Question

If 9, 6, p are in arithmetic progression, 9, 6, q are in geometric progression and 9, 6, are in harmonic progression, then what is the value of (4p – 6q + r)?

Answer 1

As 9,6,p are in arithmetic progression,
12=9+p=> p=3.
As 9,6,q are in geometric progression,
${6}^{2} = 9 \times q = > 36 \div 9 = q = > q = 4$
As 9,6,r is in harmonic progression
$\frac{2}{6} = \frac{1}{9} + \frac{1}{r}$
$= > \frac{1}{3} - \frac{1}{9} = \frac{1}{r}$
Hence r=9/2. Hence value of
(4p-6q+r)=>12-24+4.5 => -7.5 or -15/2 is the answer. Hope it helps you. Mark me as Brainliest.