Question

If the 3rd and the 6th terms of a H.P. are 1/3 and 1/26 respectively, find the 27th term of the H.P.​

Answer 1

Answer: $x_{27} = \frac{1}{187}$

Step-by-step explanation:

Note: If $\frac{1}{x} , \frac{1}{x+d} , \frac{1}{x+2d} ...$ are in HP then, $x, x+d, x+2d...$ are in AP.

Also,$x_{n} = x + (n-1)d$

Given:

3rd term of an HP: $\frac{1}{3}$ = $\frac{1}{x+2d}$

• $x+2d = 3$ (in AP)...(1)

6th term of an HP: $\frac{1}{26} = \frac{1}{x+5d}$

• $x + 5d = 26$ (in AP)...(2)

Now, subtract equation (1) from (2)

$3d = 23$

• $d = \frac{23}{3}$

Now, put $d = \frac{23}{3}$ in equation (1)

$x + 2 * \frac{23}{3} = 3$

$x + \frac{46}{3} = 3$

$x = 3-\frac{46}{3}$

$x = \frac{9-46}{3}$

• $x = \frac{-37}{3}$

Now, we need to first find the 27th term of the AP, and then HP

$x_{n} = x + (n-1)d$

$x_{27} = x + 26d$

$x_{27} = \frac{-37}{3} + [26 * \frac{23}{3}]$

$x_{27} = \frac{-37}{3} + \frac{598}{3}$

$x_{27} = \frac{561}{3}$

$x_{27} = 187$ (In AP)

But,

AP = $\frac{1}{HP}$

• $x_{27} = \frac{1}{187}$ (In HP)

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