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JEE Mathematics 2025 Question with Solution

Let TrT_r be the rthr^{th} term of an A.P. If for some mm, Tm=125T_m = \frac{1}{25}, T25=120T_{25} = \frac{1}{20}, and ∑r=125Tr=13\sum_{r=1}^{25} T_r = 13, then

5m∑r=m2mTr is equal to:5m \sum_{r=m}^{2m} T_r \text{ is equal to:}
  • A

    112112

  • B

    142142

  • C

    126126

  • D

    9898

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The rthr^{th} term of the A.P. is Tr=a+(r−1)dT_r = a + (r-1)d.

We are given

  • Tm=125T_m = \frac{1}{25}
  • T25=120T_{25} = \frac{1}{20}
  • ∑r=125Tr=13\sum_{r=1}^{25} T_r = 13

Find: The value of 5m∑r=m2mTr5m \sum_{r=m}^{2m} T_r.

Using the term formula,

Tm=a+(m−1)d=125T_m = a + (m-1)d = \frac{1}{25}

and

T25=a+24d=120T_{25} = a + 24d = \frac{1}{20}

From the sum of first 2525 terms,

∑r=125Tr=252[2a+24d]=13\sum_{r=1}^{25} T_r = \frac{25}{2}[2a + 24d] = 13

So,

25a+300d=1325a + 300d = 13

Now subtract the two term equations:

(a+(m−1)d)−(a+24d)=125−120\left(a + (m-1)d\right) - \left(a + 24d\right) = \frac{1}{25} - \frac{1}{20}

Hence,

(m−25)d=−1100(m-25)d = -\frac{1}{100}

So,

d=1100(25−m)d = \frac{1}{100(25-m)}

Substituting in the sum equation,

25a+300(1100(25−m))=1325a + 300\left(\frac{1}{100(25-m)}\right) = 13

that is,

25a+325−m=1325a + \frac{3}{25-m} = 13

Hence,

25a=13−325−m25a = 13 - \frac{3}{25-m}

Now,

∑r=m2mTr=(2m−m+1)2[Tm+T2m]\sum_{r=m}^{2m} T_r = \frac{(2m-m+1)}{2}[T_m + T_{2m}]

Using T2m=a+(2m−1)dT_{2m} = a + (2m-1)d, this becomes

∑r=m2mTr=(m+1)[a+(m−1)d+a+(2m−1)d]\sum_{r=m}^{2m} T_r = (m+1)[a + (m-1)d + a + (2m-1)d]

So,

∑r=m2mTr=(m+1)[2a+(3m−2)d]\sum_{r=m}^{2m} T_r = (m+1)[2a + (3m-2)d]

Therefore,

5m∑r=m2mTr=5m(m+1)[2a+(3m−2)d]5m \sum_{r=m}^{2m} T_r = 5m(m+1)[2a + (3m-2)d]

Upon solving with the given values of aa and dd, we get m=8m = 8.

Substituting m=8m = 8,

5×8×(8+1)[2a+(3×8−2)d]=1265 \times 8 \times (8+1)[2a + (3\times 8 - 2)d] = 126

Therefore, the correct option is C.

Using the given relations

The solution states that the correct option is C and concludes that m=8m = 8. Using that conclusion in the required expression gives the final value as 126126.

Therefore, the required value is 126126.

Common mistakes

  • Using the wrong sum formula for the A.P. The sum from r=1r=1 to 2525 must be written as 252[2a+24d]\frac{25}{2}[2a+24d], not by replacing the last term incorrectly. Always identify the first term and the 25th25^{th} term carefully before applying the formula.

  • Writing the number of terms from r=mr=m to r=2mr=2m as mm instead of m+1m+1. Since both endpoints are included, the count is 2m−m+1=m+12m-m+1 = m+1. This changes the entire value of the required sum.

  • Using an incorrect expression for T2mT_{2m}. The term formula gives T2m=a+(2m−1)dT_{2m} = a + (2m-1)d, not a+2mda + 2md. Always substitute r=2mr=2m into Tr=a+(r−1)dT_r = a + (r-1)d exactly.

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