Verified Solution • 2025–26 Session moderate

QJEEM25Ma6352: JEE Main 2025 Mathematics - Class 11 JEEM25Ma6352 Solution: Consider an A. P. of positive integers, whose sum of the first…

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :

Multiple Choice Options

  • A 108
  • B 90
  • C 122
  • D 84

Step-by-Step Verified Solution

amp;amp;S3=3a+3 d=54amp;⇒a+d=18amp; S20=10(2a+19 d)amp;⇒10(36+17 d)amp;⇒1600lt;10(36+17 d)lt;1800amp;⇒160lt;36+17 dlt;180amp;⇒124lt;17 dlt;144amp;⇒7517lt;dlt;8817amp; Common difference will be natural number amp;amp;⇒d=8⇒a=10amp;⇒a11=10+10×8=90\begin{aligned}&\begin{aligned}& \mathrm{S}_3=3 \mathrm{a}+3 \mathrm{~d}=54 \\& \Rightarrow \mathrm{a}+\mathrm{d}=18 \\& \mathrm{~S}_{20}=10(2 \mathrm{a}+19 \mathrm{~d}) \\& \Rightarrow 10(36+17 \mathrm{~d}) \\& \Rightarrow 1600<10(36+17 \mathrm{~d})<1800 \\& \Rightarrow 160<36+17 \mathrm{~d}<180 \\& \Rightarrow 124<17 \mathrm{~d}<144 \\& \Rightarrow 7 \frac{5}{17}<\mathrm{d}<8 \frac{8}{17}\end{aligned}\\&\text { Common difference will be natural number }\\&\begin{aligned}& \Rightarrow d=8 \Rightarrow a=10 \\& \Rightarrow a_{11}=10+10 \times 8=90\end{aligned}\end{aligned}

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