Download App

STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
Question
 If $\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+1012}\right) $ $ -\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots+\frac{1}{2024 \cdot 2023}\right) $

$ =\frac{1}{2024}, $ then $\alpha$ is equal to-

✓

Answer

d
$ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ - $ \left\{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right)\right\}=\frac{1}{2024} $

$\Rightarrow $ $ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ $ -\left\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\ldots+\frac{1}{2023}\right. $ $ \left.-\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2022}\right)\right\}=\frac{1}{2024} $

$\Rightarrow $ $ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ $ -\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2023}\right) $ $ \frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{1011}\right)=\frac{1}{2024} $

$\Rightarrow $ $ \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012} $ = $ \frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023} $

$\Rightarrow $ $ \alpha=1011$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The largest natural number $n$ such that $3^{ n }$ divides $66 !$ is $............$.
→
If $ai + 6j - k$ and $7i - 3j + 17k$ are perpendicular vectors, then the value of $a $ is
→
In a tournament, a team plays $10$ matches with probabilities of winning and losing each match as $\frac{1}{3}$ and $\frac{2}{3}$ respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $\mathrm{P}(|\mathrm{x}-\mathrm{y}| \leq 2)$ is $\mathrm{p}$, then $3^9 \mathrm{p}$ equals....................
→
The sum of the coefficients of even power of $x$ in the expansion of ${(1 + x + {x^2} + {x^3})^5}$ is
→
If the sum $\frac{3}{1^2} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + ...... + $ up to $20$ terms is equal to $\frac{k}{{21}}$, then $k$ is equal to
→
If the coefficients of $x^{7}$ in $\left(x^{2}+\frac{1}{b x}\right)^{11}$ and $x^{-7}$ in $\left(x-\frac{1}{b x^{2}}\right)^{11}, b \neq 0$, are equal, then the value of $b$ is equal to:
→
Let $P$ be a point in the first octant, whose image $Q$ in the plane $x+y=3$ (that is, the line segment $P Q$ is perpendicular to the plane $x+y=3$ and the mid-point of $P Q$ lies in the plane $x+y=3$ ) lies on the $z$-axis. Let the distance of $P$ from the $x$-axis be $5$ . If $R$ is the image of $P$ in the xy-plane, then the length of $P R$ is. . . . . .
→
${99^{th}}$ term of the series $2 + 7 + 14 + 23 + 34 + .....$ is
→
In the graph of the function$\sqrt 3 \sin x + \cos x$ the maximum distance of a point from $x-$ axis is
→
The remainder when $428^{2024}$ is divided by $21$ is ............
→