\frac{\sqrt{5} + 1}{2}) + β^{n - 1} (\frac{1 - \sqrt{5}}{2})]$

$= α^{n - 1} α + β^{n - 1} β = α^{n} + β^{n}$

$Thus option (2) is correct.$

$(3) a_{1} + a_{2} + \dots + a_{n}$

$= \frac{α^{1} - β^{1}}{α - β} + \frac{α^{2} - β^{2}}{α - β} + \frac{α^{3} - β^{3}}{α - β} + \dots + \frac{α^{n} - β^{n}}{α - β}$

$= \frac{1}{\sqrt{5}} [(α + α^{2} + α^{3} + \dots + α^{n}) - (β + β^{2} + β^{3} + \dots + β^{n})]$

$= \frac{1}{\sqrt{5}} [\frac{α (1 - α^{n})}{1 - α} - \frac{β (1 - β^{n})}{1 - β}]$

$= \frac{1}{\sqrt{5}} [\frac{α (1 - β) (1 - α^{n}) - β (1 - α) (1 - β^{n})}{(1 - α) (1 - β)}]$

$= \frac{1}{\sqrt{5}} [\frac{(α - α β) (1 - α^{n}) - (β - α β) (1 - β^{n})}{(1 - α) (1 - β)}]$

$= \frac{1}{\sqrt{5}} [\frac{(α + 1) (1 - α^{n}) - (β + 1) (1 - β^{n})}{1 - (α + β) + α β}]$

$= \frac{1}{\sqrt{5}} [\frac{α^{2} (1 - α^{n}) - β^{2} (1 - β^{n})}{1 - 1 - 1}] [using α^{2} - α - 1 = 0, β^{2} - β - 1 = 0]$

$= \frac{1}{\sqrt{5}} [\frac{(α^{2} - β^{2}) - (α^{n + 2} - β^{n + 2})}{- 1}]$

$= [\frac{- (α - β) (α + β)}{α - β} + \frac{(α^{n + 2} - β^{n + 2})}{α - β}] = - 1 + α_{n + 2}$

$Thus option (3) is correct.$

$(4) \sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}}$ $= \sum_{n = 1}^{\infty} \frac{α^{n} + β^{n}}{10^{n}} [using b_{n} = α^{n} + β^{n} from(b)]$

$= \sum_{n = 1}^{\infty} {(\frac{α}{10})}^{n} + {(\frac{β}{10})}^{n}$ $= \frac{\frac{α}{10}}{1 - \frac{α}{10}} + \frac{\frac{β}{10}}{1 - \frac{β}{10}}$

$= \frac{α}{10 - α} + \frac{β}{10 - β}$ $= \frac{α (10 - β) + β (10 - α)}{(10 - α) (10 - β)}$

$= \frac{10 (α + β) - 2 α β}{100 - 10 (α + β) + α β}$ $= \frac{10 + 2}{100 - 10 - 1} = \frac{12}{89}$

$Thus option (4) is incorrect.$

Topic Question Set

Q 2 :

Let $α$ and $β$ be the roots of $x^{2} - x - 1 = 0,$ with $α > β .$ For all positive integers $n$ , define $a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1$

$b_{1} = 1$ and $b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2$

Then which of the following options is/are correct? [2019]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : Let 75...57⏞r denote the (r+2) digit number where the first and the last digits are 7 and the remaining r digits are 5. Consider the sum S=77+757+7557+⋯+75...57⏞98. If S=75...57⏞99+mn, where m and n are natural numbers less than 3000, then the value of m+n is [2023]

Q 2 : Let α and β be the roots of x2-x-1=0, with α>β. For all positive integers n, define an=αn-βnα-β, n≥1 b1=1 and bn=an-1+an+1, n≥2 Then which of the following options is/are correct? [2019]

Want Us to Email you About Special Offers & Updates?

Q 2 :

Let $α$ and $β$ be the roots of $x^{2} - x - 1 = 0,$ with $α > β .$ For all positive integers $n$ , define $a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1$

$b_{1} = 1$ and $b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2$

Then which of the following options is/are correct? [2019]