Why does telescope to ?

Writing out the terms: , every intermediate term cancels out, leaving only .

JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Sequences and Series: JEE Main Mathematics Question with Solution

Q: Why is the numerator split as ?

Because , which matches the numerator directly without introducing extra minus signs.

\mathrm{a}_{\mathrm{n}}=\frac{-2}{4 \mathrm{n}^{2}-16 \mathrm{n}+15}

, then

\mathrm{a}_{1}+\mathrm{a}_{2}+\cdots+\mathrm{a}_{25}

is equal to:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{52}{147}

\frac{49}{138}

\frac{50}{141}

\frac{51}{144}

Step-by-step solutionView

Correct answer

Factor the quadratic denominator into $(2n-5)(2n-3)$ , decompose $a_n$ into partial fractions as $\frac{1}{2n-3} - \frac{1}{2n-5}$ , and evaluate the telescoping sum to get $\frac{50}{141}$ .

Option analysis

Why each option works or fails

A ·

\frac{52}{147}

Summing 26 terms instead of 25 terms gives $\frac{1}{3} + \frac{1}{2(26)-3} = \frac{1}{3} + \frac{1}{49} = \frac{52}{147}$ . Check the upper summation limit: for $n=25$ , the remaining positive term is $\frac{1}{2(25)-3} = \frac{1}{47}$ , not $\frac{1}{49}$ .

B ·

\frac{49}{138}

Summing only 23 terms or confusing the index offset results in an upper boundary of $\frac{1}{43}$ , giving $\frac{1}{3} - \left(-\frac{1}{43}\right) = \frac{46}{129}$ or similar incorrect offsets like $\frac{49}{138}$ . Carefully track index cancellation: $a_n = \frac{1}{2n-3} - \frac{1}{2n-5}$ . The surviving terms after 25 additions are precisely $-\frac{1}{2(1)-5} = \frac{1}{3}$ and $\frac{1}{2(25)-3} = \frac{1}{47}$ .

C ·

\frac{50}{141}

None. The partial fraction decomposition $\frac{1}{2n-3} - \frac{1}{2n-5}$ telescopes, leaving $\frac{1}{3} + \frac{1}{47} = \frac{50}{141}$ . This is the correct answer.

D ·

\frac{51}{144}

Summing 24 terms instead of 25 terms leaves $\frac{1}{3} + \frac{1}{2(24)-3} = \frac{1}{3} + \frac{1}{45} = \frac{16}{45} = \frac{51.2}{144}$ or an off-by-one error at the boundary resulting in $\frac{51}{144}$ . Ensure the evaluation includes all terms from $n=1$ through $n=25$ inclusive.

Reviewed route

Solution

StepWorking

01given

General term is $a_n = \frac{-2}{4n^2 - 16n + 15}$ and we need to find $S = \sum_{n=1}^{25} a_n$ .

02goal

Evaluate $\sum_{n=1}^{25} a_n$ in exact fractional form.

03approach

Factor the quadratic denominator $4n^2 - 16n + 15 = (2n - 3)(2n - 5)$ . Then decompose $a_n$ into partial fractions to set up a telescoping sum.

04execute

Factor the denominator: $4n^2 - 16n + 15 = (2n - 3)(2n - 5)$ . Notice the difference between the factors: $(2n - 3) - (2n - 5) = 2$ . Therefore, $-2 = (2n - 5) - (2n - 3)$ . Rewrite $a_n$ : $a_n = \frac{(2n - 5) - (2n - 3)}{(2n - 3)(2n - 5)} = \frac{1}{2n - 3} - \frac{1}{2n - 5}$ .

