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BITSAT 2022 Paper

Section: Mathematics

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Question : 95 of 130

Marks: +1, -0

Let

a_1, a_2, \dots, a_{40}

h_1, h_2, \dots, h_{10}

a_1=h_1=2

a_{10}=h_{10}=3

a_4 h_7

2
3
5
6

Solution: 👈: Video Solution

Let d be the common difference of the

\text{AP}

. Then,

a_{10} =3\Rightarrow a_1 +9d=3

\Rightarrow 2+9d=3\Rightarrow d=\frac{1}{9}

\therefore a_4=a_1+3d=2+\frac{1}{3}=\frac{7}{3}

Let

D

be the common difference of

\frac{1}{h_1}, \frac{1}{h_2}, \dots, -\frac{1}{h_{10}}

.

Then,

h_{10}=3

\Rightarrow \frac{1}{h_{10}} = \frac{1}{3} \Rightarrow \frac{1}{2} +9D=\frac{1}{3}

\Rightarrow 9D=-\frac{1}{6}\Rightarrow D=-\frac{1}{54}

\therefore \frac{1}{h_7}=\frac{1}{h_1}+6D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}

\Rightarrow h_7=\frac{18}{7}

\therefore a_4h_7=\frac{7}{3}\times\frac{18}{7}=6

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