First slide

Means - A.M/G.M/H.M

Question

$Let a_{1}, a_{3}, \dots . a_{10} be in A.P, and h_{1}, h_{2}, \dots h_{10} be in H.P. If a_{1} = h_{1} = 2 a$ $a_{10} = h_{10} = 3, then a_{4} h_{7} is$

Moderate

A

2
B

3
C

5
D

6

Solution

$Let 'd' is common difference of A.P ∴ 3 = a_{10} = 2 + 9 d \Rightarrow d = \frac{1}{9}, a_{4} = 2 + 3 d = \frac{7}{3}$
$Let 'D' is common difference of \frac{1}{h_{1}}, \frac{1}{h_{2}}, \dots \cdot \frac{1}{h_{10}} ∴ \frac{1}{3} = \frac{1}{h_{10}} = \frac{1}{2} + 9 D \Rightarrow D = \frac{- 1}{54}$
$\frac{1}{h_{7}} = \frac{1}{2} + 6 (- \frac{1}{54}) = \frac{7}{18}, a_{4} h_{7} = \frac{7}{3} \times \frac{18}{7} = 6$

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