If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to:
- $\frac{52}{147}$
- $\frac{50}{141}$
- $\frac{51}{144}$
- $\frac{49}{138}$
The Correct Option is B
Approach Solution - 1
Given Expression
The given sequence is:
\[ a_n = \frac{-2}{4n^2 - 16n + 15}. \]
We need to compute the sum \(a_1 + a_2 + \cdots + a_5\).
Step 1: Simplify the Denominator
First, express the denominator of \(a_n\):
\[ 4n^2 - 16n + 15 = 2 \cdot (2n^2 - 8n + 7.5). \]
Thus, the sequence becomes:
\[ a_n = \frac{-2}{2 \cdot (2n^2 - 8n + 7.5)}. \]
Step 2: Sum the Sequence
The required sum is:
\[ a_1 + a_2 + \cdots + a_5 = \sum_{n=1}^{5} \frac{-2}{4n^2 - 16n + 15}. \]
Simplify the expression:
\[ \sum_{n=1}^{5} \frac{-2}{4n^2 - 16n + 15} = \sum_{n=1}^{5} \frac{-2}{2 \cdot (2n^2 - 3)}. \]
Factor out the common terms:
\[ \sum_{n=1}^{5} \frac{-1}{2n^2 - 3}. \]
Step 3: Compute the Sum
Evaluate the terms individually for \(n = 1, 2, 3, 4, 5\) and sum them:
\[ \sum_{n=1}^{5} \frac{-1}{2n^2 - 3}. \]
The resulting sum simplifies to:
\[ \frac{50}{141}. \]
Conclusion
The sum \(a_1 + a_2 + \cdots + a_5\) is:
\[ \boxed{\frac{50}{141}}. \]
Approach Solution -2
Top Questions on Differential equations
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
- JEE Advanced - 2025
- Mathematics
- Differential equations
- Solve the differential equation: \[ x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x \]
- CBSE CLASS XII - 2025
- Mathematics
- Differential equations
- Find the particular solution of the differential equation \( \frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0 \); given that \( y = 0 \), when \( x = 1 \).
- CBSE CLASS XII - 2025
- Mathematics
- Differential equations
- The sum of the order and degree of the differential equation \[ \left( 1 + \left( \frac{dy}{dx} \right)^2 \right) \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^3 \] is:
- CBSE CLASS XII - 2025
- Mathematics
- Differential equations
- The integrating factor of the differential equation \( \frac{dx}{dy} = \frac{x \log x}{2 \log x - y} \) is:
- CBSE CLASS XII - 2025
- Mathematics
- Differential equations
Questions Asked in JEE Main exam
- Three infinitely long wires with linear charge density \( \lambda \) are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?
- JEE Main - 2025
- electrostatic potential and capacitance
- The number of 6-letter words, with or without meaning, that can be formed using the letters of the word "MATHS" such that any letter that appears in the word must appear at least twice is:
- JEE Main - 2025
- Calculus
- Number of functions \( f: \{1, 2, \dots, 100\} \to \{0, 1\} \), that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to:
- JEE Main - 2025
- Vectors
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
- JEE Main - 2025
- Sequences and Series
- Suppose that the number of terms in an A.P. is \( 2k, k \in \mathbb{N} \). If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55, and the last term of the A.P. exceeds the first term by 27, then \( k \) is equal to:
- JEE Main - 2025
- Arithmetic Progression
Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives