Arithmetic and geometric progressions

Algebra Level 2

Let $a,4,b$ are the first three terms of an arithmetic progression , and
$a,3,b$ are the first three terms of a geometric progression .

Compute $\dfrac 1a + \dfrac1b$ .

\frac34

4

\frac12

\frac98

\frac89

\frac14

2

\frac43

1 solution

Hung Woei Neoh
Jun 12, 2016

Given an AP $a,4,b,\ldots$

$d=b-4=4-a\\ \implies a+b = 8$

Given a GP $a,3,b,\ldots$

$r=\dfrac{b}{3} = \dfrac{3}{a}\\ \implies ab=9$

$\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{a+b}{ab} = \boxed{\dfrac{8}{9}}$