Question

For H.P. 2, 4, 6, ……….. , find the value of \[{{a}_{9}}\]
(A) \[\dfrac{1}{18}\]
(B) \[\dfrac{1}{32}\]
(C) \[\dfrac{1}{6}\]
(D) \[\dfrac{1}{12}\]

Answer 1

Hint: The first term, second term, and third term of the given sequence are 2, 4, and 6 respectively. Now, get the common difference of the given sequence by subtracting the first term from second term. The HP sequence is, \[\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{6},............,\] . Use the formula for the \[{{n}^{th}}\] term of H.P, \[{{T}_{n}}=\dfrac{1}{\text{first term}+\left( n-1 \right)\text{common difference}}\] and calculate the \[{{9}^{th}}\] term by putting the calculated value of common difference and \[n=9\] . Now, solve it further and get the \[{{9}^{th}}\] term of the required H.P.

Complete step by step answer:
According to the question, we have a sequence,
2, 4, 6, ……….. , ………………………………………(1)
Here, in the given sequence, we get
The first term = 2 ………………………………………….(2)
The second term = 4 ………………………………………….(3)
The third term = 6 ……………………………….………(4)
Subtracting equation (2) from equation (3), we get
The common difference between first term and second term = \[4-2\] = 2 ………………………………………..(5)
Similarly, subtracting equation (3) from equation (4), we get
The common difference between first term and second term = \[6-4\] = 2 ………………………………………..(6)
From equation (5) and equation (6), we have the value of the common difference of a given sequence.
The value of the common difference = 2 ……………………………………(7)
The HP sequence is, \[\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{6},............,\] .
We know the formula for the \[{{n}^{th}}\] term of Harmonic progression, \[{{T}_{n}}=\dfrac{1}{\text{first term}+\left( n-1 \right)\text{common difference}}\] ……………………………………….(8)
Now, from equation (2), equation (7), and equation (8), we get
 The \[{{n}^{th}}\] term of the given HP, \[{{T}_{n}}=\dfrac{1}{2+\left( n-1 \right)2}\] ……………………………………………..(9)
We have to find the \[{{9}^{th}}\] term for H.P of the given sequence.
On putting \[n=9\] in equation (9), we get
 \[\Rightarrow {{T}_{9}}=\dfrac{1}{2+\left( 9-1 \right)2}\]
\[\Rightarrow {{T}_{9}}=\dfrac{1}{18}\] ………………………………………………(10)
Therefore, the \[{{9}^{th}}\] term of the H.P. is \[\dfrac{1}{18}\] .

So, the correct answer is “Option A”.

Note: Since HP is the reciprocal of the AP series. So, we can also solve this question by using the formula of AP for \[{{n}^{th}}\] term, \[{{T}_{n}}\] = First term + \[\left( n-1 \right)\] common difference. Now, use this formula and calculate \[{{9}^{th}}\] term of the A.P. For, \[{{9}^{th}}\] term of HP, get the reciprocal of the \[{{9}^{th}}\] term of the A.P.