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

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.