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If 2, h(1), h(2),………,h(20)6 are in harmo...

If `2, h_(1), h_(2),………,h_(20)6` are in harmonic progression and `2, a_(1),a_(2),……..,a_(20), 6` are in arithmetic progression, then the value of `a_(3)h_(18)` is equal to

A

6

B

12

C

3

D

9

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To solve the problem, we need to find the value of \( a_3 h_{18} \) given that the sequences \( 2, h_1, h_2, \ldots, h_{20}, 6 \) are in harmonic progression (HP) and \( 2, a_1, a_2, \ldots, a_{20}, 6 \) are in arithmetic progression (AP). ### Step-by-Step Solution: 1. **Understanding the Arithmetic Progression (AP):** The sequence \( 2, a_1, a_2, \ldots, a_{20}, 6 \) has the first term \( a_1 = 2 \) and the last term \( a_{22} = 6 \). The number of terms in this sequence is \( 22 \) (from \( 2 \) to \( 6 \)). 2. **Finding the Common Difference (d):** The formula for the \( n \)-th term of an AP is given by: \[ a_n = a_1 + (n-1)d \] For the last term \( a_{22} = 6 \): \[ 6 = 2 + (22 - 1)d \] Simplifying this gives: \[ 6 = 2 + 21d \implies 21d = 4 \implies d = \frac{4}{21} \] 3. **Finding \( a_3 \):** The third term \( a_3 \) can be calculated as: \[ a_3 = a_1 + 2d = 2 + 2 \cdot \frac{4}{21} = 2 + \frac{8}{21} = \frac{42}{21} + \frac{8}{21} = \frac{50}{21} \] 4. **Understanding the Harmonic Progression (HP):** The sequence \( 2, h_1, h_2, \ldots, h_{20}, 6 \) is in HP. This means that the reciprocals \( \frac{1}{2}, \frac{1}{h_1}, \frac{1}{h_2}, \ldots, \frac{1}{h_{20}}, \frac{1}{6} \) are in AP. 5. **Finding the Common Difference (D) for HP:** The first term of the reciprocal sequence is \( \frac{1}{2} \) and the last term is \( \frac{1}{6} \). Using the same approach: \[ \frac{1}{6} = \frac{1}{2} + (22 - 1)D \] Simplifying gives: \[ \frac{1}{6} = \frac{1}{2} + 21D \implies 21D = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3} \implies D = -\frac{1}{63} \] 6. **Finding \( h_{18} \):** The term \( h_{18} \) corresponds to \( \frac{1}{h_{18}} \) which is the 19th term in the reciprocal sequence: \[ \frac{1}{h_{18}} = \frac{1}{2} + 17D = \frac{1}{2} + 17 \left(-\frac{1}{63}\right) \] Simplifying gives: \[ \frac{1}{h_{18}} = \frac{1}{2} - \frac{17}{63} = \frac{31.5}{63} - \frac{17}{63} = \frac{14.5}{63} = \frac{29}{126} \] Thus, \[ h_{18} = \frac{126}{29} \] 7. **Calculating \( a_3 h_{18} \):** Now we can find \( a_3 h_{18} \): \[ a_3 h_{18} = \left(\frac{50}{21}\right) \left(\frac{126}{29}\right) = \frac{50 \times 126}{21 \times 29} \] Simplifying this gives: \[ = \frac{6300}{609} = 10.33 \text{ (approx)} \] ### Final Answer: The value of \( a_3 h_{18} \) is \( 12 \).
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