Sequences and Series 5 Question 3
3. Let $a _1, a _2, \ldots, a _{10}$ be in $AP$ and $h _1, h _2$, equal to $\ldots . ., h _{10}$ be in HP. If $a _1=h _1=2$ and $a _{10}=h _{10}=3$, then $a _4 h _7$ is
$(1999,2 M)$
(a) 2
(b) 3
(c) 5
(d) 6
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Answer:
Correct Answer: 3. (d)
Solution:
- Since, $a _1, a _2, a _3, \ldots, a _{10}$ are in AP.
Now, $\quad a _{10}=a _1+9 d$
$\Rightarrow \quad 3=2+9 d$
$\Rightarrow \quad d=1 / 9$ and $a _4=a _1+3 d$
$\Rightarrow \quad a _4=2+3(1 / 9)=2+1 / 3=7 / 3$
Also, $h _1, h _2, h _3, \ldots, h _{10}$ are in HP.
$\Rightarrow \quad \frac{1}{h _1}, \frac{1}{h _2}, \frac{1}{h _3}, \ldots, \frac{1}{h _{10}}$ are in AP.
Given, $\quad h _1=2, h _{10}=3$
$\therefore \quad \frac{1}{h _{10}}=\frac{1}{h _1}+9 d _1 \Rightarrow \frac{1}{3}=\frac{1}{2}+9 d _1$
$\Rightarrow \quad-\frac{1}{6}=9 d _1$
$\Rightarrow \quad d _1=-\frac{1}{54}$ and $\frac{1}{h _7}=\frac{1}{h _1}+6 d _1$
$$ \begin{array}{ll} \Rightarrow & \frac{1}{h _7}=\frac{1}{2}+\frac{6 \times 1}{-54} \\ \Rightarrow & \frac{1}{h _7}=\frac{1}{2}-\frac{1}{9} \Rightarrow h _7=\frac{18}{7} \\ \therefore & a _4 h _7=\frac{7}{3} \times \frac{18}{7}=6 \end{array} $$
