Let a1, a2,…,a10 be in A.P and h1, h2.....h10 be in H.P. If a1=h1=2 anda10=h10=3, then a4h7 is

Solution:

Let $d$ be the common difference of the A.P…., then

$\begin{array}{l} 3 = a_{10} = 2 + 9 d \Rightarrow d = 1 / 9 \\ ∴ a_{1} = 2 + 3 d = 7 / 3 \end{array}$

Next, let $D$ be the common difference of the A.P.

$\frac{1}{h_{1}}, \frac{1}{h_{2}}, \dots, \frac{1}{h_{10}}$ Then

$\begin{array}{l} \frac{1}{3} = \frac{1}{h_{10}} = \frac{1}{2} + 9 D \Rightarrow D = - \frac{1}{54} \\ ∴ \frac{1}{h_{7}} = \frac{1}{h_{1}} + 6 D = \frac{7}{18} \Rightarrow h_{7} = \frac{18}{7} \end{array}$

Hence, $a_{4} h_{7} = (7 / 3) (18 / 7) = 6 .$

Let $a_{1}, a_{2}, \dots, a_{10}$ 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 $a_{4} h_{7}$ is

Solution:

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