Question

Let $a_1,a_2----,a_{10}$ be in A.P and $h_1,h_{2,}---,h_{10}$ be in H.P if $a_1=h_1=2$ and $a_{10}=h_{10}=3$ , then $a_4\times h_{7}is-$

• A. 2
• B. 3
• C. 5
• D. 6

Answer 1

The correct option is D 6

Given, $a_1,a_2,...\dots ..,a_{10}$ be in A.P.

$h_1,h_2,...\dots .,h_{10}$ be in H.P

If $a_1=h_1=2$ and $a_{10}=h_{10}=3$

$h^{th}$ term of A.P$=a+(n-1)d$

$a_{10}=3=a+9d=2+9d$

$\Rightarrow d=\frac{1}{9}$

Now, $a_4=a+3d=2+3\frac{1}{9}=\frac{7}{3}$

Similarly, $\frac{1}{3}=\frac{1}{2}+(10-1)d$

$\Rightarrow d=-\frac{1}{54}$

Now, $h_7=\frac{1}{\frac{1}{2}+6(-\frac{1}{54})}$

$=\frac{1}{\frac{1}{2}-\frac{1}{9}}$

$=\frac{18}{7}$

Now $a_4\times h_7=\frac{7}{3}\times \frac{18}{7}=8$.