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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2022 · Mathematics · Sequences and Series

In a sequence of 21 terms, the first 11 terms are in AP with common...

In a sequence of 2121 terms, the first 1111 terms are in APAP with common difference 22 and the last 1111 terms are in GPGP with common ratio 22. If the middle term of APAP be equal to the middle term of the GPGP, then the middle term of the entire sequence is

Options

  1. A−1031-\frac{10}{31}
  2. B1031\frac{10}{31}
  3. C3231\frac{32}{31}
  4. D−3132-\frac{31}{32}

Correct answer

A. −1031-\frac{10}{31}

Step-by-step solution

Since, the first 1111 terms are in APAP, d=2∴a11=a+10d=a+20d=2\therefore \quad a_{11}=a+10 d=a+20 The middle term of APAP is T6=a+5d=a+10T_6=a+5 d=a+10 For the next 1111 terms in GPr=2∴GPr=2\therefore The middle term of GPGP is b(2)5b(2)^5 where, bb is the first term of a GPGP which is the last term of APb(2)5=(a+20)32APb(2)^5=(a+20) 32 According to the given condition, ⇒a+10=(a+20)32⇒31a=10−640a=−63031\begin{aligned} \Rightarrow \quad a+10 & =(a+20) 32 \\ \Rightarrow \quad 31 a & =10-640 \\ a & =-\frac{630}{31} \end{aligned}\thereforeMiddletermofentiresequenceisMiddle term of entire sequence is11thtermterm∴T11=−63031+10×d=−63031+10×2=−1031\begin{aligned} \therefore \quad T_{11} & =\frac{-630}{31}+10 \times d \\ & =\frac{-630}{31}+10 \times 2=\frac{-10}{31}\end{aligned}$
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