05execute

Let $V_n = \frac{1}{2n - 3}$ . Then $V_{n-1} = \frac{1}{2(n-1) - 3} = \frac{1}{2n - 5}$ . So $a_n = V_n - V_{n-1}$ , forming a direct telescoping sum: $\sum_{n=1}^{25} a_n = \sum_{n=1}^{25} (V_n - V_{n-1}) = V_{25} - V_0$ . Evaluate the terms: $V_{25} = \frac{1}{2(25) - 3} = \frac{1}{47}$ . $V_0 = \frac{1}{2(0) - 3} = -\frac{1}{3}$ . Hence, the sum is: $S = \frac{1}{47} - \left(-\frac{1}{3}\right) = \frac{1}{47} + \frac{1}{3} = \frac{3 + 47}{47 \times 3} = \frac{50}{141}$ .

✓verify

Check for $N=1$ : $a_1 = \frac{-2}{4 - 16 + 15} = \frac{-2}{3}$ . Formula gives $V_1 - V_0 = -1 - (-\frac{1}{3}) = -\frac{2}{3}$ . Correct. For $N=25$ : $\frac{50}{141} \approx 0.3546$ . Option (2) matches precisely.

Hints that build this answer step by step

How does the denominator $4n^2 - 16n + 15$ factor?
$(2n - 5)(2n - 3)$
What is the partial fraction decomposition of $a_n = \frac{-2}{(2n-5)(2n-3)}$ ?
$\frac{1}{2n - 3} - \frac{1}{2n - 5}$
When telescoping $\sum_{n=1}^{25} \left(\frac{1}{2n-3} - \frac{1}{2n-5}\right)$ , which terms remain uncanceled?
$-\frac{1}{2(1)-5} + \frac{1}{2(25)-3}$
What is the evaluated value of $\frac{1}{3} + \frac{1}{47}$ ?
$\frac{50}{141}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the numerator

-2

split as

(2n - 5) - (2n - 3)

Because $(2n - 5) - (2n - 3) = -5 - (-3) = -2$ , which matches the numerator directly without introducing extra minus signs.

Why does

\sum_{n=1}^N (V_n - V_{n-1})

telescope to

V_N - V_0

Writing out the terms: $(V_1 - V_0) + (V_2 - V_1) + \dots + (V_N - V_{N-1})$ , every intermediate term $V_1, V_2, \dots, V_{N-1}$ cancels out, leaving only $V_N - V_0$ .

Answer

Factor the quadratic denominator into $(2n-5)(2n-3)$ , decompose $a_n$ into partial fractions as $\frac{1}{2n-3} - \frac{1}{2n-5}$ , and evaluate the telescoping sum to get $\frac{50}{141}$ .

Why each option works or fails

A: $\frac{52}{147}$ - Summing 26 terms instead of 25 terms gives $\frac{1}{3} + \frac{1}{2(26)-3} = \frac{1}{3} + \frac{1}{49} = \frac{52}{147}$ . Check the upper summation limit: for $n=25$ , the remaining positive term is $\frac{1}{2(25)-3} = \frac{1}{47}$ , not $\frac{1}{49}$ .
B: $\frac{49}{138}$ - Summing only 23 terms or confusing the index offset results in an upper boundary of $\frac{1}{43}$ , giving $\frac{1}{3} - \left(-\frac{1}{43}\right) = \frac{46}{129}$ or similar incorrect offsets like $\frac{49}{138}$ . Carefully track index cancellation: $a_n = \frac{1}{2n-3} - \frac{1}{2n-5}$ . The surviving terms after 25 additions are precisely $-\frac{1}{2(1)-5} = \frac{1}{3}$ and $\frac{1}{2(25)-3} = \frac{1}{47}$ .
C · correct: $\frac{50}{141}$ - None. The partial fraction decomposition $\frac{1}{2n-3} - \frac{1}{2n-5}$ telescopes, leaving $\frac{1}{3} + \frac{1}{47} = \frac{50}{141}$ . This is the correct answer.
D: $\frac{51}{144}$ - Summing 24 terms instead of 25 terms leaves $\frac{1}{3} + \frac{1}{2(24)-3} = \frac{1}{3} + \frac{1}{45} = \frac{16}{45} = \frac{51.2}{144}$ or an off-by-one error at the boundary resulting in $\frac{51}{144}$ . Ensure the evaluation includes all terms from $n=1$ through $n=25$ inclusive.

