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JEE Mathematics 2025 Question with Solution

Let Sn=12+16+112+120+…S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots up to nn terms. If the sum of the first six terms of an A.P. with first term −p-p and common difference pp is 2026S2025\sqrt{2026 S_{2025}}, then the absolute difference between the 20th and 15th terms of the A.P. is:

  • A

    2525

  • B

    9090

  • C

    2020

  • D

    4545

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

  • Sn=12+16+112+120+…S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots up to nn terms
  • An A.P. has first term −p-p and common difference pp
  • Sum of first six terms of the A.P. is 2026S2025\sqrt{2026 S_{2025}}

Find: The absolute difference between the 20th and 15th terms of the A.P.

The given series has general term

Tn=1n(n+1)T_n = \frac{1}{n(n+1)}

So,

S2025=∑n=120251n(n+1)S_{2025} = \sum_{n=1}^{2025} \frac{1}{n(n+1)}

Using partial fractions,

1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}

Hence,

S2025=(11−12)+(12−13)+⋯+(12025−12026)S_{2025} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{2025} - \frac{1}{2026}\right)

This is a telescoping sum, so

S2025=1−12026S_{2025} = 1 - \frac{1}{2026}

Now,

2026S2025=2026(1−12026)=2025=45\sqrt{2026 S_{2025}} = \sqrt{2026\left(1 - \frac{1}{2026}\right)} = \sqrt{2025} = 45

For the A.P. with first term a=−pa = -p and common difference d=pd = p, the sum of first six terms is

S6=62[2(−p)+5p]S_6 = \frac{6}{2}\left[2(-p) + 5p\right] S6=3(−2p+5p)=9pS_6 = 3(-2p + 5p) = 9p

Therefore,

9p=459p = 45 p=5p = 5

The 20th term is

a20=−p+19p=18pa_{20} = -p + 19p = 18p

The 15th term is

a15=−p+14p=13pa_{15} = -p + 14p = 13p

So the absolute difference is

∣a20−a15∣=∣18p−13p∣=5p=25|a_{20} - a_{15}| = |18p - 13p| = 5p = 25

Therefore, the absolute difference between the 20th and 15th terms is 2525. The correct option is A.

Use term difference directly

Given: The A.P. has first term −p-p and common difference pp.

Find: ∣a20−a15∣|a_{20} - a_{15}|

First compute pp from the given sum. The series is telescoping:

S2025=∑n=120251n(n+1)=1−12026S_{2025} = \sum_{n=1}^{2025} \frac{1}{n(n+1)} = 1 - \frac{1}{2026}

So,

2026S2025=2025=45\sqrt{2026 S_{2025}} = \sqrt{2025} = 45

Also, sum of first six terms of the A.P. is

9p9p

Hence,

9p=45⇒p=59p = 45 \Rightarrow p = 5

Now use the fact that in any A.P., the difference between the 20th and 15th terms is

a20−a15=(20−15)d=5da_{20} - a_{15} = (20-15)d = 5d

Here d=p=5d = p = 5, so

∣a20−a15∣=5p=25|a_{20} - a_{15}| = 5p = 25

Therefore, the correct option is A.

Common mistakes

  • Assuming the general term is incorrect. The sequence 12,16,112,120\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20} follows 1n(n+1)\frac{1}{n(n+1)}, not a standard A.P. or G.P. Identify the denominator pattern before summing.

  • Not using telescoping after writing 1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. If you do not cancel consecutive terms, the sum becomes unnecessarily difficult. Write out the first few expanded terms to see the cancellation.

  • Using the wrong sum formula for the first six terms of the A.P. Here a=−pa=-p and d=pd=p, so S6=62[2a+5d]=9pS_6 = \frac{6}{2}[2a+5d] = 9p. Substituting signs incorrectly gives the wrong value of pp.

  • Finding the 20th and 15th terms separately and making a sign error. A safer method is a20−a15=(20−15)d=5pa_{20}-a_{15}=(20-15)d=5p. Use the common difference directly to avoid mistakes.

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