Jambmaths question:
Three consecutive term of a geometric progression are given as n – 2, n, n + 3. Find the common ratio
Option A:
$\tfrac{1}{4}$
Option B:
$\tfrac{1}{2}$
Option C:
$\tfrac{2}{3}$
Option D:
$\tfrac{3}{2}$
Jamb Maths Solution:
$\begin{align} & The\text{ }geometric\text{ }sequence\text{ }is\text{ }a\text{ }given\text{ }as\text{ }n2,n,n+3 \\ & Common\text{ }ratio=r=\frac{2nd\text{ }term}{1st\text{ }term}=\frac{3rd\text{ }term\text{ }}{2nd\text{ }term} \\ & r=\frac{n}{n-2}=\frac{n+3}{n} \\ & {{n}^{2}}=(n-2)(n+3) \\ & {{n}^{2}}={{n}^{2}}+n-6 \\ & n=6 \\ & r=\frac{6}{6-2}=\frac{6}{4}=\frac{3}{2} \\\end{align}$
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