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BITSAT 2022 Paper

Section: Mathematics

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Question : 121 of 130

Marks: +1, -0

In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2 . If the middle term of AP be equal to the middle term of the GP, then the middle term of the entire sequence is

$-\;\frac{10}{31}$
$\;\frac{10}{31}$
$\;\frac{32}{31}$
$-\;\frac{31}{32}$

Solution: 👈: Video Solution

Since, the first

11

terms are

Ap, d=2

\therefore a_{11} = a+10d = a+20

The middle term of

AP

is

T_6 = a+5d = a+10

For the next

11

terms in

GP

r=2

\therefore

The middle term of

GP

is

b(2)^5

where,

b

is the first term of a

GP

which is the last term of

AP

b(2)^5 = (a+20)32

According to the given condition,

\Rightarrow a+10 = (a+20)32

\Rightarrow 32a = 10 - 640

a = -\frac{630}{31}

\therefore

Middle term of entire sequence is

11\text{th}

term

\therefore T_{11} = \frac{-630}{31} + 10 \times d

= \frac{-630}{31} + 10 \times 2 = \frac{-10}{31}

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