Question

the first and second terms of a H.P are 1/3 and 1/5 respectively, find the 9th term.

Answer 1

reciprocal of HP is in AP
3 , 5 , 7 , ....... 3+(n-1)2
3 + (9-1)2
3+16=19
ans=1/19

Answer 2

The 9th term of the HP is 1/19.

Given:

The first and second terms of a H.P are 1/3 and 1/5 respectively.

To Find:

The 9th term of the HP.

Solution:

To solve this problem we will use the concept that harmonic progression is the inverse of arithmetic progression.

Now the first term of the harmonic progression is given to be 1/3.

⇒ The first term of arithmetic progression will be its inverse, i.e. $a_{1}$ = 3

The second term of the harmonic progression is given to be 1/5.

⇒ The first term of arithmetic progression will be its inverse, i.e. $a_{2}$ = 5.

We know that difference between every consecutive term of an arithmetic progression is constant and is referred to as the common difference.

Here, the common difference d = $a_{2}$ - $a_{1}$ = 5  - 3 = 2

The $n^{th}$ term of an AP, $a_{n}$ = $a_{1}$ + (n - 1)d

∴ The 9th term the AP, $a_{9}$ = $a_{1}$ + (9 - 1)d = 3 + 8.2 = 3 + 16 = 19.

Sine HP is the inverse of AP, 9th term the HP = 1/$a_{9}$ = 1/19

∴ The 9th term of the HP is 1/19.

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