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Sequences and Series question

2025 · 24 Jan · Shift 1 · Q41
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  5. /2025 · 24 Jan · Shift 1 · Q41

Sequences and Series question

2025 · 24 Jan · Shift 1 · Q41

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let Sn=12+16+112+120+…S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldotsSn​=21​+61​+121​+201​+… upto nnn terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is 2026 S2025\sqrt{2026 \mathrm{~S}_{2025}}2026 S2025​​, then the absolute difference betwen 20th 20^{\text {th }}20th  and 15th 15^{\text {th }}15th  terms of the A.P. is
  1. A
    20
  2. B
    45
  3. C
    90
  4. D
    25
View written solutionFree

Correct answer: D

  1. Identify the general term of the series SnS_nSn​

The given series is

Sn=12+16+112+120+⋯S_n=\frac12+\frac16+\frac1{12}+\frac1{20}+\cdotsSn​=21​+61​+121​+201​+⋯

Notice the denominators:

2,6,12,20,…2,6,12,20,\ldots2,6,12,20,…

These are of the form

1⋅2,  2⋅3,  3⋅4,  4⋅5,…1\cdot 2,\; 2\cdot 3,\; 3\cdot 4,\; 4\cdot 5,\ldots1⋅2,2⋅3,3⋅4,4⋅5,…

So the rrr-th term is

1r(r+1)\frac{1}{r(r+1)}r(r+1)1​

Hence,

Sn=∑r=1n1r(r+1)S_n=\sum_{r=1}^n \frac{1}{r(r+1)}Sn​=r=1∑n​r(r+1)1​
  1. Simplify the sum using telescoping

We use

1r(r+1)=1r−1r+1\frac{1}{r(r+1)}=\frac{1}{r}-\frac{1}{r+1}r(r+1)1​=r1​−r+11​

Thus,

Sn=∑r=1n(1r−1r+1)S_n=\sum_{r=1}^n \left(\frac{1}{r}-\frac{1}{r+1}\right)Sn​=r=1∑n​(r1​−r+11​)

This telescopes:

Sn=(1−12)+(12−13)+⋯+(1n−1n+1)S_n=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)Sn​=(1−21​)+(21​−31​)+⋯+(n1​−n+11​)

So,

Sn=1−1n+1=nn+1S_n=1-\frac{1}{n+1}=\frac{n}{n+1}Sn​=1−n+11​=n+1n​

Therefore,

S2025=20252026S_{2025}=\frac{2025}{2026}S2025​=20262025​
  1. Compute 2026 S2025\sqrt{2026\,S_{2025}}2026S2025​​
2026 S2025=2026⋅20252026=20252026\,S_{2025}=2026\cdot \frac{2025}{2026}=20252026S2025​=2026⋅20262025​=2025

Hence,

2026 S2025=2025=45\sqrt{2026\,S_{2025}}=\sqrt{2025}=452026S2025​​=2025​=45
  1. Use the sum of first six terms of the A.P.

The A.P. has first term a=−pa=-pa=−p and common difference d=pd=pd=p.

Sum of first 6 terms is

S6=62[2a+(6−1)d]S_6=\frac{6}{2}\left[2a+(6-1)d\right]S6​=26​[2a+(6−1)d]

Substitute a=−pa=-pa=−p and d=pd=pd=p:

S6=3[2(−p)+5p]=3(3p)=9pS_6=3\left[2(-p)+5p\right]=3(3p)=9pS6​=3[2(−p)+5p]=3(3p)=9p

Given that

S6=2026 S2025=45S_6=\sqrt{2026\,S_{2025}}=45S6​=2026S2025​​=45

So,

9p=45  ⟹  p=59p=45 \implies p=59p=45⟹p=5
  1. Find the 20th and 15th terms of the A.P.

The nnn-th term of an A.P. is

an=a+(n−1)da_n=a+(n-1)dan​=a+(n−1)d

Here,

an=−p+(n−1)p=(n−2)pa_n=-p+(n-1)p=(n-2)pan​=−p+(n−1)p=(n−2)p

Thus,

a20=(20−2)p=18pa_{20}=(20-2)p=18pa20​=(20−2)p=18p a15=(15−2)p=13pa_{15}=(15-2)p=13pa15​=(15−2)p=13p

Their absolute difference is

∣a20−a15∣=∣18p−13p∣=5p=5⋅5=25|a_{20}-a_{15}|=|18p-13p|=5p=5\cdot 5=25∣a20​−a15​∣=∣18p−13p∣=5p=5⋅5=25
  1. Final answer
25\boxed{25}25​

So the correct option is D.

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