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

BITSAT 2022 · Mathematics · Sequences and Series

Let a_1, a_2, a_ 40 be in AP and h_1, h_2, . h_ 10 be in HP. If...

Let a1,a2,……a40a_1, a_2, \ldots \ldots a_{40} be in AP and h1,h2,….h10h_1, h_2, \ldots. h_{10} be in HP. If a1=h1=2a_1=h_1=2 and a10=h10=3a_{10}=h_{10}=3, then a4h7a_4 h_7 is

Options

  1. A2
  2. B3
  3. C5
  4. D6

Correct answer

D. 6

Step-by-step solution

Let dd be the common difference of the AP. Then, a10=3⇒a1+9d=3⇒2+9d=3⇒d=19∴a4=a1+3d=2+13=73\begin{aligned} & a_{10}=3 \Rightarrow a_1+9 d=3 \\ \Rightarrow & 2+9 d=3 \Rightarrow d=\frac{1}{9} \\ \therefore & a_4=a_1+3 d=2+\frac{1}{3}=\frac{7}{3} \end{aligned} Let DD be the common difference of 1h1,1h2,….−1h10\frac{1}{h_1}, \frac{1}{h_2}, \ldots.-\frac{1}{h_{10}}. Then, h10=3\quad h_{10}=3⇒1h10=13⇒12+9D=13⇒9D=−16⇒D=−154∴1h7=1h1+6D=12−19=718⇒h7=187∴a4h7=73×187=6\begin{array}{ll} \Rightarrow & \frac{1}{h_{10}}=\frac{1}{3} \Rightarrow \frac{1}{2}+9 D=\frac{1}{3} \\ \Rightarrow & 9 D=-\frac{1}{6} \Rightarrow D=-\frac{1}{54} \\ \therefore & \frac{1}{h_7}=\frac{1}{h_1}+6 D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18} \\ \Rightarrow & h_7=\frac{18}{7} \\ \therefore & a_4 h_7=\frac{7}{3} \times \frac{18}{7}=6 \end{array}$$
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