CAT 2025 Lesson : Progressions - HP Concepts

If the $2^{nd}$and $3^{rd}$ terms of an HP are $9$ and $6$ respectively, then what is the $6^{th}$ term?

Solution

As HP is the reciprocal of terms in an AP, let the terms in this HP be $\dfrac{1}{a},\dfrac{1}{(a + d)},\dfrac{1}{(a + 2d)},......$

$2^{nd}$ term = $\dfrac{1}{(a + d)} = 9$ ⇒ $a + d = \dfrac{1}{9}$ -----(1)

$3^{rd}$ term = $\dfrac{1}{(a + 2d)} = 6$ ⇒ $a + 2d = \dfrac{1}{6}$ -----(2)

Subtracting $(1)$ from $(2)$ ⇒ $d = \dfrac{1}{6} - \dfrac{1}{9} = \dfrac{1}{18}$

Substituting in $(1)$ ⇒ $a = \dfrac{1}{9} - \dfrac{1}{18} = \dfrac{1}{18}$

$6^{th}$term of HP = $\dfrac{1}{(a + 5d)} = \dfrac{1}{\left( \dfrac{1}{18} + \dfrac{5}{18} \right)}=\dfrac{18}{6}=3$

Answer: $3$