Question

$a_1,a_2,...,a_{10}$ are in A.P., $h_1,h_2,...,h_{10}$ are in H.P. . If $a_1=h_1=2,a_{10}=h_{10}=3$, then $a_4,h_7=$

• A. $1$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $6$

Answer 1

The correct option is D $6$

Since $A_1,A_2,....,A_{10}$ are in AP...

and $A_1=2and{A}_{10}=3.$.

Therefore ,

$A_{1}0=A_1+(10-1)d$, d being the common difference...

Which gives d$=\frac{1}{9}$...

Therefore,

$A_4=A_1+(4-1)$$\times \frac{1}{9}$

$=\frac{7}{3}$

Again ,

$H_1,H_2,...,H_{10}areinHP$, so $1/H_1,1/H_2,......,1/H_{10}$ are in AP...

And $H_1=2and{H}_{10}=3$

Therefore, $\left(1/H_{1}0\right)=\left(1/H_1\right)+(10-1)d$,

which gives d$=\frac{-1}{54}$

So $\left(1/H_7\right)=\left(1/H_1\right)+(7-1)$$\times \frac{-1}{54}$

Which gives H_7 $=\frac{18}{7}$...

So, $A_4\ast H_7=6$, which is the required answer....