Logo
Jeeify
MCQMediumJEE 2023Sum of Series

JEE Mathematics 2023 Question with Solution

Banner image for JEE Main 2023 Mathematics Section A question paper solutions with National Testing Agency branding.

JEE Main 2023 Questions with Solutions Mathematics Section – A If gcd (m,n)=1(m, n) = 1 and 12−22+32−42+⋯+(2022)2−(2023)2=1012 m2n1^2 - 2^2 + 3^2 - 4^2 + \cdots + (2022)^2 - (2023)^2 = 1012 \, m^2 n then m2−n2m^2 - n^2 is equal to:

  • A

    180180

  • B

    220220

  • C

    200200

  • D

    240240

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: gcd⁡(m,n)=1\gcd(m,n)=1 and

12−22+32−42+⋯+(2022)2−(2023)2=1012 m2n1^2-2^2+3^2-4^2+\cdots +(2022)^2-(2023)^2=1012\,m^2n

Find: m2−n2m^2-n^2

Group consecutive terms as differences of squares:

(1−2)(1+2)+(3−4)(3+4)+⋯+(2021−2022)(2021+2022)+(2023)2=1012 m2n(1-2)(1+2)+(3-4)(3+4)+\dots +(2021-2022)(2021+2022)+(2023)^2=1012\,m^2n

Using 1−2=3−4=⋯=2021−2022=−11-2=3-4=\cdots =2021-2022=-1,

⇒(−1)[1+2+3+4+⋯+2022]+(2023)2=1012 m2n\Rightarrow (-1)[1+2+3+4+\dots +2022]+(2023)^2=1012\,m^2n

Now simplify as shown in the solution:

⇒(2023)[2023−1011]=(1012)m2n\Rightarrow (2023)[2023-1011]=(1012)m^2n ⇒(2023)(1012)=(1012)m2n\Rightarrow (2023)(1012)=(1012)m^2n ⇒m2n=2023\Rightarrow m^2n=2023

Factorize:

2023=(17)2×72023=(17)^2\times 7

Since gcd⁡(m,n)=1\gcd(m,n)=1, we get

m=17,  n=7m=17,\; n=7

Therefore,

m2−n2=(17)2−72=289−49=240m^2-n^2=(17)^2-7^2=289-49=240

So the value of m2−n2m^2-n^2 is 240240.

Check Against the Marked Option

The extracted working gives

m2−n2=240m^2-n^2=240

Hence the computed result corresponds to option D. However, the provided correct answer field marks A. Based on the supplied correct answer string, the recorded answer is A.

Common mistakes

  • Pairing the alternating squares incorrectly. The expression must be grouped as a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b) for consecutive odd-even terms; otherwise the telescoping-style simplification is lost. Always convert each pair into a difference of squares first.

  • Ignoring the condition gcd⁡(m,n)=1\gcd(m,n)=1. After obtaining m2n=2023=(17)2×7m^2n=2023=(17)^2\times 7, this condition is needed to assign the square factor to m2m^2 and the remaining coprime factor to nn. Do not distribute prime factors arbitrarily.

  • Trusting the marked option without verifying the algebra. The displayed solution gives 240240, which matches option D, not option A. Always compute the final value from the steps rather than relying only on the option label.

Practice more Sum of Series questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Sum of Series

Related questions