Question:

If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to:

Updated On: Mar 20, 2025

$\frac{52}{147}$
$\frac{50}{141}$
$\frac{51}{144}$
$\frac{49}{138}$

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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:

\[ \boxed{\frac{50}{141}}. \]

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Approach Solution -2

a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}

then

a_{1} + a_{2} + \dots\dots\dots a_{25}

\Rightarrow n = 1 \sum 25 a_{n} = \sum \frac{- 2}{4 n ^{2} - 16 n + 15}

= \sum \frac{- 2}{4 n ^{2} - 6 n - 10 n + 15}

= \sum \frac{- 2}{2 n ( 2 n - 3 ) - 5 ( 2 n - 3 )}

= \sum \frac{- 2}{( 2 n - 3 ) ( 2 n - 5 )}

= \sum \frac{1}{2 n - 3} - \frac{1}{2 n - 5}

= \frac{1}{47} - \frac{1}{( - 3 )}

= \frac{50}{141}

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Questions Asked in JEE Main exam

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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