Step-by-step solution

given: General term is $a_n = \frac{-2}{4n^2 - 16n + 15}$ and we need to find $S = \sum_{n=1}^{25} a_n$ .
goal: Evaluate $\sum_{n=1}^{25} a_n$ in exact fractional form.
approach: Factor the quadratic denominator $4n^2 - 16n + 15 = (2n - 3)(2n - 5)$ . Then decompose $a_n$ into partial fractions to set up a telescoping sum.
execute: Factor the denominator: $4n^2 - 16n + 15 = (2n - 3)(2n - 5)$ . Notice the difference between the factors: $(2n - 3) - (2n - 5) = 2$ . Therefore, $-2 = (2n - 5) - (2n - 3)$ . Rewrite $a_n$ : $a_n = \frac{(2n - 5) - (2n - 3)}{(2n - 3)(2n - 5)} = \frac{1}{2n - 3} - \frac{1}{2n - 5}$ .
execute: Let $V_n = \frac{1}{2n - 3}$ . Then $V_{n-1} = \frac{1}{2(n-1) - 3} = \frac{1}{2n - 5}$ . So $a_n = V_n - V_{n-1}$ , forming a direct telescoping sum: $\sum_{n=1}^{25} a_n = \sum_{n=1}^{25} (V_n - V_{n-1}) = V_{25} - V_0$ . Evaluate the terms: $V_{25} = \frac{1}{2(25) - 3} = \frac{1}{47}$ . $V_0 = \frac{1}{2(0) - 3} = -\frac{1}{3}$ . Hence, the sum is: $S = \frac{1}{47} - \left(-\frac{1}{3}\right) = \frac{1}{47} + \frac{1}{3} = \frac{3 + 47}{47 \times 3} = \frac{50}{141}$ .
verify: Check for $N=1$ : $a_1 = \frac{-2}{4 - 16 + 15} = \frac{-2}{3}$ . Formula gives $V_1 - V_0 = -1 - (-\frac{1}{3}) = -\frac{2}{3}$ . Correct. For $N=25$ : $\frac{50}{141} \approx 0.3546$ . Option (2) matches precisely.

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

Factor the quadratic denominator into $(2n-5)(2n-3)$ , decompose $a_n$ into partial fractions as $\frac{1}{2n-3} - \frac{1}{2n-5}$ , and evaluate the telescoping sum to get $\frac{50}{141}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

We think you should solve this next

Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

Let $a_1, a_2, a_3, \dots$ be an A.P. If $a_7 = 3$ , the product $a_1 a_4$ is minimum and…

Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix…

Students also ask

Sequences and Series: JEE Main Mathematics Question with Solution

Factor the quadratic denominator into (2n−5)(2n−3)(2n-5)(2n-3)(2n−5)(2n−3), decompose ana_nan​ into partial fractions as 12n−3−12n−5\frac{1}{2n-3} - \frac{1}{2n-5}2n−31​−2n−51​, and evaluate the telescoping sum to get 50141\frac{50}{141}14150​.

01given

02goal

03approach

04execute

05execute

✓verify

Let g:R→Rg: R \rightarrow Rg:R→R be a non constant twice differentiable such that…

Let a1,a2,a3,…a_1, a_2, a_3, \dotsa1​,a2​,a3​,… be an A.P. If a7=3a_7 = 3a7​=3, the product a1a4a_1 a_4a1​a4​ is minimum and…

Let y=f(x)y=f(x)y=f(x) represent a parabola with focus (−12,0)\left(-\frac{1}{2}, 0\right)(−21​,0) and directrix…

Students also ask

Factor the quadratic denominator into $(2n-5)(2n-3)$ , decompose $a_n$ into partial fractions as $\frac{1}{2n-3} - \frac{1}{2n-5}$ , and evaluate the telescoping sum to get $\frac{50}{141}$ .

Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

Let $a_1, a_2, a_3, \dots$ be an A.P. If $a_7 = 3$ , the product $a_1 a_4$ is minimum and…

Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix…