JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of the series \(20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots .\) upto \(n ^{ th }\) term is \(488\) and the \(n^{\text {th }}\) term is negative, then
- A \(n ^{\text {th }}\) term is \(-4 \frac{2}{5}\)
- B \(n =41\)
- C \(n^{\text {th }}\) term is \(-4\)
- D \(n =60\)
Answer & Solution
Correct Answer
(C) \(n^{\text {th }}\) term is \(-4\)
Step-by-step Solution
Detailed explanation
\(S =\frac{100}{5}+\frac{98}{5}+\frac{96}{5}+\frac{94}{5}+\ldots . . n\) \(S _{ n }=\frac{ n }{2}\left(2 \times \frac{100}{5}+( n -1)\left(-\frac{2}{5}\right)\right)=188\) \(n (100- n +1)=488 \times 5\) \(n ^{2}-101 n +488 \times 5=0\) \(n =61,40\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- \(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}} = \) . . . .JEE Mains 2017 Medium
- A line drawn through the point \(P(4, 7)\) cuts the circle \(x^2 + y^2\, = 9\) at the points \(A\) and \(B\). Then \(PA· PB\) is equal toJEE Mains 2017 Hard
- lf \(2 + 3i\) is one of the roots of the equation \(2x^3 -9x^2 + kx- 13 = 0,\) \(k \in R,\) then the real root of this equationJEE Mains 2015 Hard
- The value of the integral \(\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x\) is :JEE Mains 2023 Hard
- lf \(f(x)\) is a differentiable function in the interval \((0,\infty )\) such that \(f(1) = 1\) and \(\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1,\) for each \(x > 0,\) then \(f (\frac {3}{2})\) is equal toJEE Mains 2016 Hard
- The value of \(\sum \limits_{ r =0}^{20}{ }^{50- r } C _{6}\) is equal toJEE Mains 2020 Hard
More PYQs from JEE Mains
- Let \(\vec \alpha \, = \,3\hat i\, + \hat j\) and \(\vec \beta \, = \,2\hat i\, - \hat j + 3\hat k.\) If \(\vec \beta \, = \,{\vec \beta _1} - {\vec \beta _2},\) where \({\vec \beta _1}\) is parallel to \(\vec \alpha \) and \(\vec \beta_2 \) is perpendicular to \(\vec \alpha ,\) then \({\vec \beta _1} \times {\vec \beta _2}\) is equal toJEE Mains 2019 Hard
- If a point \(P (\alpha, \beta, \gamma)\) satisfying \((\alpha \ \beta \ \gamma)\left(\begin{array}{ccc}2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8\end{array}\right)=\left(\begin{array}{llll}0 & 0 & 0\end{array}\right)\) lies on the plane \(2 x+4 y+3 z=5\), then \(6 \alpha+9 \beta+7 \gamma\) is equal to:JEE Mains 2023 Medium
- If for \(\theta \in\left[-\frac{\pi}{3}, 0\right]\), the points \((x, y)=\left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)\) lie on \(x y+\alpha x+\beta y+\gamma=0\), then \(\alpha^2+\beta^2+\gamma^2\) is equal to :JEE Mains 2025 Medium
- Let \(A\) be a \(3\times3\) matrix such that \(A\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&2&3 \\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
1&0&0 \\
0&1&0
\end{array}} \right]\) Then \(A^{-1}\) isJEE Mains 2014 Hard - If \(\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1\), where \(\lambda, \mu \in \mathbb{R}\), then \(\lambda+\mu\) is equal toJEE Mains 2025 Medium
- Let the solution curve \(y = y ( x )\) of the differential equation, \(\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{d y}{d x}=x+\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] y\) pass through the points \((1,0)\) and \((2 \alpha, \alpha), \alpha>0\). Then \(\alpha\) is equal toJEE Mains 2022 Hard